Cap 06

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 1. \

Joint FBDs:

Joint B: FAB 800 lb FBC = = 15 8 17 so

FAB = 1500 lb T W FBC = 1700 lb C W

Joint C:

FAC Cx 1700 lb = = 8 15 17 FAC = 800 lb T W

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell Š 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 2.

Joint FBDs: Joint B: ΣFx = 0:

1 4 FAB − FBC = 0 5 2

ΣFy = 0:

1 3 FAB + FBC − 4.2 kN = 0 5 2

7 FBC = 4.2 kN 5

so

Joint C:

FAB =

ΣFx = 0:

12 2 kN 5

FBC = 3.00 kN C !

FAB = 3.39 kN C !

4 12 (3.00 kN) − FAC = 0 5 13 FAC =

13 kN 5

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

FAC = 2.60 kN T !

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 3.

Joint FBDs:

Joint B: FAB FBC 450 lb = = 12 13 5 FAB = 1080 lb T W

so

FBC = 1170 lb C W

Joint C: ΣFx = 0:

3 12 FAC − (1170 lb ) = 0 5 13 FAC = 1800 lb C W

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 4.

Joint FBDs:

Joint D:

FCD FAD 500 lb = = 8.4 11.6 8 FAD = 725 lb T W FCD = 525 lb C W

Joint C: ΣFx = 0:

FBC − 525 lb = 0 FBC = 525 lb C W

This is apparent by inspection, as is FAC = C y

ΣFx = 0:

8.4 3 (725 lb) − FAB − 375 lb = 0 11.6 5

Joint A:

FAB = 250lb T W ΣFy = 0:

FAC −

4 8 (250 lb) − (725 lb) = 0 5 11.6 FAC = 700 lb C W

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 5.

FBD Truss: ΣFx = 0 :

Cx = 0

By symmetry: C y = D y = 6 kN

Joint FBDs: Joint B: ΣFy = 0:

− 3 kN +

1 FAB = 0 5

FAB = 3 5 = 6.71 kN T W

Joint C:

ΣFx = 0:

ΣFy = 0:

Joint A:

ΣFx = 0:

ΣFy = 0:

2 FAB − FBC = 0 5

FBC = 6.00 kN C W

3 FAC = 0 5

FAC = 10.00 kN C W

6 kN −

6 kN −

4 FAC + FCD = 0 5

FCD = 2.00 kN T W

 1  3  − 2 3 5 kN  + 2  10 kN  − 6 kN = 0 check 5   5 

By symmetry:

FAE = FAB = 6.71 kN T W FAD = FAC = 10.00 kN C W FDE = FBC = 6.00 kN C W

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 6.

FBD Truss:

ΣM A = 0:

(10.2 m ) C y + ( 2.4 m )(15 kN ) − ( 3.2 m )( 49.5 kN ) = 0 C y = 12.0 kN

Joint FBDs: Joint FBDs: Joint C: FBC F 12 kN = CD = 7.4 7.4 8 FBC = 18.50 kN C W FCD = 18.50 kN T W

Joint B:

ΣFX = 0:

4 7 (18.5 kN) = 0 FAB − 5 7.4 FAB = 21.875 kN;

ΣFy = 0:

FAB = 21.9 kN C W

3 2.4 (21.875 kN) − 49.5 kN + (18.5 kN) + FBD = 0 5 7.4 FBD = 30.375 kN;

FBD = 30.4 kN C W

Joint D: ΣFx = 0: −

4 7 FAD + (18.5 kN ) + 15 kN = 0 5 7.4

FAD = 40.625 kN;

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

FAD = 40.6 kN T W

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 7.

Joint FBDs:

Joint E: FBE FDE 3 kN = = 5 4 3 FBE = 5.00 kN T W FDE = 4.00 kN C W

Joint B: ΣFx = 0:

− FAB +

4 (5 kN) = 0 5 FAB = 4.00 kN T W

ΣFy = 0:

FBD − 6 kN −

3 (5 kN) = 0 5 FBD = 9.00 kN C W

Joint D:

ΣFy = 0:

3 FAD − 9 kN = 0 5 FAD = 15.00 kN T W

ΣFx = 0: FCD −

4 (15 kN) − 4 kN = 0 5 FCD = 16.00 kN C W

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 8.

Joint FBDs:

Joint B: FAB = 12.00 kips C W

By inspection:

FBD = 0 W

Joint A:

FAC FAD 12 kips = = 5 13 12 FAC = 5.00 kips C W FAD = 13.00 kips T W

Joint D: ΣFx = 0: FCD −

12 (13 kips) − 18 kips = 0 13 FCD = 30.0 kips C W

ΣFy = 0:

5 (13 kips) − FDF = 0 13 FDF = 5.00 kips T W

Joint C:

ΣFx = 0: 30 kips −

12 FCF = 0 13 FCF = 32.5 kips T W

ΣFy = 0: FCE − 5 kips −

5 (32.5 kips) 13 FCE = 17.50 kips C W

Joint E:

by inspection:

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

FCF = 0 W

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 9.

FCH = 0 !

First note that, by inspection of joint H:

and

FCG = 0 !

then, by inspection of joint C: and

Joint D:

FBC = FCD FBG = 0 !

then, by inspection of joint G: and

Joint FBDs:

FDH = FGH

FFG = FGH FBF = 0 !

then, by inspection of joint B: and

FAB = FBC

FCD FDH 10 kips = = 12 13 5 Joint A:

FCD = 24.0 kips T !

so

FDH = 26.0 kips C !

FAB = FBC = 24.0 kips T !

and, from above:

FGH = FFG = 26.0 kips C !

FAF F 24 kips = AE = 5 4 41

FAF = 30.0 kips C !

Joint F:

FAE = 25.6 kips T !

ΣFx = 0: FEF −

12 (26 kips) = 0 13 FEF = 24.0 kips C !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 10.

FBD Truss: ΣFx = 0: H x = 0 By symmetry: A y = H y = 4 kips FAC = FCE and FBC = 0 W

by inspection of joints C and G :

FEG = FGH and FFG = 0 W also, by symmetry FAB = FFH , FBD = FDF , FCE = FEG and FBE = FEF

Joint FBDs: Joint A:

FAB FAC 3 kips = = 5 4 3 FAB = 5.00 kips C W

so

FAC = 4.00 kips T W FFH = 5.00 kips C W

and, from above, and

Joint B:

FCE = FEG = FGH = 4.00 kips T W

4 4 10 (5 kips) − FBE − FBD = 0 5 5 109

ΣFx = 0: ΣFy = 0:

3 3 3 FBD + FBE = 0 ( 5 kips ) − 2 − 5 5 109

so

FBD = 3.9772 kips, FBE = 0.23810 kips FBD = 3.98 kips C W

or

Joint E:

FBE = 0.238 kips C W FDF = 3.98 kips C W

and, from above,

FEF = 0.238 kips C W ΣFy = 0 :

FDE − 2

3 (0.23810 kips) = 0 5 FDE = 0.286 kips T W

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 11.

FBD Truss: ΣFx = 0:

Ax = 0

ΣM G = 0:

3a Ay − 2a (3 kN) − a (6 kN) = 0

A y = 4 kN by inspection of joint C,

FAC = FCE and FBC = 0 W

by inspection of joint D,

FBD = FDF and FDE = 6.00 kN C W

Joint FBDs: Joint A:

FAC FAB 4 kN = = 21 29 20 FAB = 5.80 kN C W FAC = 4.20 kN C W from above,

Joint B:

ΣFy = 0:

20 20 ( 5.80 kN ) − 3 kN − FBE = 0 29 29 FBE =

ΣFx = 0:

FCE = 4.20 kN C W

29 20

FBE = 1.450 kN T W

21  29  kN  − FBD = 0  5.80 kN + 29  20  FBD = 5.25 kN C W FDF = 5.25 kN C W

from above,

Joint F:

ΣFx = 0:

5.25 kN −

21 FEF = 0 29 FEF = 7.25 kN T W

ΣFy = 0:

FFG −

20 (7.25 kN) − 1 kN = 0 29 FFG = 6.00 kN C W

by inspection of joint G,

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

FEG = 0 W

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 12.

FBD Truss:

ΣFx = 0:

Ax = 0

By symmetry: A y = B y = 4.90 kN FAB = FEG , FAC = FFG , FBC = FEF

and

FBD = FDE , FCD = FDF

5 4 FAC − FAB = 0 5 29 2 3 FAC − FAB + 4.9 kN = 0 5 29

ΣFx = 0:

Joint FBDs: Joint A:

ΣFy = 0:

FAC = 2.8 29 kN

FAC = 15.08 kN T

FAD = 17.50 kN C Joint B:

ΣFx = 0: ΣFy = 0:

4 1 FBC = 0 (17.5 kN − FBD ) − 5 2 3 1 FBC − 2.8 kN = 0 (17.5 kN − FBD ) + 5 2 FBD = 15.50 kN C

FBC = 1.6 2 kN; Joint C:

ΣFy = 0:

FBC = 2.26 kN C

4 1 1.6 2 kN − FCD − 5 2

ΣFx = 0: FCF

2 (2.8 29 kN) = 0 29 FCD = 9.00 kN T 1 3 5 1.6 2 kN + (9 kN) − (2.8 29 kN) = 0 + 5 2 29 FCF = 7.00 kN T

(

(

)

)

from symmetry,

FEG = 17.50 kN C FFG = 15.08 kN T FEF = 2.26 kN C

FDE = 15.50 kN C FDF = 9.00 kN T

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 13.

FBD Truss:

ΣFx = 0: A x = 0 ΣM A = 0: (8 m) Gy − (4 m)(4.2 kN) − (2m)(2.8 kN) = 0

G y = 2.80 kN ΣFy = 0:

Ay − 2.8 kN − 4.2 kN + 2.8 kN = 0

A y = 4.2 kN Joint FBDs:

5 4 FAC − FAB = 0 5 29 2 3 FAC − FAB + 4.2 kN = 0 5 29

ΣFx = 0:

Joint A: ΣFy = 0:

FAB = 15.00 kN C !

FAC = 12.92 kN T !

FAC = 2.4 29 Joint B: ΣFx = 0: ΣFy = 0:

4 1 FBC = 0 (15.00 kN − FBD ) − 5 2 3 1 FBC − 2.8 kN = 0 (15.00 kN − FBD ) + 5 2 FBD = 13.00 kN C !

FBC = 1.6 2 kN, Joint C:

ΣFy = 0:

(

4 FCD − 5

FBC = 2.26 kN C !

)

2 1 2.4 29 kN − (1.6 2 kN) = 0 29 2

FCD = 8.00 kN T ! ΣFx = 0:

FCF +

3 (8.00 kN ) − 5 +

5 (2.4 29 kN) 29

1 (1.6 2 kN) = 0 2

FCF = 5.60 kN T !

continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

By inspection of joint E,

FDE = FEG

and

FEF = 0 !

Joint F: ΣFy = 0: ΣFx = 0:

4 FDF − 5

2 FFG = 0 29 3 5 − 5.6 kN − FDF + FFG = 0 5 29 FDF = 4.00 kN T !

FFG = 1.6 29 kN

Joint G:

ΣFx = 0:

4 FEG − 5

FFG = 8.62 kN T !

5 (1.6 29 kN) = 0 29

FEG = 10.00 kN C ! from above (joint E)

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

FDE = 10.00 kN C !

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 14.

FBD Truss:

ΣFx = 0: A x = 0 ΣM A = 0:

4a H y − 3a (1.5 kN) − 2a (2 kN) − a (2 kN) = 0

ΣFy = 0:

Ay − 1 kN − 2 kN − 2 kN − 1.5 kN − 1 kN + 3.625 kN = 0

Joint FBDs: Joint A:

A y = 3.875 kN

FAB FAC 2.625 kN = = 1 29 26

FAB = 15.4823 kN,

FAB = 15.48 kN C !

FAC = 14.6597 kN,

FAC = 14.66 kN T !

By inspection of joint C: Joint B:

H y = 3.625 kN

ΣFy = 0:

FCE = FAC = 14.66 kN T,

2 (15.4823 kN − FBD ) − 2 kN = 0 29

FBD = 10.0971 kN, ΣFx = 0:

FBC = 0 !

FBD = 10.10 kN C !

5 (15.4823 kN − 10.0971 kN ) − FBE = 0 29

FBE = 5.0000 kN,

FBE = 5.00 kN C !

continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Joint D:

FDF = 10.0971 kN,

By symmetry:

ΣFy = 0:

FDF = 10.10 kN C !

 2  10.0971 kN  − 2 kN = 0 − FDE + 2   29 

FDE = 5.50 kN T !

FGH FFH 2.625 kN = = 1 26 29

Joint H:

By inspection of joint G: ΣFx = 0 :

Joint F:

FFH = 14.1361 kN

FFH = 14.14 kN C !

FGH = 13.3849 kN

FGH = 13.38 kN T !

FEG = FGH = 13.38 kN T

FEF +

and

FFG = 0 !

2 (10.0971 kN − 14.1361 kN ) = 0 29

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

FEF = 3.75 kN C !

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 15.

FBD Truss:

ΣFx = 0: A x = 0 ΣM A = 0:

8a ( J y − 1 kN) − 7a (1 kN) − 6a (2.8 kN) − 4a (4.5 kN) − 2a (4 kN) − a(1 kN) = 0

J y = 6.7 kN ΣFy = 0:

− 1 kN − 1 kN + 6.7 kN = 0

Joint FBDs: Joint A:

Ay − 1 kN − 1 kN − 1.4 kN − 4.5 kN − 2.8 kN

ΣFy = 0:

5 kN −

A y = 6.0 kN

3 5 13 FAB = 0, FAB = kN 3 13

FAB = 6.01 kN C ! ΣFx = 0:

Joint B:

ΣFx = 0:

FAC −

2 13

 5 13  kN  = 0,   3 

FAC = 3.33 kN T !

 2  5 13 kN − FBC − FBD  = 0,  13  3 

5 13 kN 3  3  5 13 kN + FBC − FBD  − 1 kN = 0,  13  3  4 13 FBD − FBC = kN 3 3 13 kN FBD = FBD = 5.41 kN C !

2 1 13 kN FBC = FBC = 0.601 kN C !

6 FBC + FBD =

ΣFy = 0:

continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

FEF = 0 !

By inspection of joint F: Joint E:

FDE = FEG

By symmetry: ΣFy = 0:

2

1 FDE − 4.5 kN = 0, 17

FDE =

9 17 kN 4 FDE = 9.28 kN C !

ΣFx = 0:

2 3 4 9   13 kN  − 17 kN  = 0   13  2 17  4  

FDF +

Joint D:

FDF = 6.00 kN T ! ΣFy = 0:

3 3 1 9   13 kN  − 1.4 kN − FCD − 17 kN  = 0   13  2 17  4   FCD = 0.850 kN T !

Joint C:

ΣFy = 0:

0.850 kN −

FCG = ΣFx = 0:

FCI −

3 13

 13  3 kN  − FCG = 0   5  6 

1.75 kN 3

FCG = 0.583 kN C !

 10 4  1.75 2  13  kN  − kN  − kN = 0   5 3 13  6   3

FCI = 3.47 kN !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 16. \

FBD Truss:

ΣFx = 0: ΣM A = 0:

Ax = 0 8a( J y − 1 kN) − 7a(1 kN) − 6a(2.8 kN) − 4a(4.5 kN) − 2a(4 kN) − a(1 kN) = 0

J y = 6.7 kN ΣFy = 0:

Ay − 1 kN − 1 kN − 1.4 kN − 4.5 kN − 2.8 kN −1 kN − 1 kN + 6.7 kN = 0

Joint FBDs: Joint J:

ΣFy = 0:

(6.7 − 1) kN −

3 FHJ = 0, 13

A = 6.0 kN FHJ = 1.9 13 kN FHJ = 6.85 kN C !

ΣFx = 0:

Joint H:

ΣFx = 0: ΣFy = 0:

2 (1.9 13 kN) − FIJ = 0, 13 2 ( FGH + FHI − 1.9 13 kN) = 0 13

3 ( FHI − FGH + 1.9 13 kN) − 1 kN = 0 13 26 13 kN, FGH = FGH = 6.25 kN C !

15

FHI = Joint I:

ΣFx = 0:

3.80 kN −

FGI −

13 kN, 6

FHI = 0.601 kN C !

 2  13 kN  − FCI = 0  13  6 

FCI = ΣFy = 0:

FIJ = 3.80 kN T !

10.4 kN, 3

 3  13 kN  = 0,   13  6 

FCI = 3.47 kN T !

FGI = 0.500 kN T !

continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

FEF = 0 !

By inspection of joint F, By symmetry FDE = FEG Joint E: ΣFy = 0:

 1  FEG  − 4.5 kN = 0, 2 17  

FEG =

9 17 kN 4 FEG = 9.28 kN C !

Joint G:

ΣFy = 0:

3  26 3 1 9  1  13 kN  − kN − FCG − 17 kN    5 13  15 17  4  2  − 2.8 kN = 0 FCG = −

ΣFx = 0:

1.75 kN 3

FCG = 0.583 kN C !

4 9 4  1.75   17 kN  − FFG −  −   5 3  17  4  −

2  26  13 kN  = 0  13  15  FFG = 6.00 kN T !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 17.

FBD Truss:

By load symmetry,

ΣFx = 0:

θ = tan −1

Joint FBDs:

A y = H y = 1600 lb

Ax = 0

6.72 ft = 16.2602° 23.04 ft

ΣFy′ = 0:

(1600 lb − 400 lb) cosθ − FAC sin θ = 0

FAC = 4114.3 lb

Joint A:

ΣFx = 0:

FAC cos 2θ − FAB cosθ = 0 FAB = 3613.5 lb

ΣFx′ = 0: Joint B:

Joint C:

FBC = 0.768 kips C !

FCD sin 2θ − (768 lb) cosθ = 0 FCD = 1371.4 lb

ΣFx′′ = 0:

FBD = 3.84 kips C !

FBC − (800 lb) cosθ = 0 FBC = 768.00 lb

ΣFy′′ = 0:

FAB = 3.61 kips C !

3613.5 lb − FBD + (800 lb)sin θ = 0 FBD = 3837.5 lb

ΣFy′ = 0:

FAC = 4.11 kips T !

FCD = 1.371 kips T !

FCE + (1371.4 lb) cos2θ − (768 lb)sin θ − 4114.3 lb = 0 FCE = 2742.9 lb

FCE = 2.74 kips T !

continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Joint E: ΣFy = 0:

(2742.9 lb)sin 2θ − FDE cosθ = 0

FDE = 1536.01 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

FDE = 1.536 kips C !

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 18.

FBD Truss: A y = H y = 1600 lb

By load symmetry, ΣFx = 0:

Ax = 0

θ = tan −1

6.72 ft = 16.2602° 23.04 ft

Joint FBDs: Joint H:

ΣFy = 0:

1600 lb − 400 lb − FFH sin θ = 0 FFH = 4285.7 lb,

( 4285.7 lb ) cosθ

ΣFx = 0:

Joint F:

− FGH = 0

FGH = 4114.3 lb

ΣFy′ = 0:

FGH = 4.11 kips T

FFG − ( 800 lb ) cosθ = 0 FFG = 768.0 lb

ΣFx′ = 0:

FFH = 4.29 kips C

FFG = 0.768 kips C

FDF + ( 800 lb ) sin θ − 4285.7 lb = 0 FDF = 4061.7 lb

FDF = 4.06 kips C

Joint G: ΣFy = 0:

FDG sin 2θ − (768 lb) cosθ = 0 FDG = 1371.4 lb

ΣFx = 0:

FDG = 1.371 kips T

4114.3 lb − (1371.4 lb ) cosθ − FEG − (768 lb) sin θ = 0 FEG = 2742.9 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

FEG = 2.74 kips T

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Chapter 6, Solution 19.

FBD Truss:

ΣFx = 0: ΣM L = 0:

Ax = 0

(1.5 m ) (6.6 kN) + (3.0 m)(2.2 kN) + (5.5 m)(6 kN) − (11 m) Ay = 0

A y = 4.5 kN

Joint FBDs: Joint A:

ΣFy = 0:

4.5 kN − 6 kN − 2.2 kN − 6.6 kN + L = 0

L = 10.3 kN 4.5 kN FAC F = = AB , 1 2 5

FAC = 9.00 kN T !

FAB = 4.5 5,

Joint C:

FAB = 10.06 kN C !

9 kN FBC F = = CE , 16 5 281

FBC =

45 kN 16

FBC = 2.81 kN C !

FCE =

Joint B:

ΣFx = 0: ΣFy = 0:

Solving:

9 281, 16

FCE = 9.43 kN T !

2  16 FBD + (4.5 5) kN  + FBE = 0   5 265 1  FBD + (4.5 5) kN  −  5

3 45 kN = 0 FBE + 16 265

72 5 kN, 11 45 265 kN, FBE = 176 FBD = −

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

FBD = 14.64 kN C !

FBE = 4.16 kN T !

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Joint E:

ΣFx = 0:

16   9  FEG −  16 281  

FEG = ΣFy = 0:

 16  45   281  kN  − 265  kN = 0  265  176   

9 281 kN, 11

FEG = 13.72 kN T !

5 9 9  281 kN − 281 kN  +  16 281  11 

3  45  265 kN   265  176 

+ FDE = 0, FDE = −

Joint D:

ΣFx = 0:

ΣFy = 0: Solving:

45 kN, 22

FDE = 2.05 kN C !

2  72 10  FDG = 0 5 kN  +  FDF + 11 5 101 

1  72 1 45  FDG + 5 kN  − kN = 0  FDF + 11 22 5 101  7.5 101 kN, 22

FDG = 3.43 kN T !

FDF = − 8.25 5 kN,

FDF = 18.45 kN C !

FDG =

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 20.

FBD Truss:

ΣFx = 0: ΣM L = 0:

Ax = 0

(1.5 m ) (6.6 kN) + (3.0 m)(2.2 kN) + (5.5 m)(6 kN) − (11 m) Ay = 0

A y = 4.5 kN

Joint FBDs:

ΣFy = 0:

4.5 kN − 6 kN − 2.2 kN − 6.6 kN + L = 0

Joint L:

L = 10.3 kN 10.3 kN FKL FJL = = , 1 2 5

FKL = 20.6 kN T !

FJL = 10.3 5 kN,

FJL = 23.0 kN C !

20.6 kN FJK F = = IK , 16 5 281

Joint K:

FJK =

51.5 kN 8

FJK = 6.44 kN C !

FIK =

Joint J:

ΣFx = 0:

ΣFy = 0:

Solving:

−

10.3 281 kN, 8

(

FIK = 21.6 kN T!

)

2 16 FHJ + 10.3 5 kN − FIJ = 0 5 265

(

)

1 FHJ + 10.3 5 kN − 5

FHJ = − FIJ = −

3 51.5 kN = 0 FIJ − 6.6 kN + 8 265

112 5 kN, 11 1.3 265 kN, 88

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

FHJ = 22.8 kN C !

FIJ = 0.240 kN C !

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Joint I:

16  10.3 16  1.3   281 kN  − 265 kN  = 0  − FGI +  8 88 281  265   

ΣFx = 0:

FGI = ΣFy = 0:

ΣFx = 0:

ΣFy = 0:

−

10.4 kN, 88

1  112 1 10.4  FGH − 2.2 kN − 5 kN  − kN = 0  FFH + 11 88 5 101  FFH = − 8.25 5 kN, FGH = −

34 101 kN, 88

FFH = 18.45 kN C !

FGH = 3.88 kN C !

FDF = 8.25 5 kN

By symmetry: ΣFy = 0:

FHI = 0.1182 kN T !

2  112 10  FGH = 0 5 kN  −  FFH + 11 5 101 

Solving:

Joint F:

FGI = 21.3 kN T !

5  112 10.3  281 kN − 281 kN  + FHI  8 281  88  3  1.3  265 kN  = 0 −  265  88  FHI =

Joint H:

112 281 kN, 88

(

)

2 8.25 5 kN − 6 kN − FFG = 0 5

FFG = 10.50 kN T !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 21.

Joint FBDs:

FDF F 500 lb = EF = 916 916 480

Joint F:

FDF = FEF = 954.17 lb FDF = 954 lb T !

or

FEF = 954 lb C !

Joint D:

FBD FDE 954.17 lb = = 884 240 916 FBD = 920.84 lb

FDE = 250.00 lb FBD = 921 lb T !

or

FDE = 250 lb C !

By symmetry of joint A vs. joint F FAB = 954 lb T !

FAC = 954 lb C !

Joint B: ΣFx = 0:

920.84 lb −

884 4 ( 954.17 lb ) + FBE = 0 916 5 FBE = 0 !

ΣFy = 0:

FBC −

240 ( 954.17 lb ) = 0 916

FBC = 250.00 lb

or

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

FBC = 250 lb C !

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Joint C:

ΣFx = 0:

884 ( 954.17 lb ) − FCE = 0 916 FCE = 920.84 lb

ΣFy = 0:

FCH −

240 (954.17 lb) − 250.00 lb = 0 916

FCH = 500.00 lb ΣFx = 0:

FCE = 921 lb C !

or

920.84 lb −

or

FCH = 500 lb C !

884 4 ( 954.17 lb ) − FHE = 0 916 5 FHE = 0 !

Joint E: ΣFy = 0:

FEJ −

240 (954.17 lb) − 250.00 lb = 0 916

FEJ = 500.00 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

or

FEJ = 500 lb C !

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Chapter 6, Solution 22.

Joint FBDs:

FJL F 500 lb = KL = 1212 1212 480

Joint L:

FJL = FKL = 1262.50 lb FJL = 1263 lb T !

FKL = 1263 lb C !

Comparing joint G to joint L:

FGH = 1263 lb T !

FGI = 1263 lb C !

Joint J:

ΣFx = 0:

1188 (1262.50 lb ) − FHJ = 0 1212 FHJ = 1237.50 lb,

ΣFy = 0:

FJK −

240 (1262.50 lb ) − 500 lb = 0 1212

FJK = 750.00 lb,

Joint H:

ΣFx = 0:

FHJ = 1238 lb T !

1237.50 lb −

FJK = 750 lb C !

1188 4 (1262.50 lb ) + FHK = 0 1212 5 FHK = 0 !

ΣFy = 0:

FHI −

240 3 (1262.50 lb) − 500 lb − (0) = 0 1212 5 FHI = 750 lb C !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Joint I:

ΣFx = 0:

1188 (1262.50 lb ) − FIK = 0 1212 FIK = 1237.50 lb,

ΣFy = 0:

FIN −

240 (1262.50 lb ) − 750.00 lb = 0 1212

FIN = 1000.00 lb, Joint K:

ΣFx = 0:

FIK = 1238 lb C !

1237.50 lb −

FIN = 1000 lb C !

1188 4 (1262.50 lb ) − FKN = 0 1212 5 FKN = 0 !

ΣFy = 0:

FKO −

240 (1262.50 lb) − 750.00 lb = 0 1212 FKO = 1000 lb C !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 23.

FBD Truss:

ΣFx = 0:

ΣM A = 0:

Ax = 0

(12 m ) (M y − 1 kN) − (2.4 m + 4.8 m + 6 m + 8.4 m + 10.8 m) (1.5 kN) = 0

M y = 5.05 kN ΣFy = 0:

Ay − 2(1 kN) − 5(1.5 kN) + M y = 0

A y = 4.45 kN

Joint FBDs:

ΣFy = 0:

4.45 kN − 1 kN −

Joint A: ΣFx = 0:

ΣFx = 0:

Joint C: ΣFy = 0:

ΣFy = 0:

Joint B:

FAC −

5 FAB = 0, 13

12 (8.97 kN ) = 0, 13

FAC = 8.28 kN T !

12 FCE − 8.28 kN = 0, 13

FCE = 8.97 kN T !

5 (8.97 kN) − FBC = 0, 13

FBC = 3.45 kN C !

5 5 (8.97 kN ) − 1.5 kN + 3.45 kN − FBD = 0 13 13 FBD = 14.04 kN

ΣFx = 0:

FAB = 8.97 kN C !

FBD = 14.04 kN C !

12 12 (8.97 kN) − (14.04 kN) + FBE = 0 13 13 FBE = 4.68 kN

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

FBE = 4.68 kN T !

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Joint E:

ΣFx = 0:

6 12 FEH − 4.68 kN − (8.97 kN) = 0 13 37

FEH = 13.1388 kN ΣFy = 0:

FDE −

5 (8.97 kN ) − 13

or

1 (13.1388) = 0 37

FDE = 5.6100 kN Joint D:

ΣFx = 0:

ΣFy = 0:

FEH = 13.14 kN T !

or

FDE = 5.61 kN T !

12 12 1 (14.04 kN ) − FDG − FDH = 0 13 13 2

5 5 (14.04 kN ) − FDG − 1.5 kN − 5.61 kN 13 13 1 + FDH = 0 2 Solving:

FDG = 8.60 kN C !

FDH = 7.10 kN C !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 24.

FBD Truss:

ΣFx = 0:

Ax = 0

ΣM A = 0:

(12 m ) (M y − 1 kN) − (2.4 m + 4.8 m + 6 m + 8.4 m + 10.8 m) (1.5 kN) = 0

M y = 5.05 kN

Joint FBDs:

ΣFy = 0:

Ay − 2(1 kN) − 5(1.5 kN) + M y = 0

A y = 4.45 kN

Joint M: ΣFy = 0:

5.05 kN − 1 kN −

5 FKM = 0, 13

ΣFx = 0:

12 (10.53 kN) − FLM = 0, 13

ΣFx = 0:

9.72 kN −

FKM = 10.53 kN C !

FLM = 9.72 kN T !

6 FJL = 0, 37

FJL = 1.62 37

Joint L:

FJL = 9.85 kN T ! ΣFy = 0:

ΣFx = 0:

Joint K:

ΣFy = 0:

1 (1.62 37 kN) − FKL = 0, 37 12 FIK − 13

FKE = 1.620 kN C !

24 12 FJK − (10.53 kN) = 0 13 577

5 5 (10.53 kN) − FIK − 13 13

1 FJK 577

− 1.5 kN + 1.62 kN = 0 Solving:

FIK = 10.8136 kN,

FIK = 10.81 kN C !

FJK = 0.26205 kN,

FJK = 0.262 kN T !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Joint J:

ΣFx = 0:

24 (0.26205 kN) − 577

6 ( FHJ − 9.85 kN) = 0 37

FHJ = 10.1154 kN, ΣFy = 0:

1 (10.1154 kN ⋅ 9.85 kN) − 37

FHJ = 10.12 kN T !

1 (0.26205 kN) − FIJ = 0 577

FIJ = 0.054541 kN,

Joint I:

ΣFx = 0: ΣFy = 0: Solving:

Joint G:

12 24 ( FGI − 10.8136 kN ) + FHI = 0 13 25 5 7 (10.8136 kN − FGI ) + FHI − 1.5 kN + 0.05454 kN = 0 13 25 FGI = 8.6029 kN,

FGI = 8.60 kN C !

FHI = 2.1257 kN,

FHI = 2.13 kN C !

By symmetry:

ΣFy = 0:

FIJ = 54.5 N C !

FOG = 8.60 kN C !

 5 2   (8.6029 kN) − 1.5 kN − FGH = 0  13  FGH = 5.12 kN T !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 25.

FBD Truss:

ΣFx = 0: 180 lb − Ax = 0 A x = 180 lb ΣM A = (12 ft ) G y − ( 6 ft )( 480 lb ) − ( 2 ft )(180 lb ) − (8 ft)(120 lb) = 0, ΣFy = 0:

G y = 350 lb

Ay − 480 lb − 120 lb + 350 lb = 0

A y = 250 lb

Joint FBDs: Joint A:

ΣFx = 0:

1 FAC − 180 lb = 0, FAC = 180 2 lb 2

FAC = 255 lb T ! ΣFy = 0:

Joint B:

ΣFx = 0: ΣFy = 0:

Solving:

1 (180 2 lb) + 250 lb − FAB = 0, 2

FAB = 430 lb C !

4 2 FBD + FBC = 0 5 5 3 1 430 lb − FBD + FBC = 0 5 5 180 lb −

FBC = 590 5 lb,

FBC = 1319 lb T !

FBD = 1700 lb C !

Joint C:

ΣFy = 0:

5 1 1 (590 5 lb) − (180 2 lb) = 0 FCD − 41 5 2

FCD = 154 41 lb, ΣFx = 0:

FCD = 986 lb T !

4 2 (154 41 lb) − (590 5 lb) 41 5 1 − (180 2 lb) = 0 12 FCE = 744 lb T !

FCE +

By inspection of joint G:

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

FFG = 350 lb C and FEG = 0 !

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Joint F: FDF F 350 = EF = ; 5 1 20

FDF = 1750 lb C !

FEF = 700 5 lb,

FEF = 1565 lb T !

Joint E: ΣFy = 0:

5 1 (700 5 lb) = 0 FDE − 120 lb − 41 5

FDE = 164 41 lb,

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

FDE = 1050 lb T !

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Chapter 6, Solution 26.

Joint FBDs: Joint A:

ΣFy = 0:

9 FAC − 1.8 kips = 0, 41 40 FAB − 8.20 kips = 0, 41

ΣFx = 0:

DE =

ΣFy = 0:

FBD = 8.00 kips C !

FBC = 0.6 kips,

and then

Note:

FAB = 8.00 kips C !

FBD = FAB

By inspection of joint B,

Joint C:

FAC = 8.20 kips T !

FBC = 0.600 kips C !

9.2 (4.5 ft) = 2.07 ft, 20

CE = 4.14 ft = 2 DE

1 FCD − 0.6 kips = 0, 5

FCD = 0.6 5 kips FCD = 1.342 kips T !

ΣFx = 0:

8.2 kips +

(

)

2 0.6 5 kips − FCE = 0 5

FCE = 9.20 kips C !

Joint D:

ΣFy = 0:

(

)

40 2 0.6 5 kips = 0 ( FDG − 8.2 kips ) − 41 5

FDG = 9.43 kips T ! ΣFx = 0:

FDE −

(

)

9 1 0.6 5 kips = 0 ( 9.43 kips − 8.2 kips ) − 41 5

FDE = 0.330 kips C !

Joint E:

ΣFy = 0: ΣFx = 0:

5 FEG − 0.33 kips = 0, 13

FEG = 0.858 kips T !

12 ( 0.858 kips ) + 9.2 kips − FEF = 0 13 FEF = 9.992 kips,

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

FEF = 9.99 kips C !

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Joint F:

By vertical symmetry FFG = FFH ΣFx = 0:

4  9.992 kips − 2  FFG  = 0, 5  

FFG = 5.995 kips FFG = FFH = 6.00 kips C !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 27.

P6.14 Starting with

ABC, add, in order, joints E, D, F, G, H ∴simple truss

P6.15

Starting with

DEF, add, in order, G, C, B, A, I, H, J ∴simple truss

P6.23

Starting with

ABC, add, in order, E, F, D, H, G, I, J, K, L, M ∴simple truss

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 28.

P6.21 Starting with ABC, add, in order, joints E, D, F, H, J, K, L, I, G, N, P, Q, R, O, M, S, T ∴simple truss

P6.25 Starting with

ABC, add, in order, joints D, E, F, G ∴simple truss

P6.29 Starting with

ABD, add, in order, joints H, G, F, E, I, G, J ∴simple truss

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 29.

ΣFx = 0:

Fx = 0

Then, by inspection of joint F,

FFG = 0

Then, by inspection of joint G,

FGH = 0

By inspection of joint J,

FIJ = 0

Then, by inspection of joint I,

FHI = 0 FEI = 0

Then, by inspection of joint E,

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

FBE = 0

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Chapter 6, Solution 30.

By inspection of joint D,

FDI = 0

By inspection of joint E,

FEI = 0

Then, by inspection of joint I,

FAI = 0

By inspection of joint F,

FFK = 0

By inspection of joint G,

FGK = 0

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell Š 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 31. \

By inspection of joint C:

FBC = 0 W

Then, by inspection of joint B:

FBE = 0 W

Then, by inspection of joint E:

FDE = 0 W

By inspection of joint H :

FFH = 0 W and FHI = 0 W

By inspection of joint Q :

FOQ = 0 W and FQR = 0 W

By inspection of joint J :

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell Š 2007 The McGraw-Hill Companies.

FLJ = 0 W

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Chapter 6, Solution 32.

By inspection of joint C :

FBC = 0 W

By inspection of joint G :

FFG = 0 W

Then, by inspection of joint F :

FFE = 0 W

By inspection of joint I :

FIJ = 0 W

By inspection of joint M :

FMN = 0 W

Then, by inspection of joint N :

FKN = 0 W

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell Š 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 33.

By inspection of joint C:

FBC = 0 !

By inspection of joint M :

FLM = 0 !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell Š 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 34.

By inspection of joint A:

FAF = 0 W

By inspection of joint C :

FCH = 0 W

By inspection of joint E :

FDE = FEI = 0 W

By inspection of joint L :

FGL = 0 W

By inspection of joint N :

FIN = 0 W

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell Š 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 35. (a)

By inspection of joint H :

FCH = 0 W

Then, by inspection of joint C :

FCG = 0 W

Then, by inspection of joint G :

FBG = 0 W

Then, by inspection of joint B :

FBF = 0 W

(b) By inspection of joint J : Then, by inspection of joint I :

FIJ = 0 W FEI = 0 W FHI = 0 W

Then, by inspection of joint E : also,

ΣFx = 0;

FBE = 0 W

Fx = 0

So, by inspection of joint F :

FFG = 0 W

And by inspection of joint G :

FGH = 0 W

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 36.

FBD Truss:

ΣFz = 0:

Bz = 0

ΣM x = 0:

− ( 0.47 m ) C y + ( 0.08 m )( 940 N ) = 0

C y = 160 Nj

Joint FBDs: Joint D: where FAD = FAD

FBD = FBD = FBD

FCD = FCD ΣFy = 0:

ΣFx = 0: ΣFz = 0:

4 FBD − 940 N = 0, 21 10 80 − FAD − FCD − 89 101 1 39 − FAD + FCD − 89 101 Solving:

− 0.80 m i − 0.08 m k

( 0.8)

2

= FAD

2

+ ( 0.08 ) m

−10i − k 101

− 0.8 m i − 0.16 m j − 0.2 m k

( 0.8 m )2 + ( 0.16 m )2 + ( 0.2 m )2 − 20i + 4 j − 5k 21 − 0.8 m i + 0.39 m k

( 0.8 m )

2

+ ( 0.39 m )

FBD = 4935 N,

2

= FCD

− 80i + 39k 89

FBD = 4.94 kN T !

20 ( 4935 N ) = 0 21 5 ( 4935 N ) = 0 21

FAD = − 590 101 N,

FAD = 5.93 kN C !

FCD = 1335 N,

FCD = 1.335 kN T !

continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Joint C:

FBC = FBC

0.16 m j − 0.59 m k

( 0.16 m )2 + ( 0.59 m )2

= FBC

16 j − 59k 3737

16 FBC + 160 N = 0 3737

ΣFy = 0:

FBC = −10 3737 N ΣFz = 0:

− FAC −

(

FBC = 611 N C !

)

59 39 −10 3737 N − (1335 N ) = 0 89 3737

FAC = 5.00 N T !

Joint B:

FAB = FAB

ΣFy = 0:

− 0.16 m j + 0.12 m k

( 0.16 m ) −

4 FAB + 5

2

+ ( 0.12 m )

2

(

− 4 j + 3k 5

= FAB

)

16 4 10 3737 N − ( 4935 N ) = 0 21 3737

FAB = 975 N C !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 37.

FBD Truss:

ΣFz = 0:

Bz + 987 N = 0 B z = − 987 N k

ΣM x = 0:

− ( 0.47 m ) C y − ( 0.16 m )( 987 N ) + ( 0.08 m )( 940 N ) = 0

C y = −176.0 N j

Joint FBDs: Joint D: FAD = FAD FBD = FBD FCD = FCD

ΣFy = 0:

ΣFx = 0: ΣFz = 0:

− 0.8 m i − 0.08 m k

( 0.8 m )2 + ( 0.08 m )2

FAD ( −10i − k ) 101

=

− 0.8 m i − 0.16 m j − 0.2 m k

( 0.8 m )

2

2

+ ( 0.16 m ) + ( 0.2 m )

− 0.8 m i + 0.39 m k

( 0.8 m )2 + ( 0.39 m )2

=

2

=

FCD ( − 80i + 39k ) 89

4 FBD − 940 N = 0, FBD = 4935 N, 21 10 80 20 − FAD − FCD − ( 4935 N ) = 0 89 21 101 1 39 5 − FAD + FCD − ( 4935 N ) + 987 N = 0 89 21 101

Solving:

FBD ( − 20i + 4 j − 5k ) 21

FCD = − 534 N, FAD = − 422 101 N,

FBD = 4.94 kN T !

FCD = 534 N C !

FAD = 4.24 kN C !

continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Joint C:

FBC = FBC

0.16 m j − 0.59 m k

( 0.16 m )

2

+ ( 0.59 m )

2

=

FBC (16 j − 59k ) 3737

16 FBC − 176 N = 0, FBC = 11 3737 N 3737

ΣFy = 0:

FBC = 672 N T ! ΣFz = 0:

− FAC −

(

)

59 39 11 3737 N + ( 534 N ) = 0 89 3737

FAC = − 415 N,

FAC = 415 N C !

Joint B: FAB = FAB

ΣFy = 0:

− 0.16 m j + 0.12 m k

( 0.16 m )2 + ( 0.12 m )2 −

4 FAB − 5

(

=

FAB ( − 4 j + 3k ) 5

)

16 4 11 3737 N − ( 4935 N ) = 0 21 3737

FAB = −1395 N, FAB = 1.395 kN C !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 38.

FBD Truss: ΣM BD = 0: ΣM x = 0:

ΣFz = 0:

( 4 ft )( 50 lb ) + ( 4 ft )( Ez ) = 0 E z = − ( 50 lb ) k ( 4 ft )( 300 lb ) + ( 4 ft )( − 50 lb ) + ( 4 ft )( Dz ) = 0 D z = − ( 250 lb ) k

Bz − 50 lb − 250 lb = 0 B z = ( 300 lb ) k

ΣM Bz = 0:

( 2 ft )( 300 lb ) − ( 4 ft ) ( C y ) = 0

C y = (150 lb ) j

AB = AC = 20 ft AD = AE = 6 ft CD = 4 2 ft

ΣFx = 0: ΣFy = 0:

B x = − ( 50 lb ) i

Bx + 50 lb = 0 By + 150 lb − 300 lb = 0

B y = (150 lb ) j

FAE = FAE

=

2 ft i − 4 ft j + 4 ft k

( 2 ft )2 + ( 4 ft )2 + ( 4 ft )2

FAE ( 2i − 4 j + 4k ) 3

Joint FBDs: Joint E:

ΣFz = 0: ΣFx = 0:

2 FAE − 50 lb = 0, 3 1 FDE + ( 75 lb ) = 0, 3

FAE = 75.0 lb T !

FDE = − 25 lb FDE = 25.0 lb C !

ΣFy = 0:

− FCE −

2 ( 75 lb ) = 0, 3

FCE = − 50 lb FCE = 50.0 lb C !

continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

− 2 ft i − 4 ft j + 4 ft k

FAD = FAD

( 2 ft )

Joint D:

ΣFx = 0:

2

+ ( 4 ft ) + ( 4 ft )

− 4 ft i − 4 ft j

FCD = FCD ΣFz = 0:

2

( 4 ft )

2

+ ( 4 ft )

2

=

2 ft i + 4 ft k

FAC = FAC

( 2 ft )

)

Joint B:

FAB = FAB

2

+ ( 4 ft )

2

=

FAC = 0 !

FBC − 100 lb = 0,

− 2 ft i + 4 ft k

( 2 ft )

2

ΣFz = 0:

+ ( 4 ft )

2

=

FBD = 150.0 lb C !

FAC ( i + 2k ) 5

2 FAC = 0, 5

ΣFz = 0: ΣFx = 0:

FAD ( − i − 2 j + 2k ) 3

2 FAD − 250 lb = 0, FAD = 375 lb, FAD = 375 lb T !

3 1 1 FCD − ( 375 lb ) = 0, FCD = −100 2 lb 25 lb − 3 2 FCD = 141.4 lb C !

1 2 − FBD − −100 2 lb − ( 375 lb ) = 0 3 2

FBD = − 150 lb Joint C:

=

FCD ( − i − j) 2

(

ΣFy = 0:

2

FBC = 100.0 lb T !

FAB ( − i + 2k ) 5

2 FAB + 300 lb = 0 5

FAB = −150 5 lb,

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

FAB = 335 lb C !

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 39. (a)

FBD Truss:

Bx = 0

ΣM y = 0:

Bx = 0

ΣM z = 0:

(1.7 m ) Az + ( 0.6 m )(1700 N ) = 0, − (1.7 m ) Ax − (1.125 m )(1700 N ) = 0

ΣFx = 0:

Cx − 1125 N = 0

Cx = 1125 N

C y − 1700 N = 0

C y = 1700 N

Bz − 600 N = 0

Bz = 600 N

ΣM x = 0:

ΣFy = 0: ΣFz = 0:

Az = − 600 N Ax = −1125 N

A = − (1125 Ν ) i − ( 600 N ) k

so

B = ( 600 Ν ) k C = (1125 Ν ) i + (1700 N ) j

(b) Find zero force members:

by inspection of joint B,

FAB = 0 !

by inspection of joint D,

FAD = 0 !

by inspection of joint C ,

FBC = 0 !

Joint FBDs:

Joint E:

FAE = FAE

ΣFy = 0:

−1.125 m i + 1.7 m j − 0.6 m k

(1.125 m )

2

2

+ (1.7 m ) + ( 0.6 m )

1.7 FAE − 1700 N = 0, 2.125

2

=

FAE ( −1.125i + 1.7 j − 0.6k ) 2.125

FAE = 2125 N FAE = 2.13 kN T !

ΣFx = 0:

− FBE −

1.125 ( 2125 N ) , FBE = −1125 N 2.125 FBE = 1.125 kN C !

ΣFz = 0:

− FDE −

0.6 ( 2125 N ) = 0, 2.125

FDE = − 600 N FDE = 600 N C ! continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Joint D:

FBD = FBD

ΣFz = 0:

−1.125 m i + 0.6 m k

(1.125 m )2 + ( 0.6 m )2 600 N −

0.6 FBD = 0, 1.275

=

FBD ( −1.125i + 0.6k ) 1.275 FBD = 1275 N FBD = 1.275 kN T !

ΣFx = 0:

− FCD −

1.125 (1275 N ) = 0, 1.275

FCD = −1125 N FCD = 1.125 kN C !

Joint C: By inspection:

FAC = −1700 N

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

FAC = 1.700 kN C !

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 40. (a)

(b)

To check for simple truss, start with ABDE and add three members at a time which meet at a single new joint, successively adding joints G, F, H and C. ∴ This is a simple truss !

There are six reaction force components, Ax, Ay, Az, By, Bz, and Gy, no two of which are colinear, so all can be determined with the six equilibrium equations. Motion is prevented by the constraints. ∴ Truss is completely constrained, and statically determinate. !

FBD Truss:

ΣM AB = 0:

( 9.60 ft ) G y + (10.08 ft )( 240 lb ) = 0

G y = − 252 lb ΣM z = 0:

G y = − 252 lb j

(11.00 ft ) ( By − 252 lb ) − (10.08 ft )( 275 lb ) = 0 B y = 504 lb j

ΣFx = 0: ΣM y = 0:

− Ax + 275 lb = 0,

A x = − 275 lb i

− ( 9.6 ft )( 275 lb ) − (11.0 ft )( Bz ) = 0

Bz = − 240 lb,

B z = − 240 lb k

ΣFy = 0:

− Ay + 504 lb − 252 lb = 0,

A y = − 252 lb j

ΣFz = 0:

Az − 240 lb + 240 lb = 0

Az = 0

Determine zero force members: By inspection of joint C :

FBC = FCD = FGC = 0

By inspection of joint F :

FBF = FEF = FFG = 0

By inspection of joint A:

with Az = 0,

By inspection of joint H :

FDH = 0

FAD = 0

And finally, considering joint D knowing that FAD = FCD = FDH = 0, and recognizing that the three remaining members are not co-planar, they must also carry zero load, FBD = FDE = FGH = 0.

The only load bearing members are thus AB, AE, BE, BG, EG, EH, GH. continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Joint FBDs

FAB = 275 lb T,

By inspection: from above:

Joint A:

FAE = 252 lb T !

FDE = 0 !

FEF = 0 !

FEH = 240 lb C !

By inspection of joint H

ΣFy = 0:

− 252 lb −

10.08 FBE = 0 1492 FBE = − 373 lb,

Joint E:

ΣFz = 0:

240 lb −

FBE = 373 lb C !

9.6 FEG = 0 14.6 FEG = 365 lb T !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 41. (a)

(b)

To check for simple truss, start with ABDE and add three members at a time which meet at a single new joint, successively adding joints G, F, H and C. ∴ This is a simple truss !

There are six reaction force components, Ax, Ay, Az, By, Bz, and Gy, no two of which are colinear, so all can be determined with the six equilibrium equations. Motion is prevented by the constraints. ∴ Truss is completely constrained, and statically determinate. !

FBD Truss:

ΣM AB = 0:

( 9.60 ft ) G y + (10.08 ft )( 240 lb ) = 0

G y = − 252 lb, ΣM z = 0:

G y = − 252 lb j

(11.00 ft ) ( By − 252 lb ) − (10.08 ft )( 275 lb ) = 0 B y = 504 lb j

ΣFx = 0: ΣM y = 0:

− Ax + 275 lb = 0,

Ax = − 275 lb i

− ( 9.6 ft )( 275 lb ) − (11.0 ft )( Bz ) = 0

Bz = − 240 lb,

B z = − 240 lb k

ΣFy = 0:

− Ay + 504 lb − 252 lb = 0,

A y = − 252 lb j

ΣFz = 0:

Az − 240 lb + 240 lb = 0

Az = 0

Determine zero force members: By inspection of joint C :

FBC = FCD = FGC = 0

By inspection of joint F :

FBF = FEF = FFG = 0

By inspection of joint A:

with Az = 0,

By inspection of joint H :

FDH = 0

FAD = 0

And finally, considering joint D knowing that FAD = FCD = FDH = 0, and recognizing that the three remaining members are not co-planar, they must also carry zero load, FBD = FDE = FGH = 0.

The only load bearing members are thus AB, AE, BE, BG, EG, EH, GH. continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Now, by inspection of joint H,

FGH = 275 lb C !

FCG = 0 !

from above,

FDG = 0 !

FFG = 0 !

FBD Joint G:

FBG = FBG

−10.08j + 9.6k 13.92

FEG = FEG

−11i + 9.6k 14.6

ΣFx = 0: ΣFy = 0:

275 lb − −

11 FEG = 0, 14.6

10.08 FBG − 252 lb = 0, 13.92

FEG = 365 lb T !

FBG = − 348 lb FBG = 348 lb C !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 42.

FBD Truss: ΣFx = 0:

Kx = 0

(

)

6a K y − 125 lb − 5a ( 250 lb )

ΣM A = 0:

− 4a ( 250 lb ) − 3a ( 375 lb ) − 2a ( 500 lb ) − a ( 500 lb ) = 0 K y = 937.5 lb

Ay − 3 ( 250 lb ) − 2 ( 500 lb )

ΣFy = 0:

− 375 lb − 125 lb + 937.5 lb = 0 A y = 1312.5 lb

FBD Section ABEC:

( 2 ft ) FCF + ( 4 ft )( 500 lb )

ΣM E = 0:

+ ( 8 ft )( 250 lb − 1312.5 lb ) = 0 FCF = 3250 lb, ΣFy = 0:

FCF = 3.25 kips T W

1312.5 lb − 250 lb − 2 ( 500 lb ) −

FEF = 62.5 5 lb, ΣFx = 0:

3250 lb +

(

1 FEF = 0 5

FEF = 139.8 lb T W

)

2 62.5 5 lb − FEG = 0 5

FEG = 3375 lb,

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

FEG = 3.38 kips C W

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 43.

FBD Truss: ΣFx = 0: ΣM A = 0:

Kx = 0

(

)

6a K y − 125 lb − 5a ( 250 lb ) − 4a ( 250 lb ) − 3a ( 375 lb ) − 2a ( 500 lb ) − a ( 500 lb ) = 0 K y = 937.5 lb

ΣFy = 0:

Ay − 3 ( 250 lb ) − 2 ( 500 lb ) − 375 lb − 125 lb + 937.5 lb = 0 A y = 1312.5 lb

FBD Section IJLK: ΣM z = 0:

( 4 ft )( 937.5 lb −125 lb ) − ( 2 ft ) FHJ = 0 FHJ = 1625 lb,

ΣFy = 0:

937.5 lb − 125 lb − 250 lb −

FHI = 562.5 5 lbs, ΣFx = 0:

1.625 kips +

(

FHJ = 1.625 kips C W 1 FHI = 0 5

FHI = 1.258 kips C W

)

2 0.5625 5 kips − FFI = 0 5

FFI = 2.75 kips T W

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 44.

FBD Truss: ΣFx = 0:

Ax = 0

By load symmetry: A y = L y = 840 N

ΣM E = 0:

2 24 (1.75 m ) FBD + ( 2 m )( 240 N ) 3 25 − ( 4 m )( 840 N − 120 N ) = 0 FBD =

FBD Section ABC: ΣM B = 0:

FBD = 2.14 kN C W

 1.75  m  FCE − ( 2 m )( 840 N − 120 N ) = 0   3  FCE =

ΣM A = 0:

15000 N, 7

17280 N, 7

FCE = 2.47 kN T W

( 4 m ) 

7  FBE  − ( 2 m )( 240 N ) = 0  25  FBE =

3000 N, 7

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

FBE = 429 N C W

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 45.

FBD Truss: ΣFx = 0:

Ax = 0

By load symmetry A y = L y = 840 N

0.4 2 = 3 15 3.5 7 IK = 3 = 4 24 2.3 23 IJ = = 3 30

slope: GI =

FBD Section JKL:

Resolve FIK and FGJ at L ΣM J = 0:

( 3 m ) 

7   24  FIK  − ( 0.4 m )  FIK  25 25    

+ ( 3 m )( 840 N − 120 N ) − (1 m )( 240 N ) = 0 FIK = 4210.53 N, ΣM I = 0:

FIK = 4.21 kN C W



  3.5   15  2 FGJ  −  FGJ  m   229  229   3 

(4 m) 

+ ( 4 m )( 840 N − 120 N ) − ( 2 m )( 240 N ) = 0 FGJ = 3823.0 N, ΣFx = 0:

FGJ = 3.82 kN T W

7 15 30 FIJ = 0 ( 4210.5 N ) − ( 3823.0 N ) − 25 229 1429

FIJ = 318 N T W

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 46. \

FBD Truss:

ΣFx = 0: By symmetry,

Ax = 0

A y = M y = 8.5 kN

FBD Section ABDC: ΣM D = 0:

(1 m )(1.5 kN − 8.5 kN ) + ( 0.5 m )

40 FCE = 0 41

FCE = 14.35 kN T W ΣM E = 0:

(1 m )(1.5 kN ) + ( 2 m )(1.5 kN − 8.5 kN )

+ ( 0.5 m )

40 FDF = 0, 41

FDF = 25.625 kN FDF = 25.6 kN C W

ΣFx = 0:

40 40 FDE = 0 (14.35 kN − 25.625 kN ) + 41 1721

FDE = 11.4084 kN,

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

FDE = 11.41 kN T W

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 47.

FBD Truss:

ΣFx = 0: By symmetry,

Ax = 0

A y = M y = 8.5 kN

FBD Section IMNJ:

ΣM J = 0:

( 2 m )(8.5 kN − 1.5 kN ) − (1 m )(1.5 kN ) − ( 0.5 m )

40 FGI = 0 41

FGI = 25.625 kN, ΣM G = 0:

FGI = 25.6 kN T W

( 3 m )(8.5 kN − 1.5 kN ) − ( 2 m )(1.5 kN ) 40 FHJ = 0 41 FHJ = 30.8 kN C W

− (1 m )( 3 kN ) − ( 0.5 m ) FHJ = 30.75 kN, ΣFx = 0:

40 ( 30.75 kN − 25.625 kN ) − 41

40 FGJ = 0 2441

FGJ = 6.18 kN T W

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 48.

FBD Truss:

ΣFx = 0: A x = 0 By symmetry: A y = L y = 6 kN

FBD Section:

Notes:

15 m 6 2 5 5 yD = ⋅ = m 3 2 3 2 2 yE = ⋅ 1 = m 3 3 5 yF − yD = m 6 yG = 1 m yF =

yD − yG =

2 m 3 continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

ΣM D = 0: (1 m )

6 FEG + ( 2 m )( 2 kN ) + ( 4 m )(1 kN − 6 kN ) = 0 37

FEG =

8 37 kN 3

FEG = 16.22 kN T !

 1   3  ΣM A = 0: ( 2 m )( 2 kN ) + ( 4 m )( 2 kN ) − ( 6 m )  FDG  − (1 m )  FDG  = 0  10   10 

FDG = ΣFx = 0:

4 10 kN 3

6 3 12 FEG − FDG − FDF = 0 13 37 10

FDG = 4.22 kN C !

16 − 4 −

FDF = 13 kN

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

12 FDF = 0 13

FDF = 13.00 kN C !

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 49.

FBD Truss:

ΣFx = 0: A x = 0 By symmetry: A y = L y = 6 kN

FBD Section:

Notes:

so

2 3 2 = 3

yI =

m

yH

⋅

5 5 = m 2 3

yH − yI = 1 m

continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

ΣM I = 0:

( 4 m )( 6 kN − 1 kN ) − ( 2 m )( 2 kN ) − (1 m ) 

12  FHJ  = 0  13 

FHJ = ΣM H = 0:

FHJ = 17.33 kN C !  6  FGI  = 0  37 

( 4 m )( 6 kN − 1 kN ) − ( 2 m )( 2 kN ) − (1 m )  FGI =

ΣM L = 0:

52 kN 3

8 37 kN 3

( 2 m )( 2 kN ) − ( 4 m ) FHI

FGI = 16.22 kN T ! =0

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

FHI = 1.000 kN T !

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Chapter 6, Solution 50.

FBD Truss:

Distance between loads = 1.5 m ΣFx = 0: By symmetry,

Ax = 0 A y = K y = 18 kN

FBD Section ABC:

FBD Section ABC: ΣM D = 0:

(1.5 m ) FCE + (1.5 m )( 6 kN ) − ( 3 m )(18 kN − 3 kN ) = 0 FCE = 22.5 kN T !

ΣM A = 0:

(1.8 m )

4 FCD − (1.5 m )( 6 kN ) = 0 5 FCD = 6.25 kN T !

ΣFy = 0:

18 kN − 3 kN − 6 kN −

8 4 FBD + ( 6.25 kN ) = 0 17 5 FBD = 29.8 kN C !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 51.

FBD Truss:

Distance between loads = 1.5 m ΣFx = 0: By symmetry,

Ax = 0 A y = K y = 18 kN

FBD Section ABC: FBD Section GHK: FBD section GHK: ΣM F = 0:

( 4.5 m )(18 kN − 3 kN ) − ( 3 m )( 6 kN ) − (1.5 m )( 6 kN ) − ( 2.4 m ) FEG = 0 FEG = 16.875 kN,

ΣM K = 0:

 8  FFG  = 0  73 

(1.5 m )( 6 kN ) + ( 3 m )( 6 kN ) − ( 3.6 m ) 

FFG = 8.0100 kN, ΣFx = 0:

FEG = 16.88 kN T !

15 FFH − 17

FFG = 8.01 kN T !

3 (8.0100 kN ) − 16.875 kN = 0 73

FFH = 22.3 kN C !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 52.

FBD Truss:

ΣFX = 0:

A x + 4 kips − 4 kips = 0 Ax = 0

By symmetry,

Ay = Ny = 0

FBD Section ABDC:

ΣM D = 0:

(18 ft )( 4 kips ) − ( 9 ft )

4 FCE = 0 5

FCE = 10.00 kips C ! ΣM E = 0:

(18 ft )( 4 kips ) − (12 ft )

3 FDF = 0 5

FDF = 10.00 kips C

ΣFx = 0:

4 kips +

4 (10 kips − 10 kips ) − FDE = 0 5 FDE = 4.00 kips C !

FBD Joint E:

ΣFy = 0:

4 3 FEF − ( 4 kips ) = 0 5 5 FEF = 3.00 kips T !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 53.

FBD Truss:

ΣFx = 0:

A x + 4 kips − 4 kips = 0

Ax = 0 Ay = Ny = 0

By symmetry,

FBD Section IJNM:

ΣM G = 0:

4 − ( 27 ft )( 4 kips ) + ( 9 ft ) FHI = 0 5 FHI = 15.00 kips T !

ΣM N = 0:

4 5

( 9 ft ) (15 kips ) − ( 27 ft ) FGI

=0

FGI = 4.00 kips C !

FBD Joint I:

ΣFx = 0:

3 4 ( 4 kips ) − FIJ = 0 5 5 FIJ = 3.00 kips T !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 54.

FBD Truss:

ΣFx = 0:

Ax = 0 A y = M y = 7.5 kips

By symmetry,

FBD Section IJNM:

( 9 ft )( 7.5 kips − 0.5 kips ) − ( 4.5 ft )(1 kip )

ΣM J = 0:

 4  − ( 5.25 ft )  FGI  = 0,  17 

FGI = 11.4858 kips.

FGI = 11.49 kips T !

(12 ft )( 7.5 kip − 0.5 kips ) − ( 7.5 ft )(1 kip )

ΣM G = 0:

− ( 3 ft )( 3 kips ) + ( 4.5 ft ) FHJ = 0 FHJ = 15 kips C

By inspection of joint H, ΣFy = 0:

FFH = 15.00 kips C !

1 (11.4858 kips ) + 7.5 kips − 4.5 kips 17

−

3 FGJ = 0 13

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

FGJ = 6.95 kips T !

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Chapter 6, Solution 55.

FBD Truss:

ΣFx = 0: A x = 0 By symmetry:

A y = v y = 4.5 kips

FBD Joint K: ΣFx = 0:

ΣFy = 0:

1 ( FIK − FKN ) = 0 2  1  FIK  − 1 kip = 0 2  2 

so

FIK = FKN FIK =

2 kip 2

FIK = 0.707 kip C !

FBD Section ABEIJ:

ΣFy = 0:

ΣM I = 0:

4.5 kips − 1 kip − 2 (1.5 kips ) −

1 2 3 FJL = 0, kips + 2 2 2

( 3 ft )(1.5 kips ) + ( 6 ft )(1 kip − 4.5 kips ) +

3 ( 3 ft ) ( FJM ) = 0 2

FJM =

FJL = 0 !

11 kips 3

FJM = 6.35 kips T !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 56.

FBD Truss:

ΣFx = 0: ΣM N = 0:

Ax = 0

( 4.8 ft )( 300 lb ) + ( 7.2 ft )( 750 lb ) + (11.52 ft )( 750 lb ) + (16.8 ft )(1950 lb ) + (19.2 ft )( 900 lb ) + ( 26.4 ft )(1050 lb )

(

)

+ ( 33.6 ft ) 150 lb − Ay = 0

Ay = 2925 lb ΣFy = 0:

2925 lb − 2 (150 lb ) − 2 ( 750 lb ) − 1050 lb − 900 lb − 1950 lb − 300 lb + N y = 0

FBD Section ABC:

N y = 3075 lb ΣM B = 0:

( 3.6 ft ) FCE − ( 7.2 ft )( 2925 lb − 150 lb ) = 0 FCE = 5550 lb,

ΣM A = 0:

( 7.2 ft )

FCE = 5.55 kips T !

1 2 FBE + ( 3.6 ft ) FBE − ( 7.2 ft )(1050 lb ) = 0 5 5

FBE = 525 5 lb, ΣFx = 0:

(

FBE = 1.174 kips C !

)

2 FBD − 525 5 lb + 5550 lb = 0 5

FBD = −5031.2 lb

By inspection of joint D, FDF = FBD , so

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

FDF = 5.03 kips C !

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Chapter 6, Solution 57.

FBD Truss:

ΣFx = 0:

Ax = 0

( 4.8 ft )( 300 lb ) + ( 7.2 ft )( 750 lb ) + (11.52 ft )( 750 lb ) + (16.8 ft )(1950 lb ) + (19.2 ft )( 900 lb ) + ( 26.4 ft )(1050 lb ) + ( 33.6 ft ) (150 lb − Ay ) = 0

ΣM N = 0:

Ay = 2925 lb

FBD Section KJMN:

2925 lb − 2 (150 lb ) − 2 ( 750 lb ) − 1050 lb

ΣFy = 0:

− 900 lb − 1950 lb − 300 lb + N y = 0

N y = 3075 lb

( 7.2 ft )( 3075 lb − 150 lb ) − ( 2.4 ft )( 300 lb )

ΣM J = 0:

− (1.8 ft ) FIK = 0 FIK = 11300 lb, ΣM I = 0:

FIK = 11.30 kips T

(11.52 ft )( 3075 lb − 150 lb ) − ( 6.72 ft )( 300 lb )  2  − ( 4.32 ft )( 750 lb ) + ( 3.96 ft )  FHJ  = 0  5 

ΣFx = 0:

FBD Joint I:

FHJ = −3590.5 5 lb, FHJ = 8.03 kips C !

2 12 − −3590.9 5 lb − FIJ − 11300 lb = 0 13 5

(

)

FIJ = − 4461.4 lb,

ΣFx = 0:

11300 lb −

FIJ = 4.46 kip C !

12 ( 4461.4 lb ) − FGI = 0 13

FGI = 7181.8 lb,

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

FGI = 7.18 kips T !

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Chapter 6, Solution 58.

FBD Truss: Notes: α = 20°, β = 40°, γ = 60°, δ = 80°, φ = 22.5°, θ = 45°, ψ = 60° outer members AC, CE, etc. are each 1.0 m radial members AB, CD, etc. are each 0.4 m By symmetry , Ay = By = 2.6 kN ΣFx = 0: A x = 0 FBD Section ABDC:

ΣM F = 0:

( resolve FCE at E )

 1   1   2 ( 0.4 m )  FCE sin 40° +  2 ( 0.4 m )  ( FCE cos 40° )     1   + (1 m ) sin 40° + ( 0.4 m ) ( 0.4 kN ) 2   1   − (1 m ) sin 20° + (1 m ) sin 40° + ( 0.4 m ) ( 2.6 kN ) = 0, 2  

FCE = 7.3420 kN,

FCE = 7.34 kN C !

(1 m ) cos 40° − ( 0.4 m ) sin 45° + ( 0.4 m ) sin 22.5° (1 m ) sin 40° + ( 0.4 m ) cos 45° − ( 0.4 m ) cos 22.5° (1 m ) cos 40° − ( 0.4 m ) sin 45° = 27.566° = tan −1 (1 m ) sin 40° + ( 0.4 m ) cos 45°

note σ = tan −1 and ε Then,

ΣM C = 0:

= 48.848°

( 0.4 m )  FDF cos ( 90° − 22.5° − 48.848°) − (1 m ) sin 20° ( 2.6 kN ) = 0,

FDF = 2.3464 kN, and,

ΣFx = 0:

FDF = 2.35 kN T !

FCF ( cos 27.566° ) + ( 2.3464 kN )( cos 48.848° ) − ( 7.3420 kN ) sin 40° = 0 FCF = 3.5819 kN,

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

FCF = 3.58 kN T !

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Chapter 6, Solution 59.

FBD Truss: Notes: α = 20°, β = 40°, γ = 60°, δ = 80°, φ = 22.5°, θ = 45°, ψ = 60° outer members AC, CE, etc. are each 1.0 m radial members AB, CD, etc. are each 0.4 m By symmetry, Ay = By = 2.6 kN

ΣFx = 0: A x = 0 FBD Section ABDFHGEC:

( resolve FGI at I ) ( 0.4 m ) FGI sin 80° + (1 m )( sin 80°)(1 kN )

ΣM J = 0:

+ (1 m ) sin 80° + (1 m ) sin 60°  ( 0.6 kN )

+ (1 m ) sin 80° + (1 m ) sin 60° + (1 m ) sin 40° ( 0.4 kN ) − (1 m ) sin 80° + (1 m ) sin 60° + (1 m ) sin 40° + (1 m ) sin 20° ( 2.6 kN ) = 0

FGI = 10.8648 kN,

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

FGI = 10.86 kN C !

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Note: so,

σ = tan −1

ΣM H = 0:

0.4 m − (1 m ) cos80° = 12.9443° (1 m ) sin 80°

( 0.4 m ) cos 60° ( FGJ sin12.9443° ) − ( 0.4 m ) sin 60° ( FGJ cos12.9443° ) − ( 0.4 m ) sin 60° (10.8648 kN ) sin 80° + ( 0.4 m ) cos 60° (10.8648 kN ) cos80° + ( 0.4 m ) cos 60° (1 kN ) + (1 m ) sin 60° + ( 0.4 m ) cos 60° ( 0.6 kN ) + (1 m ) sin 60° + (1 m ) sin 40° + ( 0.4 m ) cos 60° ( 0.4 kN ) − (1 m )( sin 60° + sin 40° + sin 20° ) + ( 0.4 m ) cos 60° ( 2.6 kN ) = 0

FGJ = 0.93851 kN,

FGJ = 939 N T !

FBD Joint I: By symmetry: ΣFy = 0:

FIK = 10.8648 kN

− FIJ − 1.2 kN + 2 (10.8648 kN ) cos80° = 0 FIJ = 2.5733 kN,

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

FIJ = 2.57 kN T !

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Chapter 6, Solution 60.

FBD Truss:

ΣFx = 0: A x = 0 By symmetry, Ay = Ny = 1.6 kips

FBD Section using cut a–a: ΣM D = 0:

( 9 ft ) FFH + ( 6 ft )( 0.5 kip ) − (12 ft )(1.6 kips ) = 0 FFH = 1.800 kips T !

ΣFx = 0:

1.800 kips − FDG = 0 FDG = 1.800 kips C !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 61. \

FBD Truss:

ΣFx = 0: A x = 0 By symmetry Ay = Ny = 1.6 kips

FBD Joint J: ΣFx = 0:

− FGJ − FHJ = 0, FHJ = − FGJ

By inspection of joint I, FIJ = 0.6 kip C

FBD Section using cut b-b:

3  FGJ − ( − FGJ )  − 0.6 kip − 0.5 kip + 1.6 kips = 0 5

ΣFy = 0: FGJ = −

2.5 kip, 6

ΣM J = 0:

FGJ = 0.417 kip C !

(12 ft )(1.6 kips ) − ( 6 ft )( 0.5 kip ) − ( 4.5 ft )( FHK ) − ( 4.5 ft )( FIL ) = 0

ΣFx = 0:

FIL − FHK = 0 so FIL = FHK

and

FIL = 1.800 kips C !

FHK = 1.800 kips T !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 62.

FBD Section above a-a:

ΣM G = 0:

 27  FIK  − ( 2.7 m )( 40 kN ) − ( 5.4 m )( 40 kN ) = 0  793 

( 5.9 m ) 

FIK = 57.275 kN, ΣFy = 0:

FIK = 57.3 kN C W

27 ( 57.275 kN − FGJ ) = 0 793 FGJ = 57.275 kN T

FBD Section ACIG: 27 ( 57.275 kN − 57.275 kN ) 793

ΣFy = 0: + ΣFx = 0:

18 ( FHK − FHJ ) = 0, 949 3 ( 40 kN ) − 2

FHJ = FHK

25 FHK = 0 949

FHK = 53.884 kN

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

FHK = 53.9 kN C W

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Chapter 6, Solution 63.

FBD Section ACFBD:

ΣM D = 0:

( 4.3 m )

27 FFI − ( 2.7 m )( 40 kN ) = 0 793

FFI = 26.196 kN,

ΣFy = 0:

FFI = 26.2 kN C !

27 ( 26.196 kN − FDG ) = 0 793

FDG = 26.196 kN T FBD Section ACFD: ΣFy = 0:

27 ( 26.196 kN − 26.196 kN ) 793 +

ΣFx = 0:

54 ( FEI − FEG ) = 0, 6397 −2 −

FEI = FEG

59 ( FEG ) + 2 ( 40 kN ) 6397 8 ( 26.196 kN + 26.196 kN ) = 0 793

FEG = 44.136 kN,

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

FEG = 44.1 kN T !

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Chapter 6, Solution 64.

FBD Truss:

ΣFx = 0:

ΣM F = 0:

Ax = 0

( 2.4 m )(8 kN ) + ( 5.1 m )(12 kN ) − ( 7.5 m ) Ay = 0,

A y = 10.72 kN

Since only CD can provide an upward force necessary for equilibrium, it must be in tension, and FBE = 0 FBD Section ABC:

ΣFy = 0:

10.72 kN − 12 kN +

FCD = 2.4229 kN,

ΣM C = 0:

1.68 FCD = 0 3.18 FCD = 2.42 kN T

(1.68 m ) FBD − ( 2.4 m )(10.72 kN ) = 0

FBD = 15.3143 kN,

ΣFx = 0: FCE +

FBD = 15.31 kN C

2.7 ( 2.4229 kN ) − 15.3143 kN = 0 3.18 FCE = 13.26 kN T

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 65.

FBD Truss:

ΣFx = 0:

ΣM F = 0:

Ax = 0

( 2.4 m )(8 kN ) + ( 5.1 m )( 6 kN ) − ( 7.5 m ) Ay

=0

A y = 6.64 kN

FBD Section ABC:

Since only BE can provide the downward force necessary for equilibrium, it must be in tension, so FCD = 0 ΣFy = 0:

6.64 kN − 6 kN −

FBE = 1.21143 kN, ΣM = 0:

FBE = 1.211 kN T !

(1.68 m ) FCE − ( 2.4 m )( 6.64 kN ) = 0

FCE = 9.4857 kN,

ΣFx = 0:

1.68 FBE = 0 3.18

9.4857 kN +

FCE = 9.49 kN T !

2.7 (1.21143 kN ) − FBD = 0 3.18

FBD = 10.51 kN C !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 66.

FBD Truss:

ΣFx = 0: ΣM C = 0:

Cx = 0

(8 ft )( 6 kips ) − (8 ft )( 9 kips ) + (16 ft ) G y − ( 24 ft )(12 kips ) = 0

G y = 19.5 kips FBD Section ABC:

ΣFy = 0: 19.5 kips + C y − 6 kips − 9 kips − 12 kips = 0 C y = 7.5 kips

Since only BE can provide the downward force necessary for equilibrium, it must be in tension, so CD is slack, FCD = 0 ΣFy = 0: − 6 kips + 7.5 kips −

3 FBE = 0 5 FBE = 2.50 kips T !

FBD Section FGH:

Since only EF can provide the downward force necessary for equilibrium, it must be in tension, so DG is slack, FDG = 0 ΣFy = 0: 19.5 kips − 12 kips −

3 FEF = 0 5

FEF = 12.00 kips T !

Knowing that FCD = FDG = 0, inspection of joint D gives FDE = 0 !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 67 FBD Truss:

( 32 ft ) H y − ( 24 ft )(12 kips ) − (16 ft )( 9 kips )

ΣM A = 0:

− ( 8 ft )( 6 kips ) = 0, ΣFy = 0:

H y = 15 kips

Ay − 6 kips − 9 kips − 12 kips + 15 kips = 0 A y = 12 kips

ΣFx = 0: FBD Section ABC:

Ax = 0

Since only BE can provide the downward force necessary for equilibrium, it must be in tension, so CD is slack, FCD = 0 ΣFy = 0: 12 kips − 6 kips −

3 FBE = 0 5

FBE = 10.00 kips T !

Since only EF can provide the downward force necessary for equilibrium, it must be in tension, so DG is slack, FDG = 0

FBD Section FGH:

ΣFy = 0: 15 kips − 12 kips −

3 FEF = 0 5 FEF = 5.00 kips T !

Knowing that FCD = FDF = 0, inspection of joint D gives FDE = 0 !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 68 FBD Section ABDC: Since only DE can provide the leftward force necessary for equilibrium, it must be in tension, and CF must be slack, FCF = 0

ΣFx = 0: ΣM D = 0:

20 kN −

4 FDE = 0, 5

FDE = 25.0 kN C !

( 3.2 m ) FCE − ( 2.4 m )( 20 kN ) = 0 FCE = 15.00 kN T !

ΣFy = 0: FDF − 15.00 kN −

3 ( 25 kN ) = 0 5 FDE = 30.0 kN C !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 69.

FBD Section ABFE: It is not apparent which counter is active, so we guess that FEH = 0; ΣM F = 0:

 24  FEG + 40 kN  − ( 4.8 m )( 20 kN ) = 0  745 

( 3.2 m ) 

FEG = −11.3728 kN

FEG = 11.37 kN C !

ΣM G = 0:

 24  FFH  − ( 7.2 m )( 20 kN ) − (1.3 m )( 40 kN ) = 0  745 

( 5.8 m ) 

FFH = 38.432 kN, ΣFx = 0:

20 kN −

FFH = 38.4 kN C !

13 15 ( 38.432 kN − 11.3728 kN ) − FFG = 0 17 745

FFG = 8.0604 kN,

FFG = 8.06 kN T !

Since FEG is in tension, our guess was correct. A negative answer would be impossible, indicating an incorrect guess.

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 70.

Structure (a):

Non-simple truss with r = 3, m = 16, N = 10,

so 2N > m + r ∴ partially constrained !

Structure (b):

Non-simple truss with r = 3, m = 15, N = 10,

so 2N > m + r ∴ partially constrained !

Structure (c):

Non-simple truss with r = 4, m = 16, N = 10,

so N = 2m + r

Examine Forces: FHJ = 0, FIJ = P C

By inspection of joint J, Then, by inspection of joint H, By inspection of joint I,

FFH = 0,

FGI = 0,

FGH = P C Iy = 0

Although several members carry no load with the given truss loading, they do constrain the motion of GH and IJ. Truss ABFG is a simple truss with r = 3, m = 11, N = 7 (2N = r + m) so the structure is completely constrained, and determinate. !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 71.

Structure (a): Structure (b): Structure (c):

Simple truss with r = 4, m = 16, n = 10 So m + r = 20 = 2n so completely constrained and determinate !

Compound truss with r = 3, m = 16, n = 10 so So m + r = 19 < 2n = 20

partially constrained !

Non-simple truss with r = 4, m = 12, n = 8 So m + r = 16 = 2n but must examine further, note that reaction forces A x and H x are aligned, so no equilibrium equation will resolve them.

Also consider

∴ Statically indeterminate !

For ΣFy = 0: H y = 0, but then

ΣM A ≠ 0 in FBD Truss, ∴ Improperly constrained !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 72.

Structure (a):

Simple truss (start with ABC and add joints alphabetical to complete truss), with r = 4, m = 13, n = 8

r + m = 17 > 2n = 16

so

Structure is completely constrained but indeterminate. !

Structure (b):

From FBD II:

FBD I:

FBD II:

ΣM G = 0 ⇒

Jy

ΣFx = 0 ⇒

Fx

ΣM A = 0 ⇒

Fy

ΣFy = 0 ⇒

Ay

ΣFx = 0 ⇒

Ax

ΣFy = 0 ⇒

Gy

Thus have two simple trusses with all reactions known, so structure is completely constrained and determinate. !

Structure (c):

Structure has r = 4, m = 13, n = 9 so

r + m = 17 < 2n = 18,

structure is partially constrained !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 73.

Non-Simple truss with r = 4, m = 12, N = 8, so 2N = r + m, but we must examine further:

Structure (a):

ΣFx = 0: E x = 0 By symmetry, FBD = FDF and FCD = FDG

FBD joint D:

Vertical equilibrium cannot be satisfied with FCD = FDG, Improperly constrained !

Structure (b):

Non-simple truss with r = 3, m = 10, N = 7,

so 2N = m + r ∴ Partially constrained !

Structure (c):

Non-simple truss with r = 3, m = 13, N = 8, but we must examine further

so 2N = r + m,

ΣM A = 0: 3a G y − ( a + 3a + 4a ) P = 0 Gy =

Joint H:

ΣFy = 0:

Joint F:

1 FFH − P = 0 2

FFH = P 2 T

ΣFy = 0: FFG − P −

8 P 3

Joint G:

(

)

1 P 2 =0 2

FFG = 2P C

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

ΣFy ≠ 0

Improperly constrained !

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Chapter 6, Solution 74.

Structure (a): No. of members

m = 12

No. of joints

n=8

m + r = 16 = 2n

No. of react. comps.

r =4

unks = eqns

FBD of EH:

ΣM H = 0 → FDE ; ΣFx = 0 → FGH ; ΣFy = 0 → H y

Then ABCDGF is a simple truss and all forces can be determined.

This example is completely constrained and determinate. !

Structure (b):

No. of members

m = 12

No. of joints

n=8

m + r = 15 < 2n = 16

No. of react. comps.

r =3

unks < eqns partially constrained !

Note: Quadrilateral DEHG can collapse with joint D moving downward: in (a) the roller at F prevents this action.

Structure (c):

No. of members

m = 13

No. of joints

n=8

m + r = 17 > 2n = 16

No. of react. comps.

r =4

unks > eqns completely constrained but indeterminate !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 75.

Structure (a):

Rigid truss with r = 3, m = 14, n = 8

so

r + m = 17 > 2n = 16 so completely constrained but indeterminate !

Structure (b):

Simple truss (start with ABC and add joints alphabetically), with r = 3, m = 13, n = 8

so

r + m = 16 = 2n

so completely constrained and determinate !

Structure (c):

Simple truss with r = 3, m = 13, n = 8 so r + m = 16 = 2n, but horizontal reactions ( Ax and Dx ) are collinear so cannot be resolved by any equilibrium equation.

â&#x2C6;´ structure is improperly constrained !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell Š 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 76. \

FBD member AE:

Ax = 120 lb

ΣFx = 0: − Ax + 120 lb = 0, on ABCD ΣM A = 0:

A x = 120.0 lb

!

(8 in.)( C ) − ( 2 in.)(120 lb ) = 0 C = 30 lb

ΣFy = 0: 30 lb − Ay = 0,

Ay = 30 lb

A y = 30.0 lb !

on ABCD

FBD member ABCD:

ΣFx = 0: 120 lb − Bx = 0, ΣM B = 0:

B x = 120.0 lb

!

( 6 in.) D − ( 4 in.)( 30 lb ) − ( 4 in.)( 30 lb ) − ( 2 in.)(120 lb ) = 0

D = 80.0 lb ! ΣFy = 0: 30 lb − By − 30 lb + 80 lb = 0

B y = 80.0 lb !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 77.

FBD member ABC:

Note that BD is a two-force member

(a)

ΣΜ C = 0:

(16 in.) 

4  FBD  − ( 24 in.)( 80 lb ) = 0 5 

FBD = 150 lb,

(b)

ΣFy = 0: ΣFx = 0:

Cy − Cx −

FBD = 150.0 lb

3 (150 lb ) = 0, 5

C y = 90 lb

4 (150 lb ) + 80 lb = 0, 5

Cx = 40 lb

C = 98.5 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

36.9° W

66.0° W

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Chapter 6, Solution 78.

FBD Frame:

ΣM A = 0:

( 0.25 m ) Dx − ( 0.95 m )( 480 N ) = 0 D x = 1824 N

W

E x = 1824 N

W

FBD member DF:

Note that BE is a two-force member, Ex = Ey ΣFx = 0:

−1824 N + Ex = 0,

so

ΣM D = 0:

ΣFy = 0:

E y = 1824 N

W

( 0.50 m )(1824 N ) − ( 0.75 m ) C + ( 0.95 m )( 480 N ) = 0 C = 1824 N

W

D y = 480 N

W

− D y + 1824 N − 1824 N + 480 N = 0

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 79.

FBD Frame:

ΣM A = 0:

( 0.25 m ) Dx − 400 N ⋅ m = 0 D x = 1600 N

E x = 1600 N

∴ E y = 1600 N

FBD member DF: Note BE is a two-force member, so Ex = Ey ΣFx = 0: 1600 N − E x = 0,

ΣM D = 0: − ( 0.50 m )(1600 N ) + ( 0.75 m ) C − 400 N ⋅ m = 0

C = 1600 N

ΣFy = 0: Dy − 1600 N + 1600 N = 0

Dy = 0

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 80.

FBD Ring: (a)

ΣM A = 0:

(8 in.) ( FBC cos 35° ) − (8 in.)( 6 lb ) = 0

FBC = 7.3246 lb,

(b)

ΣFx = 0:

FBC = 7.32 lb C W

Ax − ( 7.3246 lb ) cos 35° = 0 Ax = 6 lb

ΣFy = 0:

Ay + ( 7.3246 lb ) sin 35° − 6 lb = 0 Ay = 1.79876 lb

A = 6.26 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

16.69° W

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Chapter 6, Solution 81.

FBD Ring:

(a)

FBC =

(b)

(8 in.) ( FBC cos 20° − 6 lb ) = 0

ΣM A = 0:

6 lb , cos 20°

ΣFx = 0:

ΣFy = 0:

FBC = 6.39 lb C W

Ax −

Ay −

6 lb cos 20° = 0, cos 20°

Ax = 6 lb

6 sin 20° − 6 lb = 0, cos 20°

Ay = 8.1838 lb

A = 10.15 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

53.8° W

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Chapter 6, Solution 82.

FBD Frame:

ΣM A = 0:

 2  180 N  = 0  2 

( 0.2 m ) Bx − ( 0.36 m ) 

Bx = 162 2 N,

Ax = 252 2 N,

ΣM C = 0:

A x = 356 N

W

2 (180 N ) = 0 2

B y = 127.3 N

B y = 90 2 N,

FBD member ABD:

W

2 (180 N ) = 0 2

ΣFx = 0: −162 2 N + Ax −

ΣFy = 0: By −

B x = 229 N

( 0.20 m ) ( 252

)

(

2 N − ( 0.08 m ) 90 2 N

− ( 0.20 m ) D = 0,

W

)

D = 216 2 N

D = 305 N

W

ΣFx = 0: 252 2 N − 162 2 N − C x = 0 Cx = 90 2 N,

C x = 127.3 N

W

ΣFy = 0: 90 2 N + C y − 216 2 N = 0 C y = 126 2 N,

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

C y = 178.2 N

W

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Chapter 6, Solution 83.

FBD Frame:

ΣM B = 0:

( 0.2 m ) Ax − 60 N ⋅ m = 0

ΣFx = 0: 300 N − Bx = 0, ΣFy = 0: By = 0,

A x = 300 N

W

B x = 300 N

W

By = 0 W

FBD member ABD: ΣM C = 0:

ΣFy = 0:

( 0.20 m )( 300 N ) − ( 0.20 m ) Dy C y − 300 N = 0

ΣFx = 0:

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

=0

D y = 300 N

W

C y = 300 N

W

Cx = 0 W

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Chapter 6, Solution 84. (a) FBD AC:

Note: CE is a two-force member ΣM A = 0:

 FCE   FCE   + ( 2 in.)   − ( 6 in.)( 24 lb ) = 0  2   2 

(8 in.) 

FCE = 14.4 2 lb,

so

E x = 14.40 lb E y = 14.40 lb

ΣFx = 0:

− Ax + 14.4 lb = 0

A x = 14.40 lb

ΣFy = 0:

Ay − 24 lb + 14.40 lb = 0 A y = 9.60 lb

(b) FBD CE:

! ! !

!

Note AC is a two-force member ΣM E = 0:

1  4  FAC + FAC  − (1 in.)( 24 lb ) = 0 17  17 

( 3 in.) 

FAC = 1.6 17 lb,

A x = 6.40 lb A y = 1.600 lb

(

)

(

)

ΣFx = 0: E x −

4 1.6 17 lb = 0, 17

ΣFy = 0:

1 1.6 17 lb − 24 lb = 0 17

Ey +

E x = 6.40 lb

E y = 22.4 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

! ! !

!

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Chapter 6, Solution 85. (a) FBD AC:

Note: CE is a two-force member ΣM A = 0:

1  1  FCE + FCE  − ( 0.3 m )( 320 lb ) 2  2 

( 0.4 m ) 

E x = 120.0 N

FCE = 120 2 N,

E y = 120.0 N

(b) FBD CE:

(

)

1 120 2 N = 0, 2

ΣFx = 0:

− Ax +

ΣFy = 0:

Ay − 320 N +

(

A x = 120.0 N

)

1 120 2 N = 0 2

A y = 200 N

!

! ! !

Note: AC is a two-force member ΣM E = 0:

1  1  FAC + FAC  − ( 0.15 m )( 320 N ) = 0 2  2 

( 0.25 m ) 

FAC = 96 2 N,

A x = 96.0 N A y = 96.0 N

ΣFx = 0: Ex − 96.0 N = 0 ΣFy = 0:

E y − 320 N + 96.0 N = 0

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

E x = 96.0 N E y = 224 N

! ! ! !

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Chapter 6, Solution 86. (a) FBD AC:

Note: CE is a two-force member ΣM A = 0:

 1   1  − ( 8 in.)  FCE  − ( 2 in.)  FCE  + 192 lb ⋅ in. = 0  2   2 

E x = 19.20 lb

FCE = 19.2 2 lb,

E y = 19.20 lb

(b) FBD CE:

ΣFx = 0:

Ax − 19.2 lb = 0,

ΣFy = 0:

Ay − 19.2 lb = 0,

A x = 19.20 lb A y = 19.20 lb

!

! ! !

Note: AC is a two-force member ΣM E = 0:

4  1  FAE + FAE  + 192 lb ⋅ in. = 0 17  17 

( 3 in.) 

FAE = −12.8 17 lb,

Ax =

4 FAE , 17

Ay =

1 FAE , 17

ΣFx = 0: Ex − 51.2 lb = 0, ΣFy = 0:

E y − 12.80 lb = 0,

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

A x = 51.2 lb A y = 12.80 lb E x = 51.2 lb E y = 12.80 lb

! ! ! !

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Chapter 6, Solution 87. (a) FBD AC:

Note: CE is a two-force member ΣM A = 0:

 2  FCE  = 0, 120 N ⋅ m − ( 0.4 m )  2  

FCE = 150 2 N

E x = 150.0 N E y = 150.0 N

(b) FBD CE:

ΣFx = 0:

Ax − 150.0 N = 0,

ΣFy = 0:

Ay − 150.0 N = 0,

A x = 150.0 N A y = 150.0 N

!

! ! !

Note: AC is a two-force member  2  ΣM E = 0: − ( 0.25 m )  FAC  + 120 N ⋅ m = 0  2 

FAC = 240 2 N,

A x = 240 N A y = 240 N

ΣFx = 0: 240 N − Ex = 0, ΣFy = 0:

E y − 240 N = 0,

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

E x = 240 N E y = 240 N

! ! ! !

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Chapter 6, Solution 88.

FBD Frame:

Note: In analysis of entire frame, location of M is immaterial. Note also that AB, BC, and FG are two force members.

ΣM H = 0:

( 6 ft ) I y

− 180 lb ⋅ ft = 0,

ΣFy = 0:

30 lb −

5 FAB = 0, 13

ΣM C = 0:

( 2.5 ft ) 

I y = 30.0 lb

!

(a) FBD AI:

12  78 lb − FFG  = 0,  13 

ΣFx = 0: FBC −

12 78 lb − 72 lb = 0, 13

FAB = 78.0 lb

22.6° !

FGH = 72.0 lb

!

FBC = 144.0 lb

!

(b) FBD AI:

FAB = 78.0 lb As above, ΣFy = 0 yields Then:  12  ΣM C = 0: ( 2.5 ft )  78.0 lb − FFG  − 180 lb ⋅ ft = 0  13 

22.6° !

FFG = 0 ! ΣFx = 0:

FBC −

12 ( 78.0 lb ) = 0, 13

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

FBC = 72.0 lb

!

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Chapter 6, Solution 89. (a) FBD ACF:

Note: BC is a two-force member ΣM F = 0:

 1  FBC  = 0,  26 

( 0.1 m )(120 N ) − ( 0.3 m ) 

B x = 200 N

FBC = 40 26 N,

B y = 40.0 N ΣFx = 0:

(b) FBD BCE:

ΣFy = 0:

(

)

5 40 26 N − Fx = 0, 26

(

Fx = 200 N

)

1 40 26 N = 0, 26

Fy − 120 N −

Fy = 160.0 N

!

! ! !

Note ACF is a two-force member ΣM B = 0:

 1  FCF  = 0,  5 

( 0.4 m )(120 N ) − ( 0.3 m ) 

FCF = 160 5 N,

Fx = 320 N Fy = 160.0 N

ΣFx = 0: Bx − ΣFy = 0:

(

(

)

2 160 5 N = 0, 5

B x = 320 N

! ! !

)

1 160 5 N − 120 N − B y = 0 5

B y = 40.0 N

!

(c) Moving the 120 N force from D to E does not affect the reactions. The answers are the same as in part (b). !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 90. (a) FBD AF: Note: BC is a two-force member ΣM F = 0:

 1  FBC  − 48 N ⋅ m = 0,  26 

( 0.3 m ) 

FBC = 160 26 N so

B x = 800 N B y = 160.0 N

(b & c) FBD BCE:

ΣFx = 0:

Fx − 800 N = 0,

ΣFy = 0:

− Fy + 160 N = 0,

Fx = 800 N Fy = 160.0 N

! ! ! !

Note: ACF is a two-force member, and that the application point of the couple is immaterial. ΣM B = 0:

 1  FCF  − 48 N ⋅ m = 0  5 

( 0.3 m ) 

FCF = 160 5 N,

Fx = 320 N Fy = 160.0 N

ΣFx = 0: 320 N − Bx = 0, ΣFy = 0:

−160 N + By = 0,

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

B x = 320 N B y = 160.0 N

! ! ! !

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Chapter 6, Solution 91. \

First note that, when a cable or cord passes over a frictionless, motionless pulley, the tension is unchanged.

ΣM C = 0:

rT1 − rT2 = 0

(a) Replace each force with an equivalent force-couple.

(b) Cut cable and replace forces on pulley with equivalent pair of forces at A as above.

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

T1 = T2

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Chapter 6, Solution 92.

FBD Pipe 2:

ΣFx′ = 0:

NG −

3 ( 220 lb ) = 0, 5

NG = 132 lb

ΣFy′ = 0:

ND −

4 ( 220 lb ) = 0, 5

N D = 176 lb

FBD Pipe 1:

θ = 90° − 2 tan −1 ΣFx′ = 0:

3 = 16.2602° 4

N F cos (16.2602° ) − 132 lb −

3 ( 220 lb ) = 0 5

N F = 275.00 lb

ΣFy′ = 0:

N C + ( 275 lb ) cos (16.2602° ) −

4 ( 220 lb ) = 0 5

NC = 99.00 lb FBD Frame & Pipes:

ΣM A = 0:





4 5

( 48 in.) E y − 24 in. + 24 in. + (10 in.)  (220 lb) = 0 



E y = 256

ΣFy = 0:

ΣFx = 0:

2 lb, 3

Ay − 2 ( 220 lb ) + 256 Ay = 183



1 lb 3

Ax − E x = 0,

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

E y = 257 lb

A y = 183.3 lb

2 lb = 0 3

Ax = E x

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FBD BE: ΣM B = 0:

( 24 in.)  256 

2   15  lb  − (18 in.) E x +  in.  ( 275 lb ) = 0 3   4 

E x = 399.5 lb,

E x = 400 lb

From above,

A x = 400 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 93.

FBD Pipe 2:

FBD Pipe 1:

ΣFx' = 0:

NG −

3 ( 220 lb ) = 0, 5

NG = 132 lb

ΣFy′ = 0:

ND −

4 ( 220 lb ) = 0, 5

N D = 176 lb

ΣFx′ = 0: ΣFy′ = 0:

ΣM B = 0:

FBD BE:

NF − NC −

3 ( 220 lb ) − 132 lb = 0, 5

4 ( 220 lb ) = 0, 5

N F = 264 lb NC = 176 lb

( 24 in.) E y − ( 32 in.) Ex + ( 5 in.) ( 264 lb ) = 0 4Ex − 3E y = 165

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

(1)

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FBD Frame with Pipes:

ΣM A = 0: ( 48 in.) E y − (14 in.) Ex + ( 5 in.) ( 264 lb ) − ( 35 in. + 45 in.) (176 lb ) = 0

(2)

solving with (1) above: E y = 355

ΣFx = 0:

2 lb, 3

Ax − 308 lb −

E y = 356 lb

4 3 ( 264 lb ) + ⋅ 2 (176 lb ) = 0 5 5

A x = 308 lb ΣFy = 0:

Ay + 355

!

!

2 3 4 lb − ( 264 lb ) − ⋅ 2 (176 lb ) = 0 3 5 5 A y = 84.3 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

!

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Chapter 6, Solution 94.

FBD AC:

ΣM c = 0:

( 0.6 m ) Ax − ( 0.06 m )(170 N ) = 0, A x = 17.00 N

ΣFx = 0:

FBD CE:

ΣFx = 0:

−17 N +

133 N −

15 (170 N ) − Cx = 0, 17

Cx = 133 N

15 (170 N ) + Ex = 0, 17

E x = 17.00 N ΣM C = 0:

W

W

( 0.48 m )(17.00 N ) + ( 0.90 m ) E y − ( 0.45 m ) (170 N ) = 0,

E y = 75.9 N W

FBD Frame:

ΣFy = 0:

Ay − 170 N + 75.9 N = 0,

A y = 94.1 N W

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 95.

FBD Frame & pulley:

ΣM B = 0:

( 0.6 m ) Ay − ( 0.075 m )( 240 N ) = 0, A y = 30.0 N W

ΣFy = 0:

− 30.0 N + By − 240 N = 0,

B y = 270 N W ΣFx = 0:

FBD AD:

ΣM D = 0:

From above

− Ax + Bx = 0,

Ax = Bx

( 0.200 m ) Ax − ( 0.075 m ) ( 240 N ) + ( 0.30 m ) ( 30 N ) = 0 Ax = Bx ,

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

A x = 45.0 N

W

B x = 45.0 N

W

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Chapter 6, Solution 96. (a)

FBD Entire machine: ΣM A = 0:

( 95 in.) (16 kips) − ( 35 in.)(18 kips ) + (145 in.) ( 2B ) − (170 in.) ( 50 kips ) = 0, B = 26.241 kips

ΣFy = 0:

B = 26.2 kips

W

−16 kips + 2 A − 18 kips + 2 ( 26.241 kips ) − 50 kips = 0,

A = 15.76 kips

W

(b)

FBD Motor unit: ΣM D = 0:

( 30 in.) Cx + (85 in.)( 52.482 kips ) − (110 in.)( 50 kips ) = 0 Cx = 34.634 kips,

C x = 34.6 kips

W

ΣFx = 0:

Dx − 34.634 kips = 0,

D x = 34.6 kips

W

ΣFy = 0:

52.482 kips − 50 kips − Dy = 0,

D y = 2.48 kips

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

W

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Chapter 6, Solution 97. (a) FBD Entire Machine:

(145 in.) (2B) − ( 35 in.)(18 kips )

ΣM A = 0:

− (170 in.)( 50 kips ) = 0 2B = 62.966 kips, ΣFy = 0:

B = 31.5 kips

!

2 A + 62.966 kips − 18 kips − 50 kips = 0 2 A = 5.034 kips,

A = 2.52 kips

!

where A and B are single wheel forces (b)

ΣM D = 0:

FBD Motor unit:

(85 in.)( 62.966 kips ) − (110 in.)( 50 kips ) + (30 in.) Cx = 0,

Cx = 4.9297 kips C x = 4.93 kips

ΣFy = 0:

− Dy − 50 kips + 62.966 kips = 0 D y = 12.97 kips

ΣFx = 0:

!

!

Dx − C x = 0, Dx = Cx , D x = 4.93 kips

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

!

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Chapter 6, Solution 98.

FBD Frame:

( 0.625 m ) F − ( 0.75 m )( 4 kN ) − (1.25 m )( 3 kN ) = 0

ΣM A = 0:

Fx = 10.8 kN ΣFx = 0:

Ax − 10.8 kN = 0,

ΣFy = 0:

Ay − 4 kN − 3 kN = 0,

FBD BF: (I)

A x = 10.80 kN

W

A y = 7.00 kN W

FBD ABD: (II)

I:

ΣM C = 0:

( 0.375 m )(10.8 kN ) − ( 0.25 m ) Bx = 0,

II:

ΣM D = 0:

( 0.25 m )(10.8 kN + 16.2 kN ) + ( 0.5 m ) By − (1.00 m )( 7.0 kN ) = 0,

Bx = 16.2 kN,

By = 0.5 kN, ΣFx = 0:

−10.8 kN − 16.20 kN + Dx = 0,

Dx = 27 kN,

ΣFy = 0:

7.0 kN − 0.5 kN − Dy = 0,

Dy = 6.5 kN,

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

B x = 16.20 kN

B y = 500 N D x = 27.0 kN

W

W W

D y = 6.50 kN W

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Chapter 6, Solution 99.

FBD Frame:

( 0.3 m ) Dx = − ( 0.4 m ) ( 900 N ) = 0,

ΣM A = 0:

Dx = 1200 N D x = 1.200 kN

ΣFx = 0: 1.200 kN − Ax = 0, ΣFy = 0:

Ax = 1.200 kN,

A x = 1.200 kN

Ay − 900 N = 0

A y = 900 N

Note: AG is a two-force member.

( 0.2 m )  By − ( 0.4 m ) ( 900 N ) = 0,

ΣM A = 0: FBD AC:

ΣFy = 0:

900 N + 1800 N − 900 N −

( FAGx = 2400 N,

FAG = 3000 N ΣFx = 0:

−

ΣM E = 0: ΣFx = 0: FBD DF:

ΣFy = 0:

3 FAG = 0 5

FAGy = 1800 N

4 ( 3000 N ) − 1200 N + Bx = 0, 5

On GBEH:

By = 1800 N

G x = 2.40 kN

,

B x = 3.60 kN

,

) Bx = 3600 N

G y = 1.800 kN !

B y = 1.800 kN

3 FDH = 0, 5 1200 N − Ex = 0

− ( 0.2 m )

!

FDH = 0

Ex = 1200 N

Ey = 0 On GBEH:

E x = 1.200 kN H x = 0,

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

,

Ey = 0 !

Hy = 0 !

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Chapter 6, Solution 100.

FBD Frame:

( 6 in.) Dx − (13.5 in.) ( 48 lb ) − (16.5 in.) ( 20 lb ) = 0

ΣM A = 0:

Dx = 163 lb ΣFx = 0:

− A x + 20 lb + 48 lb + 163 lb = 0 Ax = 231 lb

ΣFy = 0:

Ay = 0

FBD DE: ΣM E = 0:

(19.5 in.)(163 lb ) − ( 6 in.) Bx = 0 Bx = 529.75 lb

ΣM C = 0:

( 4.5 in.) By − ( 7.5 in.)( 231 lb ) = 0, B x = 530 lb

On ABC: ΣFx = 0:

FBD ABC:

ΣFy = 0: On ABC:

By = 385 lb ,

− 231 lb + 529.75 lb − Cx = 0, C y − 385 lb = 0,

B y = 385 lb W Cx = 298.75 lb

C y = 385 lb

C x = 299 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

,

C y = 385 lb W

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Chapter 6, Solution 101.

FBD Frame:

ΣM E = 0:

( 6 in.)(84 lb ) − ( 42 in.) Ay = 0 Ay = 12.00 lb

FBD CF: (I)

I: II:

II:

ΣM D = 0: ΣM B = 0:

ΣFx = 0: ΣFy = 0:

FBD ABC: (II)

( 3.5 in.) Cx + ( 7 in.) C y − ( 5 in.) (84 lb ) = 0,

Cx + 2C y = 120 lb

(12 in.) C y − ( 3.5 in.) Cx − (12 in.) (12 lb ) = 0,

− Bx + 60.632 lb = 0, 12 lb + By + 29.684 lb = 0,

3.5 Cx − 12 C y = 144 lb

Solving:

Cx = 60.632 lb,

C y = 29.684 lb

On ABC:

C x = 60.6 lb,

C y = 29.7 lb W

Bx = 60.632 lb, By = − 41.684 lb,

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

B x = 60.6 lb

W

B y = 41.7 lb W

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Chapter 6, Solution 102.

FBD Stool:  w 1 N = ( 56 kg )  9.81  = 274.68 N 2 2 kg   ΣM A = 0:

( 0.350 m ) By − ( 0.150 m )( 274.68 N ) = 0, By = 117.72 N

FBD BE:

ΣM E = 0:

 1  FFG  = 0  2 

(1.25 m )(117.72 N ) − ( 0.15 m ) 

FFG = 98.1 2 N ΣFx = 0: ΣFy = 0:

Ex −

1 (98.1 2) N = 0, 2

− Ey +

E x = 98.1 N

W

1 (98.1 2 N) + 117.72 N = 0 2

E y = 215.82 N,

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

E y = 216 N

W

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Chapter 6, Solution 103.

FBD BC:

Note that only one of the forces B y , B′y exists.

(

)

M C B y or B′y − ( 0.15 m )

ΣM C = 0:

 2  FFG =  M B or B′y  0.15 m  C y  

(

(

So we seek a maximum of M C By or B′y

1 FFG = 0 2

)

(1)

)

( ) M C ( B′y ) = ( 0.175 m ) B′y

M C By = ( 0.125 m ) By

where FBD stool:

w = ( 56 kg ) ( 9.81 N/kg ) = 549.36 N consider contact at A′, A y = 0, either

ΣM A′ = ( 0.425 m ) By − ( 0.225 m )

w =0 2

( )

M C By = ( 0.03309 m ) w

By = 0.26471 w,

or ( M A′ ) = ( 0.475 m ) B′y − ( 0.225 m )

(i)

w =0 2

( )

B′y = ( 0.23684 ) w, M C B′y = ( 0.04145 m ) w

(ii)

consider contact at A, A′y = 0, either ΣM A = 0:

( 0.35 m ) By − ( 0.15 m )

w =0 2

( )

By = 0.21429 w, M C By = ( 0.02678 m ) w or

ΣM A = 0: B′y = 0.1875 w,

( 0.40 m )

B′y − ( 0.15 m )

(iii) w =0 2

( )

M C B′y = ( 0.03281 m ) w

(iv)

(a) The maximum is (ii), with contact at A′ and B′, and

!

 2  (b) FFG max =   ( 0.04145 m ) ( 549.36 N ) = 214.69 N,  0.15 m 

FFG max = 215 N T !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 104.

Members FBDs:

I:

ΣM A = 0:

(12.8 ft ) Bx − ( 32 ft ) By − ( 20 ft )(14 kips ) = 0

II:

ΣM C = 0:

( 7.2 ft ) Bx + ( 24 ft ) By − (12 ft )( 21 kips ) = 0 Solving: Bx = 27.5 kips, By = 2.25 kips,

I:

ΣFx = 0:

A x = 27.5 kips ΣFy = 0:

Ax − 27.5 kips = 0,

Ax = 27.5 kips,

(a)

!

Ay − 14 kips − 2.25 kips = 0, Ay = 16.25 kips,

A y = 16.25 kips (b)

B x = 27.5 kips B y = 2.25 kips

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

! ! !

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Chapter 6, Solution 105.

Member FBDs:

II:

ΣM C = 0:

( 7.2 ft ) Bx − ( 24 ft ) By − (12 ft )(14 kips ) = 0

I:

ΣM A = 0:

(12.8 ft ) Bx + ( 32 ft ) By − ( 20 ft )( 21 kips ) = 0 Solving: Bx = 28.75 kips, By = 1.675 kips,

I:

ΣFx = 0:

A x = 28.8 kips ΣFy = 0:

Ax − 28.75 kips = 0,

Ax = 28.75 kips,

(a)

!

Ay − 21 kips + 1.625 kips, Ay = 19.375 kips ,

A y = 19.38 kips (b)

B x = 28.8 kips B y = 1.625 kips

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

! ! !

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Chapter 6, Solution 106. \

FBD Frame:

ΣM A = 0:

(12 in.) Dx − ( 28 in.)( 411 lb ) = 0 D x = 959 lb

FBD DF:

Note that BF and CE are two-force members. ΣFx = 0: ΣM D = 0:

Solving:

959 lb +

4 15 FCE − FBF = 0 5 17

(12 in.) 

3   8  FCE  + ( 34.5 in.)  FBF  = 0 5   17 

FBF = 357 lb,

ΣM E = 01:

( 22.5 in.)  ( 357 lb ) − (12 in.) Dy = 0

D y = 315 lb ,

8 17

so



(a) (b)

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

D = 1009 lb

18.18°

FBF = 357 lb T

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Chapter 6, Solution 107.

FBD Frame:

(12 in.) Dx − ( 34.5 in.)( 274 lb ) = 0

ΣM A = 0:

Dx = 787.75 lb

FBD DF:

ΣM D = 0: ΣFx = 0:

Solving: ΣM E = 0:

(12 in.) 

3   8  FCE  + ( 34.5 in.)  FBF − 274 lb  = 0 5 17    

787.75 lb +

4 15 FCE − FBF = 0 5 17

FBF = 684.25 lb,

( 22.5 in.)  ( 684.25 lb ) − 274 lb  − (12 in.) Dy

Dy = 90 lb,

8 17



so

(a) (b)

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

D = 793 lb

=0

6.52°

FBF = 684 lb T

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Chapter 6, Solution 108.

FBD AC:

Note that BE is a two force member ΣFy = 0:

1 FBE − P = 0 5

FBE = 5 P ΣΜ Α = 0:

 1  5

( 0.4 m ) 

(

)

  2 5 P  + ( 0.1 m )    5

(

)

 5P  

− a P + ( 0.3 m ) C = 0

so

( a − 0.6 m ) P = ( 0.3 m ) C

since P > 0 and C ≥ 0, a – 0.6 m ≥ 0 a ≥ 0.6 m

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 109.

Member FBDs:

ΣΜ Α = 0:

( 0.6 m )

4 4  FCF − ( 0.9 m )  FDG  = 0 5 5  

4 4 ΣΜ Ε = 0: − ( 0.3 m ) FCF + ( 0.6 m ) FDG − ( 0.9 m )( 800 N ) = 0 5 5 Solving:

FCF = 9000 N,

FDG = 6000 N FCF = 9.00 kN C FDG = 6.00 kN T

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 110.

Member FBDs:

4 FDG = 0 5

ΣM A = 0:

( 0.3 m ) FBF − ( 0.9 m )

ΣM E = 0:

4 − ( 0.3 m ) FBF + ( 0.6 m ) FDG − ( 0.9 m )( 0.8 kN ) = 0 5

Solving:

FBE = − 3 kN,

FBF = − 7.2 kN so

FBF = 7.20 kN T FDG = 3.00 kN C

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 111.

Member FBDs:

ΣM A = 0:

ΣM E = 0: Solving:

( 0.6 m )

4 4 FCH − ( 0.3 m ) FBG = 0 5 5

4 4 − ( 0.9 m ) FCH + ( 0.6 m ) FBG − ( 0.9 m )( 0.8 kN ) = 0 5 5 FBG = 6 kN, FCH = 3 kN so

FBG = 6.00 kN T FCH = 3.00 kN C

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 112.

Member FBDs:

I:

ΣM J = 0:

2a C y + a Cx − a P = 0 Solving:

I:

II:

Cx =

II:

ΣM K = 0: 2a C y − a C x = 0

P P , Cy = 2 4

ΣFx = 0:

1 FBG − Cx = 0, 2

ΣFy = 0:

FAF +

ΣFx = 0:

Cx −

ΣFy = 0:

−

1  2  P P  − P + = 0,  4 2  2 

1 FDG = 0, 2

2 P, 2

FBG = 2C x =

FAF =

FDG = 2C x =

P 1  2  P  + FEH = 0, +  4 2  2 

FBG = 0.707 P T P , 4

FAF = 0.250 P T

2 P, 2

FDG = 0.707 P T

P 4

FEH = 0.250 P C

FEH = −

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 113.

Note that, if we assume P is applied to BF, each individual member FBD looks like:

Fshort = 2Flong =

so

2 Fload 3

(by moment equations about S and L).

Labeling each interaction force with the letter corresponding to the joint of application, we have: F =

2(P + E) 3

= 2B

C D = 3 2 B A C = = 3 2 E =

so

2(P + E) = 2 B = 6C = 18E 3

P + E = 27 E

P 26

E=

so D = 2E =

A = 2C = 3E F =

2 P P +  3 26 

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

A= F=

P 13 3P 13

9P 13

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Chapter 6, Solution 114.

Note that, if we assume P is applied to EG, each individual member FBD looks like

so

2Fleft = 2 Fright = Fmiddle

Labeling each interaction force with the letter corresponding to the joint of its application, we see that B = 2 A = 2F C = 2B = 2D G = 2C = 2H P + F = 2G ( = 4C = 8B = 16 F ) = 2E

From

P + F = 16 F , F =

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell Š 2007 The McGraw-Hill Companies.

P 15

so A =

P 15

D=

2P 15

H=

4P 15

E=

8P 15

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Chapter 6, Solution 115. (a) Member FBDs:

FBD II:

ΣFy = 0:

FBD I:

ΣM A = 0:

By = 0

ΣM B = 0: aF2 = 0

aF2 − 2aP = 0

F2 = 0

but F2 = 0 so P = 0

not rigid for P ≠ 0 !

(b) Member FBDs:

(

)

Note: 7 unknowns Ax , Ay , Bx , By , F1, F2 , C but only 6 independent equations.

System is statically indeterminate !

System is, however, rigid ! (c) FBD whole:

FBD right:

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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FBD I:

ΣM A = 0: 5aBy − 2aP = 0 ΣFy = 0: Ay − P +

FBD II:

FBD I:

ΣM c = 0:

2 P=0 5

By = Ay =

a 5a Bx − By = 0 2 2

ΣFx = 0: Ax + Bx = 0

2 P 5

3 P 5

Bx = 5By Ax = − Bx

B x = 2P

A x = 2P

A = 2.09P

16.70° !

B = 2.04P

11.31° !

System is rigid !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 116.

Note: In all three cases, the right member has only three forces acting, two of which are parallel. Thus the third force, at B, must be parallel to the link forces. (a) FBD whole:

ΣM A = 0:

− 2aP −

a 4 1 B + 5a B=0 4 17 17

4 B=0 17

ΣFx = 0:

Ax −

ΣFy = 0:

Ay − P +

1 B=0 17

B = 2.06 P

B = 2.06P

14.04° !

A = 2.06P

14.04° !

A x = 2P Ay =

P 2

rigid ! (b) FBD whole:

Since B passes through A,

ΣM A = 2aP = 0 only if P = 0

∴ no equilibrium if P ≠ 0

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

not rigid !

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(c) FBD whole:

ΣM A = 0: 5a

1 3a 4 B+ B − 2aP = 0 4 17 17

ΣFx = 0: Ax +

4 B=0 17

ΣFy = 0: Ay − P +

1 B=0 17

B=

17 P 4

B = 1.031P

14.04° !

A = 1.250P

36.9° !

Ax = − P Ay = P −

P 3P = 4 4

System is rigid !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 117. (a) Member FBDs:

FBD I:

ΣM A = 0: aF1 − 2aP = 0

FBD II:

ΣM B = 0: − aF2 = 0

F1 = 2P;

ΣFy = 0: Ay − P = 0

F2 = 0

ΣFx = 0: Bx + F1 = 0, Bx = − F1 = −2P

B x = 2P

ΣFy = 0: By = 0

FBD I:

Ay = P

ΣFx = 0: − Ax − F1 + F2 = 0

Ax = F2 − F1 = 0 − 2P

so B = 2P

!

A x = 2P so A = 2.24P

26.6° !

Frame is rigid ! (b) FBD left:

FBD I:

FBD II:

FBD whole:

ΣM E = 0:

a a 5a P + Ax − Ay = 0 2 2 2

ΣM B = 0: 3aP + aAx − 5aAy = 0

Ax − 5 Ay = − P Ax − 5 Ay = −3P This is impossible unless P = 0 ∴ not rigid !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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(c) Member FBDs:

FBD I: ΣFy = 0: A − P = 0

FBD II:

ΣM D = 0: aF1 − 2aA = 0

F1 = 2 P

ΣFx = 0: F2 − F1 = 0

F2 = 2 P

ΣM B = 0: 2aC − aF1 = 0

C =

ΣFx = 0: F1 − F2 + Bx = 0 ΣFx = 0: By + C = 0

F1 = P 2

A= P

!

C= P

!

Bx = P − P = 0 By = −C = − P

B= P !

Frame is rigid !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 118.

FBD ABC: ΣM C = 0: 0.045 m + ( 0.30 m ) sin 30° ( 400 N ) sin 30°  + 0.030 m + ( 0.30 m ) cos 30°  ( 400 N ) cos 30°

 12   5  − ( 0.03 m )  FBD  − ( 0.045 m )  FBD  = 0 13 13     FBD = 3097.64 N ΣFx = 0:

5 ( 3097.64 N ) − ( 400 N ) sin 30° − Cx = 0 13 C x = 991.39 N

ΣFy = 0:

12 ( 3097.64 N ) − ( 400 N ) cos 30° − C y = 0 13 C y = 2512.9 N

FBD Blade: (a)

Vertical component at D =

12 ( 3097.64 N ) 13 = 2.86 kN

(b)

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

C = 2.70 kN

68.5°

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Chapter 6, Solution 119.

FBD Blade:

ΣFy = 0:

FBD ABC:

ΣM C = 0:

3 kN −

12 FBD = 0: FBD = 3.25 kN C 13

0.045 m + ( 0.3 m ) sin 30°  P sin 30° + 0.030 m + ( 0.3 m ) cos 30° P cos 30°

− ( 0.045 m ) − ( 0.03 m )

5 ( 3.25 kN ) 13

12 ( 3.25 kN ) = 0 13

P = 0.41968 kN so ΣFx = 0:

P = 420 N

(a)

− ( 0.41968 kN ) sin 30° +

5 ( 3.25 kN ) − Cx = 0 13

C x = 1.04016 kN ΣFy = 0:

− ( 0.41968 kN ) cos30° +

12 ( 3.25 kN ) − C y = 0 13

C y = 2.6365 kN so

(b)

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

C = 2.83 kN

68.5°

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Chapter 6, Solution 120.

FBD Stamp D: ΣFy = 0:

E − FBD cos 20° = 0,

ΣM A = 0:

FBD ABC:

E = FBD cos 20°

( 0.2 m )( sin 30°)( FBD cos 20°) + ( 0.2 m )( cos 30° )( FBD sin 20° ) − ( 0.2 m ) sin 30° + ( 0.4 m ) cos15°  ( 250 N ) = 0

FBD = 793.64 N C and, from above,

E = ( 793.64 N ) cos 20° E = 746 N

(a) ΣFx = 0:

!

Ax − ( 793.64 N ) sin 20° = 0 A x = 271.44 N

ΣFy = 0:

Ay + ( 793.64 N ) cos 20° − 250 N = 0

A y = 495.78 N so

(b )

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

A = 565 N

61.3° !

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Chapter 6, Solution 121. \

FBD Stamp D: ΣFy = 0:

900 N − FBD cos 20° = 0,

FBD = 957.76 N C

(a) FBD ABC: ΣM A = 0:

( 0.2 m )( sin 30° )  ( 957.76 N ) cos 20°

+ ( 0.2 m )( cos 30° )  ( 957.76 N ) sin 20° − ( 0.2 m ) sin 30° + ( 0.4 m ) cos15° P = 0

P = 301.70 N,

P = 302 N

(b)

ΣFx = 0:

Ax − ( 957.76 N ) sin 20° = 0 A x = 327.57 N

ΣFy = 0:

− Ay + ( 957.76 N ) cos 20° − 301.70 N = 0 A y = 598.30 N

so

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

A = 682 N

61.3°

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Chapter 6, Solution 122.

FBD ABC:

( 45 mm ) sin 25° − 10 mm 

θ = sin −1  ΣM C = 0:

100 mm

= 5.1739°

( 70 mm )(110 N ) − ( 45 mm ) sin 25° FBD cos 5.1739° + ( 45 mm ) cos 25° FBD sin 5.1739° = 0

FBD = 504.50 N ΣFx = 0:

( 504.50 N ) cos 5.1739° − (110 N ) sin 25° − Cx = 0,

Cx = 455.96 N

FBD CE: ΣFx = 0:

455.96 N − Q = 0

Q = 455.96 N,

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

(a)

Q = 456 N

(b)

FBD = 540 N T

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Chapter 6, Solution 123.

Member FBDs:

θ = tan −1

= tan −1

( BC + CD ) sin 60° AF + ( AB + BC − CD ) cos 60° ( 2.5 in. + 1 in.) sin 60° 4.5 in. + (1.5 in. + 2.5 in. − 1 in.) cos 60°

θ = 26.802°

From FBD CDE: ΣM C = 0:

( 7.5 in.)( 20 lb ) − (1 in.) FDF cos ( 30° − 26.802°) = 0, FDF = 150.234 lb C

ΣFx = 0:

(150.234 lb ) cos ( 26.802°) − ( 20 lb ) sin 60° − Cx = 0, Cx = 116.774 lb

ΣFy = 0:

(150.234 lb ) sin 26.802° − ( 20 lb ) cos 60° − C y

=0

C y = 57.742 lb From FBD ABC: ΣM A = 0:

(1.5 in.) sin 60° Bx + (1.5 in.) cos 60° By − ( 4 in.) sin 60°  (116.774 lb ) + ( 4 in.) cos 60° ( 57.742 lb ) = 0,

3 Bx + By = 385.37 lb continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

(1)

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From FBD BFG: ΣM G = 0:

− (1.5 in.) (150.234 lb ) sin 26.802° + (1.5 in.) cos 30°  (150.234 lb ) cos 26.802°

− (1.5 in.) cos 30° Bx − 6 in. + (1.5 in.) sin 30°  B y = 0,

Bx = 243.32 lb,

Solving (1) and (2):

ΣFx = 0:

ΣFy = 0:

3Bx − 9 B y = 96.775 lb

By = − 36.075 lb

243.32 lb − (150.234 lb ) cos 26.802° − Gx = 0,

G x = 109.226 lb G y = 31.667 lb

G y − (150.234 lb ) sin 26.802° + 36.075 lb = 0, On G,

G = 113.7 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

16.17° !

(2)

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Chapter 6, Solution 124.

Member FBDs:

θ = tan −1

= tan −1

( BC + CD ) sin 60° AF + AB cos 75° + ( BC − CD ) cos 60° ( 2.5 in. + 1 in.) sin 60° 4.5 in. + (1.5 in.) cos 75° + ( 2.5 in. − 1 in.)  cos 60°

θ = 28.262° From FBD CDE: ΣM C = 0:

( 7.5 in.)( 20 lb ) − (1 in.) FDF cos ( 30° − 28.262°) = 0 FDF = 150.069 lb

ΣFx = 0:

(150.069 lb ) cos 28.262° − ( 20 lb ) sin 60° − Cx = 0 Cx = 114.859 lb

ΣFy = 0:

(150.069 lb ) sin 28.262° − ( 20 lb ) cos 60° − C y

=0

C y = 61.058 lb From FBD ABC: ΣM A = 0:

(1.5 in.) sin 75° Bx − (1.5 in.) cos 75° By − (1.5 in.) sin 75° + ( 2.5 in.) sin 60° (114.859 lb ) + (1.5 in.) cos 75° + ( 2.5 in.) cos 60° ( 61.058 lb ) = 0

or 1.44889 Bx − 0.38823 B y = 315.07 lb

(1) continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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From FBD BFG: ΣM G = 0:

− (1.5 in.) (150.069 lb ) sin 28.262° + (1.5 in.) cos15° (150.069 lb ) cos 28.262°  − (1.5 in.) cos15° Bx + ( 6 in.) + (1.5 in.) sin15°  B y = 0

or 1.44889 Bx − 6.3882 B y = 84.926 lb

Solving (1) and (2):

Bx = 227.74 lb,

(2)

By = 38.358 lb

ΣFx = 0:

227.74 lb − (150.069 lb ) cos 28.262° − Gx = 0,

G x = 95.560 lb

ΣFy = 0:

G y − (150.069 lb ) sin 28.262° + 38.358 lb = 0,

G y = 32.700 lb

So force on G is

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

G = 101.8 lb

18.89° !

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Chapter 6, Solution 125.

FBD Piston:

ΣFy = 0:

180 lb − FCE cos 20° = 0,

FCE = 191.552 lb C

FBD Joint C:

FBC 191.552° = sin 40° sin 40° FBC = 191.552 lb

FBD AB:

ΣM A = 0:

( 6 in.)(191.552 lb ) cos 30° − M

=0

M = 995 lb ⋅ in.

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 126.

FBD AB: ΣM A = 0:

( 6 in.) FBC cos 30° − 660 lb ⋅ in. = 0, FBC = 127.017 lb T

FBD Piston: 180 lb C cos φ

ΣFy = 0:

180 lb − FCE cos φ = 0,

FCE =

ΣFy = 0:

180 lb cos φ + (127.017 lb ) sin 30° cos φ

− FCD cos φ = 0,

FBD Joint C:

FCD cos φ = 243.51 lb ΣFx = 0:

(1)

(127.017 lb ) cos30° −

180 lb sin φ − FCD sin φ = 0 cos φ

FCD sin φ = 110.000 lb − (180 lb ) tan φ divide (2) by (1)

(2)

tan φ = 0.45173 − 0.73919 tan φ tan φ = 0.25974,

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

φ = 14.56°

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Chapter 6, Solution 127. (a) FBD E:

ΣFx = 0:

FDE cos α −160 N = 0,

where α = tan −1 FBD Joint D:

β = tan −1

FDE =

160 N cosα

38.8 mm = 22.823°, so FDE = 173.59 N C 92.2 mm 37.5 mm − 18.8 mm = 25.732° 38.8 mm

FCD cos ( 45° − 25.732° )

ΣFx = 0:

− (173.59 N ) cos ( 25.732° + 22.823° ) = 0

FCD = 121.717 N C FBD BC: ΣM B = 0:

( 0.045 m )(121.717 N ) sin 45° − M M = 3.873 N ⋅ m,

= 0, M = 3.87 N ⋅ m

(b) FBD E:

ΣFx = 0:

FDE cos α −160 N = 0,

where α = tan −1

FDE =

160 N cos α

20 mm + 35.4 mm − 26.4 mm = 16.8584° 92.2 mm + 27.8 mm − 24.3 mm so FDE = 167.185 N C

continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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FBD Joint D:

β = tan −1

27.8 mm − 24.3 mm = 7.5520° 26.4 mm

γ = tan −1

20 mm + 35.4 mm − 26.4 mm = 42.363° 24.3 mm + 7.5 mm

ΣFy′ = 0:

(167.185 N ) sin (16.8584° + 42.363°) − FCD sin ( 90° − 7.5520° − 42.363° ) = 0, FCD = 223.07 N C

FBD BC:

δ = tan −1

35.4 mm = 51.857° 27.8 mm

ΣM B = 0: M − ( 0.045 m )( 223.07 N ) sin ( 90° − 51.857° − 7.5520° ) = 0

M = 5.12 N ⋅ m !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 128. (a) FBD BC:

( 0.045 m ) FCD sin 45° − 6.00 N ⋅ m = 0

ΣM B = 0:

FCD = 188.562 N C

FBD Joint D:

α = tan −1

38.8 mm = 22.823° 92.2 mm

β = tan −1

37.5 mm − 18.8 mm = 25.732° 38.8 mm

ΣFx′ = 0:

(188.562 N ) cos ( 45° − 25.732°) − FDE cos ( 22.823° + 25.732° ) = 0, FDE = 268.92 N C

FBD E:

( 268.92 N ) cos ( 22.823°) − P = 0,

ΣFx = 0:

P = 248 N

(b) FBD BC:

δ = tan −1

35.4 mm = 51.857° 27.8 mm

β = tan −1

27.8 mm − 24.3 mm = 7.5520° 26.4 mm

ΣM B = 0:

!

( 0.045 m ) FCD sin ( 90° − 51.857° − 7.5510°) − 6.00 N ⋅ m = 0 FCD = 262.00 N C continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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FBD Joint D:

β = 7.5520°, α = tan −1

γ = tan −1

20 mm + 35.4 mm − 26.4 mm = 42.363° 24.3 mm + 7.5 mm

20 mm + 35.4 mm − 26.4 mm = 16.8584° 92.2 mm + 27.8 mm − 24.3 mm

ΣFy′ = 0:

FDE sin (16.8584° + 42.363° ) − ( 262.00 N ) sin ( 90° − 7.5520° − 42.363° ) = 0, FDE = 196.366 N C

FBD E:

ΣFx = 0:

(196.366 N )( cos16.8584°) − P = 0, P = 187.9 N

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 129.

Member FBDs:

BCD:

AB:

ΣFy = 0:

By = 0

ΣM C = 0:

( 0.128 m ) Bx − ( 0.10 m )(80 N ) = 0,

ΣM A = 0:

(0.15 m) ( 62.5 N ) − M = 0,

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

Bx = 62.5 N

M = 9.38 N ⋅ m

!

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Chapter 6, Solution 130.

FBD BCD: d = 0.24 m

ΣFy = 0:

0.100 m = 0.1875 m 0.128 m

(80 N ) sin 45° − By

=0

By = 40 2 N

ΣM C = 0:

( 0.100 m )(80 N ) cos 45° + ( 0.1875 m )( 80 N ) sin 45°

(

+ ( 0.24 m ) 40 2 N

)

− ( 0.128 m ) Bx = 0

Bx = 233.13 N FBD AB:

ΣM A = 0:

( 0.2 m ) ( 40

)

2 N − ( 0.15 m )( 233.13 N ) = 0

M = 46.3 N ⋅ m.

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 131.

FBD BD:

θ = 90° − 25° = 65° ΣM B = 0:

 2 ( 0.2 m ) cos 25° D sin 65° −15 N ⋅ m = 0

D = 45.654 N

FBD AC:

ΣM A = 0:

M A = ( 0.2 m )( 45.654 N ) = 0

M A = 9.13 N ⋅ m

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 132.

FBD BD: ΣM B = 0:

 2 ( 0.2 m ) cos 25° C −15 N ⋅ m = 0

C = 41.377 N

FBD AC: ΣM A = 0:

M A − ( 0.2 m )( 41.377 N ) sin 65° = 0

M A = 7.50 N ⋅ m

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 133.

FBD Member CD:

Noting that B is Perpendicular to CD,

ΣFx′ = 0:

Dy sin 30° − ( 80 lb ) cos 30° = 0,

D y = 138.564 lb FBD Machine:

ΣM A = 0:

10 in.

(138.564 lb ) − M = 0, sin 30° M = 2771.3 lb ⋅ in.

M = 2.77 kip ⋅ in.

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 134.

FBD CD: Since B is ⊥ to CD

ΣFx′ = 0:

Dy sin 30° − ( 80 lb ) cos 30° = 0 D y = 138.564 lb

FBD Whole:

a = 10 in. − ( 8 in.) cos 30° = 3.07180 in. b=

a = 5.32051 in. tan 30°

d = b − 4 in. = 1.32051 in. ΣM A = 0:

(10 in.)(80 lb ) + (1.32051 in.)(138.564 lb ) − M

=0

M = 983 lb ⋅ in.

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 135.

FBD ABD:

First note, by inspection, that P = W = 250 lb.

Note: that BC is a two-force member

ΣM D = 0:

( 5 in.)

3 1 FBC + ( 3 in.)

FBC − ( 9 in.)( 250 lb ) = 0 10 10

FBC = 125 10 lb T

ΣFx = 0:

Dx −

(

)

3 125 10 lb = 0 10

D x = 375 lb ΣFy = 0:

250 lb −

(

)

1 125 10 lb − Dy = 0 10

D y = 125 lb

(note, that Dy = 125 lb is evident by symmetry of the barrel) so D = 395 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

18.43°

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Chapter 6, Solution 136. \

FBD whole:

By symmetry A = B = 22.5 kN

FBD ADF:

ΣM F = 0:

( 75 mm ) CD − (100 mm )( 22.5 kN ) = 0 CD = 30 kN

ΣFx = 0: ΣFy = 0:

Fx − CD = 0 22.5 kN − Fy = 0

W

Fx = CD = 30 kN Fy = 22.5 kN so F = 37.5 kN

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

36.9° W

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Chapter 6, Solution 137.

FBD Joint A:

TB TC 2.1 kN = = 17 17 30 TB = TC = 1.19 kN

FBD BF and CF w/cables to A:

By symmetry

Ex = Fx = 1.05 kN and E y = Fy

FBD Machine:

By symmetry

H x = I x = 1.05 kN and H y = I y

ΣM D = 0:

FBD BF:

( 0.075 m ) Fy − ( 0.08 m ) (1.05 kN ) 15 (1.19 kN ) 17 8 − ( 0.16 m ) (1.19 kN ) = 0 16 Fy = 4.48 kN − ( 0.15 m )

ΣM G = 0:

( 0.075 m ) ( 4.48 kN ) + ( 0.08 m ) (1.05 kN ) + ( 0.02 m ) (1.05 kN ) − ( 0.15 m ) H y = 0

FBD FGH:

H y = 2.94 kN ΣFy = 0:

2.94 kN − G y + 4.48 kN = 0 ,

G y = 7.42 kN

ΣFx = 0:

−1.05 kN + Gx + 1.05 kN = 0 ,

Gx = 0

so

H = 3.12 kN

70.3° !

G = 7.42 kN !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 138.

FBD Machine: By inspection, P = W = 110 lb, By symmetry: ΣFy = 0:

H x = Ix,

110 lb − 2H y = 0 ,

ΣFy = 0:

H y = I y = 55 lb,

FCE = FBD

By symmetry:

FBD ABC:

Hy = Iy

4  110 lb − 2  FCE  = 0 , 5  

FCE = FBD =

550 lb C 8

 4  550    3  550   lb   + (17 in.)   lb      5  8 5  8

( 2.5 in.) 

ΣM H = 0:

− (4 in.) FFG = 0

FBD DFH:

FFG = 209.69 lb, ΣFx = 0:

209.69 lb −

FFG = 210 lb

W

3  550  lb  − H x = 0  5 8  H x = 168.440 lb so

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

H = 177.2 lb

18.08° W

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Chapter 6, Solution 139.

FBD top handle: Note CD and DE are two-force members

ΣM A = 0:

( 4 in.)

6 5 FCD − (1.5 in.)

FCD − (13.2 in.) ( 90 lb ) = 0 61 61

FCD = 72 61 lb FDE = FCD = 72 61 lb

By symmetry:

FBD Joint D: ΣFx = 0:

D−2

(

)

5 72 61 lb = 0, 61

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

D = 720 lb W

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Chapter 6, Solution 140.

FBD top handle: By symmetry the horizontal force at F must be zero ΣFx = 0: Dx = 0 ΣM F = 0:

( 0.015 m ) Fy − ( 0.185 m )(135 N ) = 0 Dy = 1665 N

FBD ABD:

ΣM B = 0:

( 40 mm ) (1665 N ) − ( 30 mm ) A = 0 A= 2220 N A= 2.22 kN W

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 141.

FBD cutter AC:

ΣM C = 0:

( 32 mm )1.5 KN − ( 28 mm ) Ay − (10 mm ) Ax

=0

 11  10 Ax + 28  Ax  = 48 kN  13  Ax = 1.42466 kN Ay = 1.20548 kN

FBD handle AD:

ΣM D = 0:

(15 mm )(1.20548 kN ) − ( 5 mm )(1.42466 kN ) − (70 mm) P = 0 P = 0.1566 kN = 156.6 N W

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 142.

FBD CD:

θ = tan − 1 ΣM D = 0:

0.85 in. − 0.3 in. = 22.932° 2.6 in. − 1.3 in.

( 3.4 in.)( 30 lb ) − ( 0.8 in.) FAC sin 22.932° + ( 0.3 in.) FAC cos 22.932° = 0 FAC = 2879.6 lb

ΣFx = 0: Dx − ( 2879.6 lb ) cos 22.932° = 0,

Dx = 2652.0 lb

ΣFy = 0: Dy − ( 2879.6 lb ) sin 22.932° + 30 lb = 0, Dy = 1092.00 lb

FBD BD: ΣM B = 0:

( 2.6 in.) F + ( 0.2 in.)(1092.00 lb ) − ( 0.85 in.)( 2652.0 lb ) = 0 F = 783 lb W

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 143.

FBD Joint B:

α = tan − 1

1 = 5.7106° 10

ΣFx′ = 0: FBC cos ( 30° + 5.7106° ) − ( 54 lb ) cos 30° = 0 FBC = 57.595 lb T

FCD = 57.595 lb

By symmetry:

FBD Joint C:

ΣFx′′ = 0: 2 ( 57.595 lb ) sin ( 5.7106° ) − P = 0 P = 11.4618 lb so

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

P = 11.46 lb

30.0° W

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Chapter 6, Solution 144.

FBD BC:

ΣM B = 0:

(1 in.) Cx − ( 2.65 in.) ( 270 lb ) = 0 Cx = 715.5 lb

FBD CF:

( 3.5 in.) P − ( 0.75 in.) FDE cos 40° + ( 0.65 in.) FDE sin 40° = 0

ΣM C = 0:

FDE = 22.333 P ΣFx = 0: C x − ( 22.333 P ) cos 40° + P sin 30° = 0 P = 43. 081 lb

P = 43.1 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

60.0˚ W

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Chapter 6, Solution 145.

FBD Handles:

ΣM C = 0: − aBx + b (100 N − 100 N ) = 0 Bx = 0

FBD top handle: ΣM A = 0:

( 28 mm ) By − (110 mm ) (100 N ) = 0 By = 392.86 N

FBD top blade:

ΣM D = 0:

( 40 mm ) ( 392.86 N ) − ( 30 mm ) E = 0 E = 523.81 N

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

E = 524 N W

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Chapter 6, Solution 146.

FBD BE:

ΣM B = 0:

(

)

746 mm C − ( 640 mm ) ( 0.25 kN ) = 0 C=

160 kN 746

FBD Blade: ΣM A = 0:

 2  160  kN      5  746

( 38 mm ) F − ( 47 mm ) 

 1  160  kN   = 0 − ( 23 mm )     5  746 F = 8.82 kN W

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 147.

FBD Boom with bucket and man:

θ = tan − 1

3 ft + ( 6 ft ) sin 35° = 62.070° ( 6 ft ) cos 35° − 1.5 ft

ΣM A = 0: ( 6 ft ) cos 35° ( FBD sin 62.070° ) − ( 6 ft ) sin 35° ( FBD cos 62.070° ) − ( 9ft ) cos 35°  (1400 lb ) − ( 20 ft ) cos 35°  ( 450 lb ) = 0

FBD = 17693 lb,

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

FBD = 6.48 kips

62.1° W

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Chapter 6, Solution 148.

FBD AED:

ΣM A = 0:

(a)

( 36 in.) (1500 lb ) − ( 25 in.) F = 0, F = 2160 lb

ΣFy = 0: Ay − 1500 lb = 0, ΣFy = 0: Ax − 2160 lb = 0,

FBD Entire machine:

ΣM B = 0:

Ay = 1500 lb

Ax = 2160 lb

(84 in.) (1500 lb ) − ( 25 in.) Gx = 0, Gx = 5040 lb

ΣM C = 0:

FBD AG:

( 24 in.) ( Gy + 1500 lb ) − (12.5 in.) ( 5040 lb − 2160 lb ) = 0, Gy = 0 W

(b)

ΣFy = 0: C y − 1500 lb + 0 = 0, C y = 1500 lb ΣFx = 0:

Cx − 2160 lb − 5040 lb = 0, Cx = 7200 lb

so on BCF,

C x = 7.20 kips

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

,

C y = 1.500 kips

W

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Chapter 6, Solution 149.

FBD boom:

θ = tan −1

Note:

( 3.2sin 24° − 1) m ( 3.2cos 24° − 0.6 ) m

θ = 44.73° (a) ΣM A = 0:

( 6.4 m ) cos 24° ( 2.3544 kN ) − ( 3.2 m ) cos 24° B sin 44.73° + ( 3.2 m ) sin 24° B cos 44.73° = 0

B = 12.153 kN

B = 12.15 kN

44.7° W

ΣFx = 0: Ax − (12.153 kN ) cos 44.73° = 0

A x = 8.633 kN (b) ΣFy = 0:

− 2.3544 kN + (12.153 kN ) sin 44.73° − Ay = 0

A y = 6.198 kN

On boom: On carriage:

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

A = 10.63 kN A = 10.63 kN

35.7° 35.7° W

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Chapter 6, Solution 150.

FBD Bucket (one side):

(a) ΣM D = 0:

( 0.8 m ) ( 2.4525 kN ) − ( 0.5 m ) FAB = 0 FAB = 3.924 kN

w = ( 250 kg ) ( 9.81 N/kg ) = 2452.5 N = 2.4525 kN

FBD link BE:

ΣM E = 0:

( 0.68 m )( 3.924 kN ) − ( 0.54 m ) 

80  FCD   89 

 39  + ( 0.35 m )  FCD  = 0,  89 

FCD = 7.682 kN FCD = 7.68 kN C W

FBD top blad

(b)

FBD Entire linkage:

ΣM G = 0:

( 2.5 m ) ( 2.4525 kN ) 1 1 FGH − ( 0.6 m ) FGH = 0, 2 2 FFH = 21.677 kN, FFH = 21.7 kN C W

+ ( 0.2 m )

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 151. \

FBD Central Gear:

By symmetry: F1 = F2 = F3 = F ΣM A = 0:

3 ( rA F ) − 15 N ⋅ m = 0,

ΣM C = 0:

rB ( F − F4 ) = 0,

F =

5 N⋅m rA

FBD Gear C:

ΣFx′ = 0:

C x′ = 0

ΣFy′ = 0:

C y′ − 2 F = 0,

F4 = F

C y′ = 2 F

Gears B and D are analogous, each having a central force of 2F FBD Spider:

ΣM A = 0:

75 N ⋅ m − 3 ( rA + rB ) 2 F = 0 25 N ⋅ m − ( rA + rB )

10 N⋅m = 0 rA

rA + rB r = 2.5 = 1 + B , rA rA FBD Outer gear:

Since

rA = 24 mm,

ΣM A = 0:

rB = 1.5 rA rB = 36.0 mm !

(a) 3 ( rA + 2rB ) F − M E = 0

3 ( 24 mm + 72 mm ) (b)

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

5 N⋅m −ME = 0 24 mm

M E = 60.0 N ⋅ m

!

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Chapter 6, Solution 152.

FBD Gear A: looking from C (a)

M A = 15 lb ⋅ ft ΣM A = 0:

rA = 4 in.

M A − P rA = 0

P=

M A 180 lb ⋅ in. = rA 4 in.

P = 45 lb

FBD Gear B: looking from F

ΣM B = 0:

M 0 − rB P = 0

M 0 = rB P = ( 2.5 in.)( 45 lb ) = 112.5 lb ⋅ in.

M 0 = 112.5 lb ⋅ in. i !

FBD ABC: looking down

(b)

( 2 in.)( 45 lb ) − ( 5 in.) C

ΣM B = 0: ΣFz = 0: FBD BEG:

=0

45 lb − B + 18 lb = 0

By analogy, using FBD DEF ΣFz = 0:

E = 63 lb k

C = 18 lb k B = −63 lb k F = 18 lb k

Gz + 63 lb − 63 lb = 0 Gz = 0

ΣFy = 0

Gy = 0

ΣM G = 0

M G − ( 6.5 in.)( 63 lb ) = 0

M G = ( 410 lb ⋅ in.) i !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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FBD CFH: ΣF = 0:

Hz = H y = 0

ΣM H = 0 M H = − ( 6.5 in.)(18 lb ) = −117 lb ⋅ in.

M G = − (117.0 lb ⋅ in.) i !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 153.

Note: The couples exerted by the two yokes on the crosspiece must be equal and opposite. Since neither yoke can exert a couple along the arm of the crosspiece it contacts, these equal and opposite couples must be normal to the plane of the crosspiece.

If the crosspiece arm attached to shaft CF is horizontal, the plane of the crosspiece is normal to shaft AC, so couple M C is along AC. FBDs shafts with yokes:

(a)

ΣM x = 0:

FBD CDE:

M C cos30° − 50 N ⋅ m = 0

M C = 57.735 N ⋅ m

FBD BC: ΣM x′ = 0: M A − M C = 0

(b)

M A = 57.7 N ⋅ m W

ΣM C = 0: M A i′ + ( 0.5 m ) Bz j′ − ( 0.5 m ) By′ k = 0 ΣF = 0: B + C = 0

so

B = 0W

C=0

FBD CDF : ΣM Dy = 0: − ( 0.6 m ) Ez + ( 57.735 N ⋅ m ) sin 30° = 0 E z = 48.1 N k ΣFx = 0: Ex = 0 ΣM Dz = 0: ( 0.6 m ) E y = 0

Ey = 0

0

ΣF = 0: C + D + E = 0

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

so E = ( 48.1 N ) k W D = −E = − ( 48.1 N ) k W

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Chapter 6, Solution 154.

Note: The couples exerted by the two yokes on the crosspiece must be equal and opposite. Since neither yoke can exert a couple along the arm of the crosspiece it contacts, these equal and opposite couples must be normal to the plane of the crosspiece.

If the crosspiece arm attached to CF is vertical, the plane of the crosspiece is normal to CF, so the couple M C is along CF.

(a)

FBD CDE: FBD BC:

ΣM x = 0: M C − 50 N ⋅ m = 0 ΣM x′ = 0: M A − M C cos 30° = 0

M C = 50 N ⋅ m M A = ( 50 N ⋅ m ) cos 30° M A = 43.3 N ⋅ m W

(b)

ΣM Cy′ = 0: M C sin 30° + ( 0.5 m ) Bz = 0 ΣM Cz = 0: − ( 0.5 m ) By = 0 ΣF = 0: B + C = 0

FBD CDE:

Bz = −

( 50 N ⋅ m )( 0.5) 0.5 m

= −50 N so B = − ( 50.0 N ) k W

By = 0

C = −B

so

C = ( 50 N ) k on BC

ΣM Dy = 0: − ( 0.4 m ) C z − ( 0.6 m ) Ez = 0

4 Ez = − ( 50 N )   = −33.3 N 6

ΣM Dz = 0: E y = 0 ΣFx = 0: Ex = 0 ΣF = 0: C + D + E = 0

so E = − ( 33.3 N ) k W

− ( 50 N ) k + D − ( 33.3 N ) k = 0

D = ( 83.3 N ) k W

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 155.

FBD Joint A: By inspection of FBD of entire device, FA = w = 1500 lb

By symmetry, FAB = FAC  1  FAC  = 0, 1500 lb − 2   5 

ΣFy = 0:

FBD CDF:

ΣFx = 0:

(

FAC = 750 5 lb

)

2 750 5 lb = 0, 5

Dx − Fx −

Dx − Fx = 1500 lb ΣFy = 0:

(

(1)

)

1 750 5 lb = 0, 5

Fy − D y +

Dy − Fy = 750 lb ΣM D = 0:

FBD EF:

(2)

 1   2  (4.6 ft)  750 5 lb  + ( 0.6 ft )  750 5 lb   5   5 

(

)

(

)

− (1 ft ) Fy − ( 3.6 ft ) Fx = 0, 3.6 Fx + Fy = 4350 lb By symmetry:

ΣM F = 0:

Ex = Dx ,

E y = Dy

Hy = Jy =

w = 750 lb, 2

(4), (5) H y = 750 lb

!

( 3 ft ) E y − ( 3.6 ft ) Ex − (1 ft ) H y + ( 0.6 ft )( 750 lb ) = 0 3.6Ex − 3E y + H y = 450 lb

ΣFx = 0:

(3)

Dx + Fx − H x = 0,

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

(6)

(7)

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Solving Eqns, (1)–(7), on EFH

Fx = 540 lb Fy = 2410 lb E x = 2040 lb E y = 3160 lb H x = 2580 lb H y = 750 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

! ! ! ! ! !

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 156.

FBD Truss:

ΣM E = 0: ( 9 ft ) Fy − ( 6.75 ft )( 4 kips ) − (13.5 ft )( 4 kips ) = 0 Fy = 9 kips ΣFy = 0: − E y + 9 kips = 0

E y = 9 kips

ΣFx = 0: − Ex + 4 kips + 4 kips = 0

E x = 8 kips

By inspection of joint E:

FEF = 8.00 kips T !

Joint FBDs: Joint F:

FEC = 9.00 kips T !

By inspection of joint B:

FAB = 0 !

FBD = 0 !

ΣFx = 0: Joint C:

4 FCF − 8 kips = 0 5

ΣFy = 0: FDF −

3 (10 kips) = 0 5

ΣFx = 0: 4 kips −

FCF = 10.00 kips C !

FDF = 6.00 kips T !

4 (10 kips ) + FCD = 0 5 FCD = 4.00 kips T !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

ΣFy = 0: FAC − 9 kips +

3 (10 kips ) = 0 5 FAC = 3.00 kips T !

Joint A:

ΣFx = 0: 4 kips −

4 FAD = 0 5

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

FAD = 5.00 kips C !

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Chapter 6, Solution 157.

FBD Truss: ΣM A = 0: ( 6 ft )  6Ly − 3 ( 0.5 kip ) − 2 (1 kip ) − 1(1 kip )  = 0 L y = 0.75 kip

FJK = 0 !

Inspection of joints K , J , and I , in order, shows that

FIJ = 0 !

FHI = 0 !

Joint FBDs:

FIK = FKL ; FHJ = FJL and FGI = FIK

and that

0.75 F F = JL = KL 1 8 5

FJL = 2.1213 kips

FKL = 1.6771 kips

FJL = 2.12 kips C !

FKL = 1.677 kips T !

FHJ = 2.12 kips C !

and, from above:

FGI = FIK = 1.677 kips T ! ΣFx = 0:

2 1 ( FFH + FGH ) − ( 2.1213 kips ) = 0 5 2

ΣFy = 0:

1 1 ( FGH + FFH ) + ( 2.1213 kips ) = 0 5 2

Solving:

FFH = 2.516 kips FGH = − 0.8383 kips

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

FFH = 2.52 kips C !

FGH = 0.838 kips T !

COSMOS: Complete Online Solutions Manual Organization System

ΣFx = 0:

2 ( FDF − 2.516 kips ) = 0 5

ΣFy = 0: FFG − 0.5 kip +

FDF = 2.52 kips C

1 ( 2 )( 2.516 kips ) = 0 5

FFG = 1.750 kips T !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 6, Solution 158.

FBD Truss:

ΣM F = 0: (10 m ) G y − ( 7.5 m )( 80 kN ) − ( 8 m )( 30 kN ) = 0 G y = 84 kN

ΣFx = 0: − Fx + 30 kN = 0

Fx = 30 kN

ΣFy = 0: Fy + 84 kN − 80 kN = 0

Fy = 4 kN FEG = 0 !

By inspection of joint G:

FCG = 84 kN C !

84 kN F F = CE = AC = 10.5 kN 8 61 29

Joint FBDs:

FCE = 82.0 kN T !

FAC = 56.5 kN C !

ΣFy = 0:

2 6 FAE + (82.0 kN ) − 80 kN = 0 5 61

FAE = 19.01312

ΣFx = 0: − FDE −

1 (19.013 kN ) + 5

FDE = 43.99 kN

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

FAE = 19.01 kN T !

5 (82.0 kN ) = 0 61

FDE = 44.0 kN T !

COSMOS: Complete Online Solutions Manual Organization System

ΣFx = 0:

1 FDF − 30 kN = 0 5

FDF = 67.082 kN

ΣFy = 0:

2 ( 67.082 kN ) − FBF − 4 kN = 0 5

FBF = 56.00 kN ΣFx = 0: 30 kN + ΣFy = 0: 56 kN −

Solving:

ΣFy = 0:

FDF = 67.1 kN T !

5 FBD − 61 6 FBD − 61

FBF = 56.0 kN C !

5 FAB = 0 29 3 FAB = 0 29

FBD = 42.956 kN

FBD = 43.0 kN T !

FAB = 61.929 kN

FAB = 61.9 kN C !

6 2 ( 42.956 N ) + ( FAD − 67.082 N ) = 0 61 5

FAD = 30.157 kN

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

FAD = 30.2 N T !

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Chapter 6, Solution 159.

FBD Truss:

ΣFx = 0: A x = 0 By load symmetry, A y = I y = 8 kips

FBD Section:

ΣM D = 0:

( 7 ft )( 2 kips − 8 kips ) + ( 3 ft ) ( FCE ) = 0 FCE = 14 kips

ΣM E = 0:

FCE = 14.00 kips T W

( 7 ft ) 1( 4 kips ) + 2 ( 2 kips − 8 kips )

( 4.5 ft ) FDF =

7 FDF = 0 51.25

8 51.25 kips 4.5

ΣFy = 0: 8 kips − 2 kips − 4 kips +

FDF = 12.73 kips C W 1.5 8 51.25 kips − 51.25 4.5

FDE = −1.692 kips

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

3 FDE = 0 58

FDE = 1.692 kips C W

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Chapter 6, Solution 160.

FBD Section:

p = 8.4 ( 2.7 m ) = 1.89 m Note: BG 12.0

ΣM A = 0:

( 2.7 m ) FFG − ( 3.6 m )(8 kN ) − ( 7.8 m )(8 kN ) − (12 m )( 4 kN ) = 0 FFG = 51.56 kN

ΣM G = 0:

FFG = 51.6 kN C W

(1.89 m ) 

12  FAB  − ( 4.2 m )( 8 kN ) − ( 8.4 m )( 4 kN ) = 0  12.3  FAB = 36.44 kN

ΣM D = 0:

FAB = 36.4 kN T W

( 4.2 m )(8 kN ) + (8.4 m )(8 kN ) − ( 8.4 m ) 

3  FAG  = 0 5  

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

FAG = 20.0 kN T W

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Chapter 6, Solution 161.

FBD Truss:

ΣM A = 0:

( 24 ft ) K x − (18 ft )( 2 kips ) − ( 36 ft )( 2 kips ) = 0 K x = 4.5 kips

FBD Section:

ΣM F = 0:

(12 ft )( 4.5 kips ) − (18 ft )( 2 kips ) − ( 36 ft )( 2 kips ) + ( 36 ft ) FEJ = 0 FEJ = 1.500 kips T W

ΣM J = 0:

(18 ft )( 2 kips ) + (12 ft )( 4.5 kips ) − ( 36 ft ) FAF

=0 FAF = 2.50 kips T W

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 162.

FBD Frame:

ΣM F = 0:

(10.8 ft ) Ay − (12 ft )( 4.5 kips ) = 0 A y = 5.00 kips

ΣFx = 0:

W

− Ax + 4.5 kips = 0 A x = 4.50 kips

FBD member ABC:

W

Note: BE is a two-force member ΣM C = 0:

ΣFx = 0:

ΣFy = 0:

(12 ft ) FBE + (10.8 ft )( 5 kips ) − (18 ft )( 4.5 kips ) = 0 FBE = 2.25 kips

W

C x = 2.25 kips

W

Cx + 2.25 kips − 4.5 kips = 0

C y − 5 kips = 0 C y = 5.00 kips

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

W

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 163.

FBD Frame:

ΣM E = 0:

( 3.75 in.) Bx + (8.75 in.)( 75 lb ) = 0

Bx = 175 lb

FBD member ACE:

ΣFx = 0:

Ex − Bx = 0

ΣFy = 0:

E y + By − 75 lb = 0

B x = 175.0 lb

W

E x = 175.0 lb

W

By = 75 lb − E y

ΣM C = 0:

Thus

− (1.25 in.)( 75 lb ) + ( 3.75 in.)(175 lb ) − ( 4.5 in.) E y = 0 E y = 125.0 lb

W

B y = 50.0 lb

W

By = 75 lb − 125 lb = −50 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 164.

FBD Frame:

Q = 65 N

P = 15 N ΣM D = 0:

( 0.25 m )( P + Q ) − (.15 m )( P + Q ) − ( 0.08 m ) Ex

Ex = 1.2 ( P + Q ) = 100 N ΣFx = 0:

(b)

Dx − Ex = 0 = Dx − 100 N (a)

ΣFy = 0:

E x = 100.0 N

=0 W

Dx = 100 N D x = 100.0 N

W

E y − Dy − 2P − 2Q = 0 E y = D y + 2 ( P + Q ) = Dy + 160 N

FBD member BF:

ΣM C = 0:

( 0.15 m )( 65 N ) − ( 0.1 m ) Dy − ( 0.04 m )(100 N ) − ( 0.25 m )(15 N ) = 0 Dy = 20 N

From above

D y = 20.0 N

W

E y = 20 N + 160 N = 180 N (b) E y = 180.0 N

W

ΣFx = 0:

(a)

− C x + 100 N = 0 (a)

ΣFy = 0:

C x = 100.0 N

W

− 65 N + C y − 20 N − 15 N = 0 (a)

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

C y = 100.0 N

W

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Chapter 6, Solution 165.

Member FBDs: FBD I:

ΣM C = 0:

R ( FBD sin 30° ) −  R (1 − cos 30° )  ( 200 N ) − R ( 100 N ) = 0

FBD = 253.6 N ΣFx = 0:

(b)

FBD = 254 N T W

− C x + ( 253.6 N ) cos 30° = 0 C x = 219.6 N

ΣFy = 0:

C y + ( 253.6 N ) sin 30° − 200 N − 100 N = 0

C y = 173.2 N

(c) so C = 280 N

ΣM A = 0:

aP − a ( 253.6 N ) cos 30°  = 0

FBD II:

(a)

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

38.3° W

P = 220 N

W

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Chapter 6, Solution 166.

FBD jaw AB:

ΣFx = 0: Bx = 0 ΣM B = 0:

( 0.01 m ) G − ( 0.03 m ) A = 0 A=

ΣFy = 0:

G 3

A + G − By = 0 By = A + G =

4G 3

FBD handle ACE:

By symmetry and FBD jaw DE: D = A = E y = By = ΣM C = 0:

G , Ex = Bx = 0, 3 4G 3

( 0.105 m )( 240 N ) + ( 0.015 m )

G 4G − ( 0.015 m ) =0 3 3 G = 1680 N W

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 167.

Note: By symmetry the vertical components of pin forces C and D are zero.

FBD handle ACF: (not to scale)

ΣFy = 0: ΣM C = 0: ΣFx = 0:

Ay = 0

(13.5 in.)( 20 lb ) − (1.5 in.) Ax

= 0 Ax = 180 lb

C − Ax − 20 lb = 0 C = (180 + 20 ) lb = 200 lb

FBD blade DE:

ΣM D = 0:

( 9 in.) E − ( 3 in.)(180 lb ) = 0 E = 60.0 lb W

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.