Cap 07

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Chapter 7, Solution 1.

FBD FRAME:

ΣM A = 0:

( 0.3 m ) Dx − ( 0.4 m )( 900 N ) = 0 ∴ D x = 1200 N

FBD JEHDF:

ΣFx = 0:

1200 N + V = 0 V = −1200 N

ΣFy = 0: F = 0 ΣM J = 0:

( 0.15 m )(1200 N ) − M

=0

M = + 180 N ⋅ m Thus,

( on JE )

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.

F =0

V = 1200 N

M = 180.0 N ⋅ m

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Chapter 7, Solution 2.

FBD FRAME:

ΣM A = 0:

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

ΣFy = 0:

FBD ABC:

Ay + Dy + 480 N = 0

(1)

Note: BE is a two-force member

ΣM B = 0:

( 0.75 m ) Ay

Then, from (1) above,

=0

Ay = 0

Dy = − 480 N D y = 480 N

FBD sect. DJ:

ΣFx = 0:

F − 1824 N = 0

ΣFy = 0:

− 480 N – V = 0

ΣM J = 0:

F = 1.824 kN

V = − 480 N V = 480 N

M = 120.0 N ⋅ m

M + ( 0.25 m )( 480 N ) = 0 M = −120 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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 3.

FBD CEF:

ΣFx = 0: C x − ΣΜ E = 0:

1 (180 N ) 2

C x = 90 2 N

( 0.2 m ) C y − ( 0.05 m ) ( 90

2 N

)

− ( 0.15 m + 0.08 m )(180 N ) = 0 C y = 126 2 N

FBD sect. CJ:

ΣFx = 0: 90 2 N − F = 0

F = 127.3 N

ΣFy = 0: −126 2 N + V = 0 ΣM J = 0:

( 0.05 m ) ( 90

)

V = 178.2 N

(

)

2 N − ( 0.10 m ) 126 2 N − M = 0

M = 11.46 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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 4.

FBD Frame:

Note: AC is a two-force member, resolve FAC at C:

ΣM E = 0: ( 0.25 m + 0.25 m )

1 FAC − ( 0.15 m )( 320 N ) = 0 2

FAC = 96 2 N

FBD sect. AB:

ΣFx = 0:

− 96 N +

ΣFy = 0: 96 N +

3 7 V + F =0 4 4

7 3 V − F =0 4 4

Solving:

V = 8.50 N

41.4°

F = 135.5 N

46.8°

4− 7  ΣM B = 0: M − ( 0.3 m )( 96 N ) +   10 m  ( 96 N ) = 0  

M = 15.799 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.

M = 15.80 N ⋅ m

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Chapter 7, Solution 5.

FBD Frame:

Note: AB is a two-force member, so Ay Ax = 12 5

ΣM C = 0:

(1)

(15 in.) Ax − ( 24 in.)( 78 lb ) = 0 A x = 124.8 lb From (1) above,

A y = 52.0 lb

FBD AJ:

ΣFx = 0:

−124.8 lb + F = 0

ΣFy = 0:

52 lb − V = 0

ΣM J = 0:

M − (10 in.)( 52 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.

F = 124.8 lb V = 52.0 lb M = 520 lb ⋅ in.

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Chapter 7, Solution 6.

FBD CD:

Note: AB is a two-force member

ΣΜ C = 0:

5   12  FAB  + ( 7.5 in.)  FAB   13   13 

(18 in.) 

− ( 24 in.)( 78 lb ) = 0 FAB = 135.2 lb

ΣFx = 0:

Cx −

12 (135.2 lb ) = 0 13

C x = 124.8 lb

ΣFy = 0:

Cy +

5 (135.2 lb ) − ( 78 lb ) = 0 13

C y = 26 lb FBD CK:

ΣFx′ = 0:

−F +

12 5 (124.8 lb ) + ( 26 lb ) = 0 13 13

F = 125.2 lb ΣFy′ = 0:

V −

5 12 (124.8 lb ) + ( 26 lb ) = 0 13 13

V = 24.0 lb ΣM K = 0:

22.6°

( 5 in.)(124.8 lb ) − (12 in.)( 26 lb ) − M

67.4°

=0

M = 312 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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 7.

FBD half-section: By symmetry,

Ay = B y =

W 2

Where W = ( 9 kg )( 9.81 N/kg ) = 88.29 N

FBD AJ:

ΣFx = 0: F = 0 ΣFy = 0:

F =0

W W − −V = 0 2 2

ΣM J = 0: M − ( r − x ) but

x=

V =0

W =0 2

2 W  = 2.406 N ⋅ m , so M = 1 −  r π 2 π 

2r

M = 2.41 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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 8.

FBD AJ:

Note: Cut is just left of contact with ground.

W = ( 9 kg )( 9.81 N/kg ) = 88.29 N

Also

x=

ΣFx = 0: ΣFy = 0:

2r

π

and

r = 0.15 m

F =0

−

ΣM J = 0: M =

F=0

W +V =0 2

V = 44.1 N

x

W −M =0 2

2

( 0.15 m ) 

π

88.29 N   2  

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.

M = 4.22 N ⋅ m

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Chapter 7, Solution 9.

SOLUTION FBD AB:

FBD sect. BJ:

ΣM A = 0:

ΣFy′ = 0:

(

)

r By − 12 lb = 0

B y = 12 lb

(12 lb ) cos 30° − (12 lb ) sin 30° − F = 0 F = 4.39 lb

ΣFx′ = 0:

(12 lb ) cos 30° + (12 lb ) sin 30° − V

=0

V = 16.39 lb ΣM J = 0:

60°

30°

( 4 in.) sin 30° (12 lb ) + ( 4 in.)(1 − cos 30° )(12 lb ) − M M = 30.4 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.

=0

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Chapter 7, Solution 10.

FBD AB:

ΣM A = 0:

ΣΜ J = 0: FBD BJ:

(

)

r By − 12 lb = 0

B y = 12 lb

( 4 in.) sin θ (12 lb ) + ( 4 in.)(1 − cosθ )(12 lb ) − M M = ( 48 lb ⋅ in.)(1 − cosθ + sin θ )

(a) to maximize M, set

=0

(1)

dM =0 dθ

dM = ( 48 lb ⋅ in.)( sin θ + cosθ ) = 0 dθ so tan θ = −1

θ = − 45°, 135° (only θ = 135° is on ring) (b)

ΣFy′ = 0:

F + (12 lb ) cosθ − (12 lb ) sin θ = 0

θ = 135°, so

F = 16.97 lb@135° ΣFx′ = 0:

(12 lb ) cosθ + (12 lb ) sin θ

θ = 135°

−V = 0

F = 16.97 lb

45°

θ = 135°, V = 0

From (1) above, M = ( 48 lb ⋅ in.)(1 − cos135° + sin135° ) = 115.88 lb ⋅ in.

M = 115.9 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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 11.

FBD Frame:

ΣM A = 0:

(16.4 in.)

24 FEC − (12.6 in.)(120 lb ) = 0 25 FEC = 93.75 lb

FBD CBD:

ΣΜ C = 0:

(16.2 in.)(120 lb ) − (13.5 in.) By

=0

By = 144 lb ΣFy = 0:

144 lb +

7 ( 93.75 lb ) − 120 lb − C y = 0 25 C y = 50.25 lb

ΣFx = 0:

Cx −

24 ( 93.75 lb ) = 0 27 C x = 90.0 lb

FBD CJ:

ΣFy′ = 0:

F + ( 50.25 lb ) cos30° − ( 90.0 lb ) sin 30° = 0

F = 1.482 lb ΣFx′ = 0:

V − ( 90 lb ) cos 30° + ( 50.25 lb ) sin 30° = 0 V = 103.1 lb

ΣM O = 0:

60°

(8.4 in.)( 50.25 lb + 1.482 lb ) − M

30°

=0

M = 435 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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 12.

FBD Frame:

ΣM A = 0:

(16.4 in.)

24 FEC − (12.6 in.)(120 lb ) = 0 25 FEC = 93.75 lb

FBD CBD: ΣM C = 0:

(16.2 in.)(120 lb ) − (13.5 in.) By

=0

B y = 144 lb

FBD DK: ΣFx′ = 0:

(144 lb − 120 lb ) sin 30° − F = 0 F = 12.00 lb

ΣFy′ = 0:

(144 lb − 120 lb ) cos 30° − V

=0

V = 20.8 lb ΣΜ O = 0:

30°

(8.4 in.)(12 lb − 144 lb ) + (11.1 in.)(120 lb ) − M

60°

=0

M = 223 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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 13.

FBD AB:

( 0.8 m )(1.8 kN ) − ( 0.24 m ) Ax = 0

ΣΜ B = 0:

A x = 6.0 kN

Geometry:

y = kx 2 ; at B, 0.24 m = k ( 0.8 m ) so

k = 0.375

2

1 m

1 2  J , y J =  0.375  ( 0.48 m ) = 0.0864 m m 

at

dy = 2kx dx

slope of parabola at J ,

dy 1  = 2  0.375  ( 0.48 m ) = 0.36 = tan θ J dx m 

θ J = 19.799°

FBD AJ: ΣFx′ = 0:

( 6 kN ) cos19.799° − (1.8 kN ) sin19.799° − F = 0 F = 6.26 kN

ΣFy′ = 0:

( 6 kN ) sin19.799° − (1.8 kN ) cos19.799° − V V = 0.3387 kN

ΣM J = 0:

19.80°

V = 339 N

( 0.48 m )(1.8 kN ) − ( 0.0864 m )( 6 kN ) − M M = 0.3456 kN ⋅ 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.

=0 70.2°

=0

M = 346 N ⋅ m

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Chapter 7, Solution 14.

FBD AB:

ΣΜ B = 0:

LP − hAx = 0, Ax =

L P h

Geometry:

h hx 2 y , = L2 L2

y = kx 2 , at B: h = kL2 , so k =

at J: y J = ka 2 =

slope =

FBD AJ:

ΣM J = 0:

ha 2 L2

dy = 2kx, dx

aP −

at J : slope =

2ha = tan −1 θ J 2 L

ha 2 LP −M =0 L2 h

 a2  M = P  a −  L  

dM =0 da a  or P 1 − 2  = 0, L 

To maximize set

Then M max

a=

L 2

2  L      L PL 2 = P −    = 2 L  4    

PL 4

at a =

L

2

M max =

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 7, Solution 15.

FBD Frame with pulley and cord:

ΣΜ Α = 0:

(1.8 m ) Bx − ( 2.6 m )( 360 N ) − ( 0.2 m )( 360 N ) = 0 B x = 560 N

FBD BE:

Note: Cord forces have been moved to pulley hub as per Problem 6.91.

ΣΜ Ε = 0:

(1.4 m )( 360 N ) + (1.8 m )( 560 N ) − ( 2.4 m ) By

=0

By = 630 N

FBD BJ:

ΣFx′ = 0:

F + 360 N −

3 4 ( 630 N − 360 N ) − ( 560 N ) = 0 5 5

F = 250 N ΣFy′ = 0:

V +

4 3 ( 630 N − 360 N ) − ( 560 N ) = 0 5 5

V = 120.0 N ΣM J = 0:

36.9°

53.1°

M + ( 0.6 m )( 360 N ) + (1.2 m )( 560 N ) − (1.6 m )( 630 N ) = 0

M = 120.0 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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 16.

FBD Frame with pulleys and cord: ΣM B = 0:

(1.8 m ) Ax − ( 2.0 m )( 360 N ) − ( 2.6 m )( 360 N ) = 0 A x = 920 N

FBD AE:

Note: Cord forces have been moved to pulley hub as per Problem 6.91. ΣM E = 0:

( 2.4 m ) Ay − (1.8 m )( 360 N ) = 0 A y = 270 N

FBD AK:

ΣFx = 0:

920 N − 360 N − F = 0

F = 560 N ΣFy = 0:

− 270 N + 360 N − V = 0

V = 90.0 N ΣM K = 0:

(1.6 m )( 270 N ) − (1.0 m )( 360 N ) − M

=0

M = 72.0 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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 17.

FBD Frame:

ΣM A = 0:

(1 m ) Dy − ( 2.675 m )( 360 N ) = 0 D y = 963 N

FBD DEF:

ΣM F = 0:

(1 m )( 963 N ) − ( 2.4 m ) Dx − ( 0.125 m )( 360 N ) = 0 D x = 3825 N

FBD DJ:

ΣFy′ = 0:

F−

12 5 ( 963 N ) − ( 382.5 N ) = 0 13 13

F = 1036 N

ΣFx′ = 0:

F = 1.036 kN

67.4°

5 12 ( 963 N ) − ( 382.5 N ) − V = 0 13 13

V = 17.31 N ΣM J = 0:

( 0.5 m )( 963 N ) − (1.2 m )( 382.5 N ) − M

22.6° =0

M = 22.5 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 7, Solution 18.

FBD AC:

Note: Cord forces moved to pulley hub as per Problem 6.91.

To determine θ the coordinates of C are xC = 2.55 m, yC =

4 ( 2.55 m ) = 3.40 m, and 3

xG = 0, yG = 3.75 m ∴ θ = tan −1

3.75 m − 3.40 m = 7.8153° 2.55 m

ΣM A = 0:

( 3.75 m ) ( 360 N ) cos 7.8153° − ( 2.55 m )( 360 N ) − ( 3 m ) FBG

=0

FBG = 139.820 N

FBD KC:

ΣFy′ = 0:

F−

4 4 ( 360 N ) − ( 360 N ) cos  7.8153° + tan −1  = 0 5 3 

F = 462.8 N ΣFx′ = 0:

− V + 139.820 N +

F = 463 N

53.1°

3 ( 360 N ) 5

4  − ( 360 N ) sin  7.8153° + tan −1  = 0 3 

V = 41.1 N ΣM K = 0:

− M − (1.5 m )(139.820 N ) −

36.9°

4 ( 2.75 m )( 360 N ) 5

4  + ( 2.75 m )( 360 N ) sin  7.8153° + tan −1  = 0 3 

M = 61.7 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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 19.

FBD Frame and pipe:

W = (10 ft )(18.5 lb/ft ) = 185 lb ΣM A = ( 24 in.) Cx − (12.6 in.)185 lb = 0 C x = 97.125 lb FBD pipe: By symmetry N E = N D = N

 21  2  N  − 185 N = 0, N = 127.738 lb  29 

ΣFy = 0: Also note

20  20 8  = in. a = r tan  tan −1  = ( 2.8 in.)

21 21 3  

FBD BC:

ΣM B = 0:

8   in.  (127.735 lb ) + (12 in.)( 97.125 lb ) 3  − (12.6 in.) C y = 0 C y = 119.534 lb

FBD CJ:

ΣFx′ = 0:

F −

21 20 ( 97.125 lb ) − (119.534 lb ) = 0 29 29

F = 152.8 lb ΣFy′ = 0:

−V −

20 21 ( 97.125 lb ) + (119.534 lb ) = 0 29 29

V = 19.58 lb ΣM C = 0:

46.4°

44.6°

M − ( 8.7 in.)(19.58 lb ) = 0 M = 170.3 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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 20.

FBD Frame:

W = (10 ft )(18.5 lb/ft ) = 185 lb ΣM C = 0:

( 24 in.) Ax − (12.6 in.)(185 lb ) = 0 A x = 97.125 lb

FBD Pipe

By symmetry N E = N D

ΣFy = 0: Also note FBD AD:

ΣM B = 0:

 21  2  N D  − 185 lb = 0,  29 

N D = 127.738 lb

20    20  8 a = r tan  tan −1  = ( 2.8 in.)   = in.

21    21  3

(12.6 in.)( 97.125 lb ) − (12.6 in.) Ay 8  −  in.  (127.738 lb ) = 0 3 

A y = 65.465 lb

FBD AK:

ΣFx′ = 0:

ΣFy′ = 0:

ΣM A = 0:

21 20 ( 97.125 lb ) + ( 65.465 lb ) − F = 0 29 29 F = 115.5 lb

44.6°

20 21 ( 97.125 lb ) − ( 65.465 lb ) + V = 0 29 29 V = 19.58 lb

46.4°

M − ( 8.7 in.)(19.58 lb ) = 0 M = 170.3 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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 21.

(a)

FBD Rod:

ΣFx = 0: ΣM D = 0:

Ax = 0

aP − 2aAy = 0

Ay =

P 2

ΣFx = 0: V = 0

FBD AJ:

ΣFy = 0:

P −F =0 2 F=

P 2

ΣM J = 0: M = 0

(b)

FBD Rod:

ΣM A = 0 4  3  2a  D  + 2a  D  − aP = 0 5  5  ΣFx = 0: ΣFy = 0:

Ax −

D=

4 5 P=0 5 14

Ay − P +

3 5 P=0 5 14

5P 14

Ax = Ay =

2P 7

11P 14 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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2 P −V = 0 7

ΣFx = 0: FBD AJ:

V = 11P −F =0 14

ΣFy = 0:

ΣM J = 0:

(c)

FBD Rod:

ΣM A = 0: ΣFx = 0: ΣFy = 0:

FBD AJ:

a

4 5P =0 5 2

Ay − P −

ΣFx = 0:

F=

11P 14

M =

2 aP 7

2P −M =0 7

a  4D    − aP = 0 2 5  Ax −

2P 7

3 5P =0 5 2

D=

5P 2

Ax = 2 P Ay =

5P 2

2P − V = 0

V = 2P ΣFy = 0:

5P −F =0 2

F= ΣM J = 0:

5P 2

a ( 2P ) − M = 0

M = 2aP

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 7, Solution 22.

(a)

FBD Rod:

ΣM D = 0:

aP − 2aA = 0 P 2

A= ΣFx = 0:

V −

P =0 2

V = FBD AJ:

ΣFy = 0:

ΣM J = 0:

F =0

M −a

P =0 2

M =

(b)

FBD Rod:

ΣM D = 0:

A=

P 2

aP −

aP 2

a4   A = 0 25 

5P 2

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

FBD AJ:

ΣFx = 0:

3 5P −V = 0 5 2

V = ΣFy = 0:

3P 2

4 5P −F =0 5 2

F = 2P M =

(c)

FBD Rod:

ΣM D = 0:

3  4  aP − 2a  A  − 2a  A  = 0 5   5  A=

ΣFx = 0:

5P 14

 3 5P  V − =0  5 14 

V = ΣFy = 0:

3 aP 2

3P 14

4 5P −F =0 5 14

F= ΣM J = 0:

2P 7

 3 5P  M − a =0  5 14 

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.

3 aP 14

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 23.

FBD Rod:

FBD JB:

ΣM A = 0:

Note: r =

2r   rB −  r − W =0 π  

r sin

φ

φ 2 =

2

x=

ΣM J = 0:

r sin 30°

2  B = 1 − W π 

60° 2 = 3r

π

180°

π

3r 1 3r r  3 − =  3−  2 2π 2 π

M +

r 2 r 3  2  1 −  W −  3 −  W  = 0 π π  3  2 2 M = 0.774Wr

on BK

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 7, Solution 24.

FBD Rod:

2r   1   2r  B − r − W =0 π   2  

ΣM A = 0:

1 1 B = 2  − W 2 π 

FBD JB:

 60°  r sin    2  = 3r r = π π 30° 180°

x = r sin 60° − r sin 30°

x=

1 3  3 − r π 2

21 1  r sin 60° + r (1 − cos 60° )  2  2 − π W  

ΣM J = 0:

−

1 3  2W −M =0  3 − r π 3 2 M = − 0.01085Wr

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.

M = 0.01085Wr

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 25.

FBD Rod:

ΣFy = 0:

Ay − W = 0 Ay = W

ΣM B = 0:

r ( Ax − W ) +

2r

π

W =0

2  A x = 1 − W π  FBD AJ:

W′ =

θ 2θ W = W π π 2

r =r

ΣM J = 0:

sin

θ

2 = 2r sin θ θ θ 2 2

2  r sin θ 1 − W − r (1 − cosθ )W π   2r  2θ θ θ W −M =0 +  sin  cos − r cosθ  θ 2 2    π 2θ   M = Wr  sin θ − 1 + cosθ − cosθ  π  

For Mmax,

dM 2 2θ   sin θ  = 0 = Wr  cosθ − sin θ − cosθ + dθ π π   or tan θ =

π −2 π − 2θ

Solving numerically:

θ = 0.48338 rod = 27.7° M max = 0.0777Wr

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 7, Solution 26.

FBD Rod:

2  1 1 2 rB − r 1 − W = 0 B = 2  − W π 2 π   1 1 1 1 1 2  − W = 0 A x =  − W ΣFx = 0: Ax − 2 π 2 2 π    1 1 1 1 1 2  − W = 0 A y =  + W ΣFy = 0: Ay − W + 2 2 π  2 π 

FBD AJ:

W′ =

ΣM A = 0:

θ 2θ W = W π π

2 2r θ r = sin θ 2 ΣM J = 0:

1 1 1 1 M + r sin θ  − W − r (1 − cosθ )  +  W π 2   2 π   2r   2θ  θ θ +  sin cos  − r cosθ   W  = 0 θ 2 2     π

 1 1   2θ M = Wr  +  (1 − sin θ − cosθ ) + cosθ  π  2 π   For M max ,  1 1   dM 2 2θ sin θ  = 0 = Wr  +  ( − cosθ + sin θ ) + cosθ − dθ π π  2 π   or tan θ =

π −2 π + 2 − 4θ

Solving numerically θ = 0.27539 rad = θ1 or θ = 1.16164 rad = θ 2 M (θ1 ) = − 0.0230Wr , M (θ 2 ) = 0.0362Wr so M max = 0.0362Wr at θ = θ 2 = 66.6°

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 7, Solution 27.

Note: α =

180° − 60° π = 60° = 2 3 r =

r

α

sin α =

3r

3 3 3 r = π 2 2π

Weight of section = W ΣFy′ = 0:

F−

ΣM O = 0:

120 4 = W 270 9

4 W cos30° = 0 9 rF − ( r sin 60° )

2 3 3 3 3 − M = r 2π 2  9

F =

2 3 W 9

4W −M =0 9

2 3 1  4 −  Wr W =  π  9   9

M = 0.0666Wr

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 7, Solution 28.

ΣFx = 0:

FBD Rod: ΣM B = 0:

FBD AJ:

rAy +

2r W 2r 2W − =0 π 3 π 3

Ay =

2W 3π

α =

Note:

Ax = 0

60° π = 30° = 2 6

Weight of segment = W F = ΣM J = 0: M =

2W 9

r

α

sin α =

( r cosα − r sin 30° )

60 2W = 270 9

r 3r sin 30° = π /6 π

2W 2W + ( r − r sin 30° ) −M =0 9 3π

 3r 3 r  3 1 3r  1  − + − +   = Wr   2 2π  9 3π  π 2  3π

M = 0.1788Wr

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 7, Solution 29.

(a) FBD Beam:

ΣM C = 0:

ΣFy = 0:

LAy − M 0 = 0

Ay =

M0 L

C=

M0 L

− Ay + C = 0

Along AB:

ΣFy = 0:

−

ΣM J = 0:

M0 −V = 0 L x

V =−

M0 +M =0 L

M0 L

M =−

straight with M = −

M0 at B 2

M0 −V = 0 L

V =−

M0 x L

Along BC:

ΣFy = 0: ΣM K = 0:

−

M +x

x  M = M 0 1 −  L  

M0 − M0 = 0 L

straight with M = (b) From diagrams:

M0 L

M0 at B 2 V

max

M = 0 at C =

M0 everywhere L

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.

max

=

M0 at B 2

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Chapter 7, Solution 30.

FBD Beam:

(a)

ΣFx = 0:

Ax = 0, by symmetry Ay = By = P

ΣFy = 0:

Along AB:

P −V = 0

V = P ΣM J = 0:

M − xP = 0

M = Px

Along BC:

ΣFy = 0:

P − P −V = 0

V =0

ΣM K = 0:

M − xP + ( x − a ) P = 0

M = Pa

Along CD:

ΣFy = 0:

P − P − P −V = 0 V = −P

ΣM L = 0:

M − xP + ( x − a ) P + ( x − L + a) P = 0

M = P ( L − x) Note: Symmetry in M diag. follows symmetry of FBD (b)

V

max

= P along AB and CD 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.

max

= Pa along BC

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 31.

FBD Section:

ΣFy = 0: − V −

1  x x  w0  = 0 2  L

V =−

(a)

1 w0 2 x 2 L

1 V ( L ) = − w0 L 2 ΣM J = 0: M +

1 1  x  x  x  w0   = 0 3  2  L 

M =− M ( L) = −

1 w0 3 x 6 L 1 w0 L2 6

(b)

V 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.

max

max

= =

1 w0 L at B 2 1 w0 L2 at B 6

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 32.

(a) FBD Beam:

ΣFx = 0:

ΣM B = 0:

Bx = 0

aP + 2aC − 3.5aP = 0 C = 1.25P

ΣFy = 0:

− P + By + 1.25P − P = 0 By = 0.75P

Along AB:

ΣFy = 0: ΣM J = 0:

−P − V = 0 M + xP = 0

V = −P M = − Px

Along BC:

ΣFy = 0: ΣM K = 0:

− P + 0.75P − V = 0

V = − 0.25P

M + xP − ( x − a )( 0.75P ) = 0 M ( 3a ) = −1.5Pa

M = − 0.75Pa − 0.25Px Along CD:

ΣFy = 0: ΣM L = 0:

V −P=0 − M − x1 P = 0

V =P M = − Px1 M (1.5a ) = −1.5Pa V

(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.

max

= P along AB and CD

M

max

= 1.5Pa at C

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 33.

(a) FBD Beam:

ΣFx = 0: ΣM C = 0:

Cx = 0

( 3.6 ft )(1 kip ) − ( 3 ft )( 4 kips ) + ( 6 ft )( 2 kips ) − ( 9.6 ft ) B = 0 B = 0.375 kip

ΣFy = 0: −1 kip + C y − 4 kips + 2 kips − 0.375 kip = 0 C y = 3.375 kips

Along AC:

ΣFy = 0:

−1 kip − V = 0

V = −1 kip ΣM J = 0:

M + x (1 kip ) = 0

M = − (1 kip ) x M ( 3.6 ft ) = 3.6 kip ⋅ ft

Along CD:

ΣFy = 0:

−1 kip + 3.375 kips − V = 0 V = 2.375 kips

ΣM K = 0:

M + x (1 kip ) − ( x − 3.6 ft )( 3.375 kips ) = 0 M = −12.15 kip ⋅ ft + ( 2.375 kips ) x M ( 6.6 ft ) = 3.525 kip ⋅ ft 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

Along EB:

ΣFy = 0:

V − 0.375 kips = 0 V = 0.375 kips

ΣM L = 0:

− M − x1 ( 0.375 kips ) = 0

M = − ( 0.375 kips ) x1

M ( 3.6 ft ) = −1.35 kip ⋅ ft Along DE:

ΣFy = 0:

V + 2 kips − 0.375 kips = 0

V = −1.625 kips Also M is linear (b)

V

max

= 2.38 kips along CD

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.

max

= 3.60 kips at C

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 34.

(a)

FBD Beam:

ΣM B = 0:

(.6 ft )( 4 kips ) + ( 5.1 ft )(8 kips ) + ( 7.8 ft )(10 kips ) − ( 9.6 ft ) Ay

=0

A y = 12.625 kips

ΣFy = 0: 12.625 kips − 10 kips − 8 kips − 4 kips + B = 0 B = 9.375 kips Along AC:

ΣFy = 0:

12.625 kips − V = 0

V = 12.625 kips ΣM J = 0:

M − x (12.625 kips ) = 0

M = (12.625 kips ) x M = 22.725 kip ⋅ ft at C Along CD:

ΣFy = 0: 12.625 kips − 10 kips − V = 0 V = 2.625 kips ΣM K = 0:

M + ( x − 1.8 ft )(10 kips ) − x (12.625 kips ) = 0 M = 18 kip ⋅ ft + ( 2.625 kips ) x M = 29.8125 kip ⋅ ft at D ( x = 4.5 ft ) 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

Along DE:

ΣFy = 0:

(12.625 − 10 − 8) kips − V

=0

V = −5.375 kips

ΣM L = 0: M + x1 ( 8 kips ) + ( 2.7 ft + x1 )(10 kips ) − ( 4.5 ft + x1 )(12.625 kips ) = 0 M = 29.8125 kip ⋅ ft − ( 5.375 kips ) x1 M = 5.625 kip ⋅ ft at E

( x1 = 4.5 ft )

Along EB:

ΣFy = 0:

V + 9.375 kips = 0

ΣM N = 0:

V = 9.375 kips

x2 ( 9.375 kip ) − M = 0

M = ( 9.375 kips ) x2 M = 5.625 kip ⋅ ft at E From diagrams: (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.

V

max

M

max

= 12.63 kips on AC = 29.8 kip ⋅ ft at D

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 35.

(a)

ΣM E = 0:

FBD Beam:

(1.1 m )( 540 N ) − ( 0.9 m ) C y + ( 0.4 m )(1350 N ) − ( 0.3 m )( 540 N ) = 0 C y = 1080 N

ΣFy = 0:

− 540 N + 1080 N − 1350 N

−540 N + E y = 0

E y = 1350 N

Along AC:

ΣFy = 0:

− 540 N − V = 0

V = −540 N ΣM J = 0:

x ( 540 N ) + M = 0

M = − ( 540 N ) x

Along CD:

ΣFy = 0: ΣM K = 0:

− 540 N + 1080 N − V = 0

V = 540 N

M + ( 0.2 m + x1 )( 540 N ) − x1 (1080 N ) = 0 M = −108 N ⋅ m + ( 540 N ) x1 M = 162 N ⋅ m at D ( x1 = 0.5 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

Along DE:

ΣFy = 0: ΣM N = 0:

V + 1350 N − 540 N = 0

V = −810 N

M + ( x3 + 0.3 m ) ( 540 N ) − x3 (1350 N ) = 0

M = −162 N ⋅ m + ( 810 N ) x3 M = 162 N ⋅ m at D ( x3 = 0.4 ) Along EB:

(b)

ΣFy = 0: ΣM L = 0:

V − 540 N = 0

V = 540 N

M + x2 ( 540 N ) = 0

M = −540 N x2

( x2

M = −162 N ⋅ m at E

= 0.3 m ) V

From diagrams 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.

max

max

= 810 N on DE

= 162.0 N ⋅ m at D and E

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 36.

(a) FBD Beam: a =1m

ΣFx = 0:

ΣFy = 0:

Bx = 0

− 1.5 kN + 2 kN − 4 kN + 5 kN − By = 0 By = 1.5 kN

ΣM B = 0:

a  4 (1.5 kN ) − 3 ( 2 kN ) + 2 ( 4 kN ) − 1( 5 kN )  − M B = 0

M B = ( 3 kN ) a = 3 kN ⋅ m

Along AC:

ΣFy = 0: ΣM J = 0: Along CD:

ΣFy = 0: ΣM K = 0:

−1.5 kN − V = 0

V = −1.5 kN

M − x (1.5 kN ) = 0

M = − (1.5 kN ) x

M (1 m ) = −1.5 kN ⋅ m

−1.5 kN + 2 kN − V = 0

V = 0.5 kN

M + x (1.5 kN ) − ( x − 1 m )( 2 kN ) = 0

M = − 2 kN ⋅ m + ( 0.5 kN ) x

M ( 2 m ) = −1 kN ⋅ 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

Along EB:

ΣFy = 0: ΣM L = 0:

V − 1.5 kN = 0

V = 1.5 kN

− M − x1 (1.5 kN ) − 3 kN ⋅ m = 0 M = − 3 kN ⋅ m − (1.5 kN ) x1,

M (1 m ) = − 4.5 kN ⋅ m

Along DE:

ΣFy = 0:

V + 5 kN − 1.5 kN = 0

V = − 3.5 kN

Also M is linear here V (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.

max

M

= 3.50 kN along DE

max

= 4.50 kN ⋅ m at E

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 37.

(a) FBD Beam:

ΣM A = 0: − (1.3 m ) (1.8 kN/m )( 2.6 m )  − (1.6 m )( 4 kN ) + ( 4 m ) B = 0 B = 3.121 kN

ΣFy = 0:

Ay − (1.8 kN/m )( 2.6 m ) − 4 kN + 3.121 kN = 0 A y = 5.559 kN

Along AC:

ΣFy = 0:

5.559 kN − (1.8 kN/m ) x − V = 0 V = 5.559 kN − (1.8 kN/m ) x

ΣM J = 0:

M +

x (1.8 km ) x  − x ( 5.559 kN ) = 0 2 M = ( 5.559 kN ) x − ( 0.9 kN/m ) x 2

Along CD:

ΣFy = 0:

5.559 kN − x (1.8 kN/m ) − 4 kN − V = 0 V = (1.559 kN ) − (1.8 kN/m ) x

ΣM K = 0: M + ( x − 1.6 m )( 4 kN ) +

x (1.8 kN/m ) x  − x ( 5.559 kN ) = 0 2

M = 6.4 kN ⋅ m + (1.559 kN ) x − ( 0.9 kN/m ) x 2 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

Along DB:

ΣFy = 0:

V + 3.121 kN = 0 V = − 3.121 kN

(b)

ΣM L = 0:

− M + x1 ( 3.121 kN ) = 0

M = ( 3.121 kN ) x1 V 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.

max

max

= 5.56 kN at A

= 6.59 kN ⋅ m at C

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 38.

(a) FBD Beam:

ΣFx = 0: ΣFy = 0:

Ax = 0 Ay + ( 2 m )( 24 kN/m ) − 48 kN − 8 kN = 0 A y = 8 kN

ΣM A = 0:

M A + (1 m )( 2 m )( 24 kN/m ) − ( 3.5 m )( 48 kN ) − ( 2 m )( 8 kN ) = 0,

M A = 152 kN ⋅ m

Along AC:

ΣFy = 0:

8 kN + x ( 24 kN ⋅ m ) − V = 0 V = 8 kN + ( 24 kN/m ) x

ΣM J = 0:

M + 152 kN ⋅ m − x ( 8 kN ) −

x ( 24 kN/m ) x = 0 2

M = (12 kN/m ) x 2 + ( 8 kN ) x − 152 kN ⋅ m

Along DB:

ΣFy = 0:

V − 8 kN = 0 V = 8 kN

ΣM K = 0:

M + x1 ( 8 kN ) = 0,

M = − ( 8 kN ) x1

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

Along CD:

ΣFy = 0: ΣM L = 0:

V − 48 kN − 8 kN = 0,

V = 56 kN

M + ( x1 − 0.5 m )( 48 kN ) + x1 ( 8 kN ) = 0 M = 24 kN ⋅ m − ( 56 kN ) x1

(b)

V

max

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.

= 56.0 kN along CD

max

= 152.0 kN ⋅ m at A

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 39.

(a) FBD Beam:

by symmetry, C y = Gy =

Cx = 0, and

1  2 (12 lb/in )(10 in ) + 2 (100 lb ) + (150 lb )  2

C y = G y = 295 lb Along AC:

ΣFy = 0:

− (12 lb/in.) x − V = 0  lb  V = − 12 x  in. 

ΣM J = 0:

M +

x (12 lb/in.) x = 0, 2

 lb  M = −  6  x2  in. 

Along CD:

ΣFy = 0:

− (12 lb/in.)(10 in.) + 295 lb − V = 0 V = 175 lb

ΣM K = 0:

M + ( x − 5 in.)(12 lb/in.)(10 in.) − ( x − 10 in.)( 295 lb ) = 0 M = − 2350 lb ⋅ in. + (175 lb ) x 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

Along DE:

ΣFy = 0: ΣM N = 0:

− (12 lb/in.)(10 in.) + 295lb − 100 lb − V = 0,

V = 75 lb

M + ( x − 16 in.)(100 lb ) − ( x − 10 in.)( 295 lb ) + ( x − 5 in.)(12 lb/in.)(10 in.) = 0 M = − 750 lb ⋅ in + ( 75 lb ) x

Complete diagrams using symmetry.

V (b)

max

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.

= 175.0 lb along CD and FG M

max

= 900 lb ⋅ in. at E

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 40.

(a) FBD Beam:

ΣM D = 0:

( 6 ft )(1 kip/ft )( 6 ft ) − ( 7.5 ft )( 2 kips/ft )( 9 ft ) − (12 ft ) Fy + (15 ft )( 33 kips ) = 0

Fy = 33 kips ΣFy = 0:

− (1 kip/ft )( 6 ft ) + Dy − ( 2 kips/ft )( 9 ft ) − 33 kips + 33 kips = 0

D y = 24 kips Along AC:

ΣFy = 0: ΣM J = 0:

− (1 kip/ft ) x − V = 0, M +

V = − (1 kip/ft ) x

x (1 kip ⋅ ft ) x = 0, 2

1  M = −  kip/ft  x 2 2 

Along CD:

ΣFy = 0: ΣM K = 0:

− (1 kip/ft )( 6 ft ) − V = 0,

V = − 6 kips

M + ( x − 3 ft )(1 kip/ft )( 6 ft ) = 0 M = 18 kip ⋅ ft − ( 6 kips ) x 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

Along DE:

ΣFy = 0: ΣM L = 0:

− 6 kips + 24 kips − V = 0,

V = 18 kips

( x − 3 ft )( 6 kips ) − ( x − 9 ft )( 24 kips ) + M

=0

M = −198 kip ⋅ ft + (18 kips ) x

Along FB:

ΣFy = 0: ΣM N = 0:

V + 33 kips = 0,

V = − 33 kips

x1 ( 33 kips ) − M = 0,

M = ( 33 kips ) x1

Along EF:

ΣFy = 0: V − ( 2 kips/ft ) x2 − 33 kips + 33 kips = 0 V = ( 2 kips/ft ) x2 ΣM O = 0:

M +

x2 ( 2 kips/ft ) x2 − ( 3 ft ) ( 33 kips ) = 0 2

M = 99 kip ⋅ ft − (1 kip/ft ) x22 V (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.

max

M

= 33.0 kips along FB max

= 99.0 kip ⋅ ft at F

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 41.

(a) FBD Beam:

( 4 m )( w) − ( 2 m )(12 kN/m ) = 0

ΣFy = 0:

w = 6 kN/m

Along AC:

− x ( 6 kN/m ) − V = 0,

ΣFy = 0:

V = − ( 6 kN/m ) x

V = −6 kN at C ( x = 1 m ) x ( 6 kN/m )( x ) = 0 2 M = − ( 3 kN/m ) x 2 M = −3 kN ⋅ m at C

ΣM J = 0:

M +

Along CD:

ΣFy = 0:

− (1 m )( 6 kN/m ) + x1 ( 6 kN/m ) − V = 0 V = ( 6 kN/m )(1 m − x1 ) ,

ΣM K = 0:

V = 0 at x1 = 1 m

x1 ( 6 kN/m ) x1 = 0 2 M = −3 kN ⋅ m − ( 6 kN ) x1 + ( 3 kN/m ) x12 M + ( 0.5 m + x1 )( 6 kN/m )(1 m ) −

M = −6 kN ⋅ m at center Finish by symmetry. (b) From diagrams:

V

( x1 = 1 m ) max

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.

= 6.00 kN at C and D

max

= 6.00 kN at center

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 42.

(a) FBD Beam:

ΣFy = 0:

(12 m ) w − ( 6 m )( 3 kN/m ) = 0 w = 1.5 kN/m

Along AC:

x (1.5 kN/m ) − V = 0,

ΣFy = 0:

V = (1.5 kN/m ) x

V = 4.5 kN at C ΣM J = 0:

M −

x (1.5 kN/m )( x ) = 0 2

M = ( 0.75 kN/m ) x 2 ,

M = 6.75 N ⋅ m at C

Along CD:

x (1.5 kN/m ) − ( x − 3 m )( 3 kN/m ) − V = 0

ΣFy = 0:

V = 9 kN − (1.5 kN/m ) x, ΣM K = 0:

V = 0 at x = 6 m

x  x − 3m M +  ( 3 kN/m )( x − 3 m ) − (1.5 kN/m ) x = 0 2 2   M = −13.5 kN ⋅ m + ( 9 kN ) x − ( 0.75 kN/m ) x 2 M = 13.5 kN ⋅ m at center

( x = 6 m)

Finish by symmetry. (b) From diagrams:

V 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.

max

max

= 4.50 kN at C and D

= 13.50 kN ⋅ m at center

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 43.

(a) FBD Beam:

ΣFy = 0:

(8 m ) w − 2 ( 6 kN ) − ( 4 m )( 5 kN/m ) = 0 w = 4 kN/m

Along AC:

ΣFy = 0:

( 4 kN/m ) x − V

=0

V = ( 4 kN/m ) x ΣM J = 0: M −

x ( 4 kN/m ) x = 0, 2

M = ( 2 kN/m ) x 2

Along CD:

( 4 kN/m ) x − 6 kN − V

ΣFy = 0:

=0

V = ( 4 kN/m ) x − 6 kN ΣM K = 0:

M + ( x − 1 m )( 6 kN ) −

x ( 4 kN/m ) x = 0 2

M = ( 2 kN/m ) x 2 − ( 6 kN ) x + 6 kN ⋅ m Note: V = 0 at x = 1.5 m where M = 1.5 kN/m Along DE:

ΣFy = 0:

( 4 kN/m )( 2 m ) − 6 kN − (1 kN/m )( x − 2 m ) − V V = 4 kN − (1 kN/m ) x

=0

 x − 2 m ΣM L = 0: M +   (1 kN/m )( x − 2 m ) + ( x − 1 m ) 6 kN 2   − ( x − 1 m )( 4 kN/m )( 2 m ) = 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

1  M = −  kN/m  x 2 + ( 4 kN ) x − 4 kN ⋅ m 2   Note: V = 0 at x = 4 m, where M = 4 kN ⋅ m Complete diagrams using symmetry. (b)

V 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.

max

max

= 4 kN at C and F = 4 kN ⋅ m at center

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 44.

(a) FBD Beam:

ΣFy = 0:

w (1.5 m ) − 2 ( 3.6 kN ) = 0 w = 4.8 kN/m

Along AC:

ΣFy = 0: ΣM J = 0:

( 4.8 kN/m ) x − V M −

= 0,

V = ( 4.8 kN/m ) x

x ( 4.8 kN/m ) x = 0, 2

M = ( 2.4 kN/m ) x 2

Along CD:

ΣFy = 0:

( 4.8 kN/m ) x − 3.6 kN − V

=0

V = ( 4.8 kN/m ) x − 3.6 kN ΣM K = 0:

M + ( x − 0.3 m )( 3.6 kN ) −

x ( 4.8 kN/m ) x = 0 2

M = 1.08 kN ⋅ m − ( 3.6 kN ) x + ( 2.4 kN/m ) x 2 Note: V = 0 at x = 0.75 m, where M = − 0.27 kN ⋅ m Complete diagrams using symmetry.

V

max

M (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.

= 2.16 kN at C and D

max

= 270 N ⋅ m at center

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 45.

FBD CE:

ΣFx = 0:

Cy = 0

ΣFy = 0:

C y − 4 kN = 0

C y = 4 kN

ΣM C = 0:

M C − ( 0.5 m )( 4 kN ) = 0 M C = 2 kN ⋅ m

ΣFx = 0:

Beam AB:

ΣFy = 0: ΣM A = 0:

Ax = 0 Ay − 4 kN − 2 kN − 1 kΝ = 0

M A − 2 kN ⋅ m − ( 0.5 m )( 4 kN ) − (1 m )( 2 kN ) − (1.5 m )(1 kN ) = 0,

Along AC:

ΣFy = 0:

7 kN − V = 0

V = 7 kN ΣM J = 0:

Ay = 7 kN

M + 7.5 kN ⋅ m − x ( 7 kN ) = 0 M = ( 7 kN ) x − 7.5 kN ⋅ 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.

M A = 7.5 kN ⋅ m

COSMOS: Complete Online Solutions Manual Organization System

Along DB:

ΣFy = 0:

V − 1 kN = 0 V = 1 kN

ΣM K = 0:

− M + x1 (1 kN ) = 0 M = (1 kN ) x1

Along CD:

ΣFy = 0: ΣM M = 0:

V − 2 kN − 1 kN = 0

V = 3 kN

M + ( x1 − 0.5 m )( 2 kN ) + x1 (1 kN ) = 0 M = 1 kN ⋅ m − ( 3 kN ) x1

Note: M exhibits a discontinuity at C, equal to 2 kN ⋅ m, the value of MC. From the diagrams,

V

max

= 7.00 kN along AC

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.

max

= 7.50 kN at A

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 46.

FBD CE or DF:

ΣFx = 0: Cx , Dx = 0 ΣFy = 0: C y − 750 N = 0,

C y = 750 N Dy = 750 N

ΣM C = 0: M C − ( 0.3 m )( 750 N ) = 0 M C = 225 N ⋅ m = M D Beam AB:

ΣM A = 0:

( 0.9 m ) Dy

− 2 ( 225 N ⋅ m ) − ( 0.3 m )( 750 N )

− ( 0.9 m )( 750 N ) − (1.2 m )( 540 N ) = 0 Dy = 2220 N ΣFy = 0: Ay − 2 ( 750 N ) − 540 N + 2220 N = 0 Ay = −180 N

A y = 180 N

ΣFy = 0: −180 N − V = 0

Along AC:

V = −180 N ΣM J = 0: M + x (180 N ) = 0

M = (180 N ) x

Along CD:

ΣFy = 0: −180 N − 750 N − V = 0, ΣM K = 0:

V = − 930 N

M − 225 N ⋅ m + ( x − 0.3 m )( 750 N ) + x (180 N ) = 0

M = 450 N ⋅ m − ( 930 N ) x

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

Along DB:

ΣFy = 0: V − 540 N = 0 V = 540 N ΣM N = 0: M + x1 ( 540 N ) = 0

M = − ( 540 N ) x1

Note: The discontinuities in M, at C and D, equal 225 N ⋅ m, M C and M D

From the diagrams

V

max

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.

= 930 N along CD

max

= 387 N ⋅ m at D

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 47.

FBD Angle:

ΣFy = 0:

T − Cy = 0 Cy = T

ΣM C = 0:

( 0.3 ft ) T

M C = ( 0.3 ft ) T

− M C = 0,

By symmetry, Dy = T , M D = ( 0.3 ft ) T ΣFy = 0:

2T − 810 lb − (100 lb/ft )( 9 ft ) = 0 T = 855 lb.

(a) Beam AB:

From above M C = M D = ( 0.3 ft )( 855 lb ) = 256.5 lb ⋅ ft ΣFy = 0: − x (100 lb/ft ) − V = 0

Along AC:

V = − (100 lb/ft ) x ΣM J = 0: M +

x (100 lb/ft ) x = 0, 2

M = − ( 50 lb/ft ) x 2

M C = 256.5 lb ⋅ ft

Along CI:

ΣFy = 0:

− (100 lb/ft ) x + 855 lb − V = 0

V = 855 lb − (100 lb/ft ) x

ΣM K = 0:

x (100 lb/ft ) x 2

M − 256.5 lb ⋅ ft +

− ( x − 3.6 ft )( 855 lb ) = 0 M = − ( 50 lb/ft ) x 2 + ( 855 lb ) x − 2821.5 lb ⋅ ft Complete diagrams using symmetry Note: Discontinuities in M, at C and D, equal M C and M D

(b)

V 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.

max

max

= 495 lb at C and D

= 648 lb ⋅ ft at C and D

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 48.

FBD Angle:

ΣFy = 0: T − C y = 0

Cy = T

ΣM C = 0: M C − ( 0.3 ft ) T = 0,

M C = ( 0.3 ft ) T

By symmetry, Dy = T and M D = ( 0.3 ft ) T

(a) Beam AB:

ΣFy = 0: 2T − (100 lb/ft )( 9 ft ) − 810 lb = 0

T = 855 lb

From above, M C = M D = ( 0.3 ft )( 855 lb ) = 256.5 lb ⋅ ft

ΣFy = 0: − (100 lb/ft ) x − V = 0

Along AC:

V = − (100 lb/ft ) x ΣM J = 0:

M +

x (100 lb/ft ) x = 0 2

M C = 256.5 lb ⋅ ft

Along CI:

ΣFy = 0:

M = ( 50 lb/ft ) x 2

855 lb − (100 lb/ft ) x − V = 0

V = 855 lb − (100 lb/ft ) x ΣM K = 0: M − 256.5 lb ⋅ ft +

x (100 lb/ft ) x 2

− ( x − 2.7 ft )( 855 lb ) = 0

M = − ( 50 lb/ft ) x 2 + ( 855 lb ) x − 2052 lb ⋅ ft Complete diagrams using symmetry Note: The discontinuities in M, at C and D, equal M C and M D (b)

V

max

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.

= 585 lb at C and D max

= 783 lb ⋅ ft at I

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 49.

Note: D passes through C, so FBD Whole:

Dy Dx

0.35 m = 0.7 0.5 m

=

ΣM H = 0: ( 0.65 m )( 200 N ) − ( 0.5 m ) ( 0.7 Dx ) + ( 0.1 m ) ( Dx ) + ( 0.25 m )( 400 N ) − ( 0.15 m )( 200 N ) = 0 D x = 800 N

D y = 560 N ΣFx = 0: 800 N − H x = 0

ΣFy = 0:

H x = 800 N

560 N − 400 N − 2 ( 200 N ) + H y = 0

H y = 240 N Beam AB with forces at D & H replaced by forces and couples at E and G. Horizontal forces not shown to avoid clutter.

ΣFy = 0:

Along AE:

− 200 N − V = 0

V = − 200 N ΣM J = 0:

x ( 200 N ) + M = 0

M = − ( 200 N ) x Along EF:

ΣFy = 0:

− 200 N + 560 N − V = 0

V = 360 N ΣM K = 0: M − 80 N ⋅ m + x ( 200 N ) − ( x − 0.15 m )( 560 N ) = 0

M = ( 360 N ) x − 4 N ⋅ m Along GB:

ΣFy = 0: V − 400 N = 0

V = 200 N ΣM L = 0:

M + x1 ( 200 N ) = 0

M = ( 200 N ) x1 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

ΣFy = 0: V + 240 N − 200 N = 0

Along FG:

V = − 40 N

(

)

M − 160 N ⋅ m + x1 − 0.15 m ( 240 N ) − x1 ( 200 ) N = 0

ΣM N = 0:

M = 124 N ⋅ m + ( 40 N ) x1 From diagrams,

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.

V

max

= 360 N along EF

M

max

= 140.0 N ⋅ m at F

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 50.

FBD AB + Pulley & Cord:

ΣM A = 0:

( 48 in.) 

5   12  D  + ( 20 in.)  D  − (100 in.)(120 lb ) = 0 13    13 

D = 325 lb

so D x = 300 lb

ΣFy = 0: − Ay + 125 lb − 120 lb = 0

, D y = 125 lb

A y = 5 lb

Neglecting the diameter of pulley G, the cord EG has slope 3/4, and tension 120 lb, E x = 96 lb

(a)

E y = 72 lb

,

Beam AB with forces at D and G replaced by forces and couples at E and F. Horizontal forces are omitted to avoid clutter.

Along AE: ΣFy = 0:

− 5 lb − V = 0,

V = − 5 lb

ΣM J = 0: x ( 5 lb ) + M = 0,

M = ( − 5 lb ) x

Along EF:

ΣFy = 0: ΣM K = 0:

− 5 lb + 197 lb − V = 0,

V = 192 lb

M + 6000 lb ⋅ in. + x ( 5 lb ) − ( x − 48 in.)(197 lb ) = 0 M = (192 lb ) x − 15456 lb ⋅ in.

Along FB: ΣFy = 0: V − 120 lb = 0, V = 120 lb ΣM L = 0: − M − x1 (120 lb ) = 0 M = − (120 lb ) x1

(b)

From diagrams,

V M

max

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.

max

= 192.0 lb along EF

= 6240 lb ⋅ in. = 520 lb ⋅ ft at E

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 51.

ΣFy = 0: Lw − 2P = 0

w=

2P L

Along AC:

ΣM J = 0: M −

x  2P   x = 0 2 L 

M =

P 2 x L

Along CD:

ΣM K = 0: M + ( x − a ) P − M =

x  2P   x = 0 2 L  P 2 x − Px + Pa L

Complete diagram using symmetry Note: M min = Pa −

1 L (center) PL at x = 2 4

Setting M max = − M min : Solving a = −

P 2 PL L2 a = − Pa + or a 2 + La − =0 4 L 4

L 2 ± L, positive root a = 2 2

(a)

Then, with L = 1.5 m,

a = 0.31066 m

(b)

and, with P = 3.6 kN,

M max =

Pa 2 L

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.

2 −1 L 2 a = 0.311 m

M

max

= 232 N ⋅ m

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 52.

ΣFy = 0: T − C y = 0

FBD Angle CE:

Cy = T

ΣM C = 0: M C − ( 0.3 ft ) T = 0, by symmetry D y = T

M C = ( 0.3 ft ) T

and M D = ( 0.3 ft ) T

ΣFy = 0: 2T − 810 lb − ( 9 ft )(100 lb/ft ) = 0

Beam AB:

T = 855 lb

From above M C = M D = 256.5 lb ⋅ ft ΣM J = 0: M −

Along AC:

x (100 lb/ft ) x = 0 2

M = − ( 50 lb/ft ) x 2

Along CI: M C = 256.5 lb ⋅ ft ΣM K = 0:

x (100 lb/ft ) x 2 − ( x − 4.5 ft + a )( 855 lb ) = 0

M − 256.5 lb ⋅ ft +

M =  − 50 x 2 + 855 ( x + a ) − 3591 lb ⋅ ft with a in ft Complete M diagram using symmetry At x = ( 4.5 − a ) ft, M min = − 50 ( 4.5 − a ) lb ⋅ ft 2

At x = 4.5 ft, M max = ( 855 a − 756 ) lb ⋅ ft Setting M max = − M min : a 2 − 26.1 a + 35.37 = 0 (a)

Solving: a = 13.05 ± 11.6160,

a < 4.5 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.

giving M

a = 1.434 ft max

= 470 lb ⋅ ft

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 53.

Replacing the 1500 N force with equivalent force and couple at D,

M C = 0:

(1.25 m ) P − (1.2 m )(1500 N ) − 1800 N ⋅ m + ( 2.4 m ) E y − ( 3.65 m ) 2P = 0

assume P in N: E y = (1500 + 2.5208 P ) N

ΣFy = 0:

C y − P − 2P − 1500 + (1500 + 2.5208 P ) = 0

C y = 0.47917 P ΣM J = 0: M + x P = 0

Along AC:

M = − Px ΣM K = 0:

Along CD:

M + x P − ( x − 1.25 m )( 0.47917 P ) = 0 M = − 0.5208 Px − 0.5990 P at x = 2.45− (left of D), M = −1.875 P at x = 2.45+ (right of D), M = 1800 − 1.875 P

ΣM E = 0: − M − (1.25 m )( 2 P ) = 0

At E:

M = − 2.5 P (a)

Setting M max = − M min : 1800 − 1.875 P = 2.5 P

P = 411.43 P = 411 N

(b)

M max = 2.5 P

M max = 1029 N ⋅ m

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.

max

= 1.029 kN ⋅ m

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 54.

Since there are no distributed loads, M is piecewise linear, and only pts A, C, and D need be considered.

At A:

ΣM A = 0: M + ( 0.75 m )( 4 kN ) + (1.75 m )(16 kN ) − (1.75 m + a )( 8 kN ) = 0 (With a in m) M = ( 8 a − 17 ) kN ⋅ m

At C:

ΣM C = M + (1 m )(16 kN ) − (1 m + a m )( 8 kN ) = 0 M = ( 8 a − 8 ) kN ⋅ m ΣM D = 0: M − ( a m )( 8 kN ) = 0

At D:

M = 8 a kN ⋅ m

(a)

Apparently M max = 8 a kN ⋅ m at D, and M min = ( 8 a − 17 ) kN ⋅ m at A Setting M max = − M min : 8 a = 17 − 8 a

a=

17 m 16 a = 1.063 m

(b)

and M max = 8 a =

17 kN ⋅ m 2

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.

M

max

= 8.50 kN ⋅ m

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 55.

ΣM A = 0: a Dy − ( 5 ft )( 500 lb ) − (10 ft )( 500 lb ) = 0 Dy =

7500 lb ⋅ ft a

Since there are no distributed loads, M is piecewise linear, so only points C and D need be considered. Assume a in ft. ΣM D = 0:

At D:

M + (10 − a ) ft  ( 500 lb ) = 0

M D = − 500 (10 − a ) lb ⋅ ft At C:

7500 lb = 0 ΣM C = 0: M + ( 5ft )( 500 lb ) − 5 − (10 − a )  ft a 37500   M C =  5000 −  lb ⋅ ft a   (a)

Apparently M max = M C and M min = M D (recall 5 < a < 10 ) Setting M C = − M D : 5000 −

37500 = 5000 − 500 a a 37500 = 500 a 2 a = 75

(b)

M

max

(

)

= 500 10 − 75 lb ⋅ ft

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 = 8.66 ft M

max

= 670 lb ⋅ ft

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 56.

ΣM A = 0: aDy − ( 5 ft )( 250 lb ) − (10 ft )( 500 lb ) = 0

Dy = With a in ft, Dy =

6250 lb ⋅ ft a

6250 lb a

Since there are no distributed loads, M is piecewise linear, so only points C and D need be considered.

ΣM D = 0:

At D:

(10 − a ) ft  ( 500 lb ) + M D = 0

M D = − 500 (10 − a ) lb ⋅ ft At C:

ΣM C = 0:

 6250  M C + ( 5 ft )( 500 lb ) − 5 − (10 − a )  ft  lb  = 0  a 

31250   M C =  3750 −  lb ⋅ ft a   (a)

Apparently M max is M C and M min is M D Setting M C = − M D : 3750 −

( 5 < a < 10 )

31250 = − 500 a + 5000 a

500 a 2 − 1250 − 31250 = 0 or a 2 − 2.5 a − 62.5 = 0 a = 1.25 ± 8.004, positive root a = 9.254 ft

a = 9.25 ft (b)

M

max

= 500 (10 − a ) lb ⋅ ft

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.

M

max

= 373 lb ⋅ ft

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 57.

ΣM J = 0: − M −

M due to distributed load:

x wx = 0 2 1 M = − wx 2 2

ΣM J = 0: − M + xw = 0

M due to counter weight:

(a) Both applied:

M = wx

M = Wx − And here M =

w 2 x 2

dM W = W − wx = 0 at x = dx w

W2 > 0 so M max ; M min must be at x = L 2w

So M min = WL −

1 2 wL . For minimum M 2

set M max = −M min , so

max

W2 1 = −WL + wL2 or W 2 + 2wLW − w2 L2 = 0 2w 2

W = −wL ± 2w2 L2 (need +)

W = (b) w may be removed

M max

(

W2 = = 2w

(

)

2 − 1 wL = 0.414wL

)

2 −1 2

2

wL2

M = Wx, M max = WL at A

Without w,

M = Wx −

With w (see part a)

M min = WL − For minimum M max , set M max ( no w ) = −M min ( with w )

WL = − WL +

M max = 0.858wL2

w 2 x , 2

M max =

W2 W at x = 2w w

1 2 wL at x = L 2

1 2 1 wL → W = wL → 2 4

With

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.

M max =

1 2 wL 4

W =

1 wL 4

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 58.

(a) FBD Beam:

ΣM C = 0:

ΣFy = 0:

LAy − M 0 = 0

− Ay + C = 0

Ay =

M0 L

C=

M0 L

M0  dV  = w = 0 . at A, and remains constant  L  dx  Moment Diag: M starts at zero at A and decreases linearly M M M   dM = V = − 0  to − 0 at B, where M jumps by M 0 to + 0 .  2 2 dx L   M M continues to decrease with slope − 0 to zero at C. L

Shear Diag: V = −

(b) From diagrams: V

max

=

M0 everywhere L

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.

max

=

M0 at B 2

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 59.

(a) and (b)

By symmetry Ay = By ΣFy = 0: 2 Ay − 2 P = 0

Ay = P

By = P

Shear Diag: V is piecewise constant with discontinuities equal to P at A, B, C and D in the direction of the loads.

Moment Diag: M is piecewise linear with slope equal to + P on AB, 0 on BC, –P on CD. M B = Pa V

max

= P along AB and CD 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.

max

= Pa along BC

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 60.

(a) and (b)

Shear Diag:  dV  = − w  starting at 0 at A, and Since w is linear, V is quadratic   dx  1 decreasing to − w0 L at B. 2

Moment Diag:  dm  = V  to M is zero at A and decreases cubically   dx 

1 1 1  2  − w0 L  L = − w0 L at B. 3 2 6  V 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.

max

max

=

=

1 w0 L at B 2

1 w0 L2 at B 6

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 61.

(a) and (b)

ΣM B = 0:

aP + 2aC y − 3.5aP = 0 C y = 1.25P

ΣFy = 0:

By − 2 P + 1.25P = 0

By = 0.75P

Shear Diag: V is piecewise constant, equal to –P from A to B, jumping up 0.75P, at B, to − 0.25P, and jumping up 1.25 P, at C, to + P. Moment Diag:

 dM  M is zero at A, decreasing linearly  = V = − P  to − Pa at B, and  dx   dM  further,  = V = − 0.25P  to − Pa − ( 0.25P )( 2a ) = −1.5Pa at C. M  dx   dM  then increases linearly  = V = P  to −1.5Pa + P (1.5a ) = 0 at D,  dx  as it must.

V

max

= P along AB and CD 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.

max

= 1.5Pa at C

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 62.

(a) and (b)

ΣM B = 0:

( 0.6 ft )( 4 kips ) + ( 5.1 ft )(8 kips ) + ( 7.8 ft )(10 kips ) − ( 9.6 ft ) Ay

=0

A y = 12.625 kips Shear Diag:

 dV  V is piecewise constant,  = 0  with discontinuities at each  dx  concentrated force. (equal to force) V

= 12.63 kips

max

Moment Diag:

 dM  = V  throughout.

M is zero at A, and piecewise linear   dx  M C = (1.8 ft )(12.625 kips ) = 22.725 kip ⋅ ft M D = 22.725 kip ⋅ ft + ( 2.7 ft )( 2.625 kips ) = 29.8125 kip ⋅ ft M E = 29.8125 kip ⋅ ft − ( 4.5 ft )( 5.375 kips ) = 5.625 kip ⋅ ft M B = 5.625 kip ⋅ ft − ( 0.6 ft )( 9.375 kips ) = 0

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.

max

= 29.8 kip ⋅ ft

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 63.

(a) and (b)

FBD Beam:

ΣM E = 0:

(1.1 m )( 0.54 kN ) − ( 0.9 m ) C y + ( 0.4 m )(1.35 kN ) − ( 0.3 m )( 0.54 kN ) = 0 C y = 1.08 kN

ΣFy = 0:

− 0.54 kN + 1.08 kN − 1.35 kN + E − 0.54 kN = 0 E = 1.35 kN

Shear Diag:

 dV  V is piecewise constant,  = 0 everywhere  with discontinuities at  dx  each concentrated force. (equal to the force) V

max

= 810 N

Moment Diag:

M is piecewise linear starting with M A = 0 M C = 0 − 0.2 m ( 0.54 kN ) = 0.108 kN ⋅ m M D = 0.108 kN ⋅ m + ( 0.5 m )( 0.54 kN ) = 0.162 kN ⋅ m M E = 0.162 kN ⋅ m − ( 0.4 m )( 0.81 kN ) = − 0.162 kN ⋅ m M B = 0.162 kN ⋅ m + ( 0.3 m )( 0.54 kN ) = 0

M

max

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.

= 0.162 kN ⋅ m = 162.0 N ⋅ m

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 64.

(1.5 m ) E y − 

1  m  ( 360 N/m )(1 m ) − (1.3 m )( 600 N ) 2 

ΣM A = 0:

− ( 2 m )( 420 N ) = 0 ΣFy = 0:

(a)

E y = 1200 N

Ay + 1200 N − ( 360 N/m )(1 m ) − 600 N − 420 N = 0 A y = 180.0 N

Shear Diag:

V jumps to Ay = 180 N at A, then decreases linearly  dV  = − w = − 360 N/m  to 180 N − ( 360 N/m )(1 m ) = −180 N at C.  dx   From C, V is piecewise constant ( w = 0 ) with jumps of − 600 N at D, + 1200 N at E, − 420 N at B. Moment Diag:

M starts at zero at A with slope

dM = V = 180 N/m, decreasing to zero dx

1 (180 N )( 0.5 m ) = 45 N ⋅ m. M is zero again 2 at C, decreasing to − (180 N )( 0.3 m ) = − 54 N ⋅ m at D. M then

at x = 0.5 m . There M =

decreases by ( 780 N )( 0.2 m ) = 156 N ⋅ m to − 210 N ⋅ m at E, and increases by ( 420 N )( 0.5 m ) = 210 N ⋅ m to zero at B. (b)

From the diagrams,

V

max

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.

= 780 N along EB

max

= 210 N ⋅ m at E

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 65.

(a)

Shear Diag:

 dV  V is zero at A with constant slope  = − w = −1 kip/ft  decreasing to  dx  − 3.6 kips at C. V then jumps 9 kips to 5.4 kips and is constant to D. Then V increases with constant slope 1.5 kips/ft for 3 ft, to 9.9 kips at B. This is also equal to By .

Moment Diag:

 dM  = V  decreasing to − 3.6 kips at C, M is zero at A, with zero slope   dx  1 where M = ( − 3.6 kips )( 3.6 ft ) , M C = − 6.48 kip ⋅ ft. M then increases 2 linearly with slope 5.4 kips to − 6.48 kip ⋅ ft + ( 5.4 kips )(1.8 ft ) = 3.24 kip ⋅ ft at D. Finally, M increases, with increasing slope, to  5.4 kips + 9.9 kips  M B = 3.24 kip ⋅ ft +   ( 3 ft ) , 2  

(b)

M B = 26.19 kip ⋅ ft.

From the diagrams,

V 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.

max

max

= 9.90 kips at B

= 26.2 kip ⋅ ft at B

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 66.

(a)

ΣM B = 0: ( 4 m ) Ay − ( 2.7 m )(1.8 kN/m )( 2.6 m ) − ( 2.4 m )( 4 kN ) = 0 A y = 5.559 kN Shear Diag:

At A, V jumps up 5.559 kN, then decreases with uniform slope of −1.8 kN/m to 2.679 kN at C. V then jumps down 4 kN to −1.321 kN, and continues with uniform slope −1.8 kN/m to − 3.121 kN at D. V is then constant to B. Note: By = 3.121 kN Moment Diag:

dM = V = 5.559 kN. The slope decreases to dx  5.559 + 2.679  kN  (1.6 m ) , 2.679 kN at C, where M =  2  

M is zero at A, with slope

M C = 6.59 kN ⋅ m. At C the slope drops to −1.321 kN and continues to  1.321 + 3.121  decrease, M D = 6.59 kN ⋅ m −   kN (1 m ) = 4.37 kN ⋅ m.

2   M then decreases with uniform slope − 3.121 kN, to zero at B. From the diagrams,

(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.

V M

max

max

= 5.56 kN at A

= 6.59 kN ⋅ m at C

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 67.

(a)

ΣFy = 0:

Ay + ( 24 kN/m )( 2 m ) − 48 kN − 8 kN = 0 A y = 8 kN

ΣM A = 0:

M A − (1 m )( 24 kN/m )( 2 m ) + ( 3.5 m )( 48 kN ) + ( 4 m )( 8 kN ) = 0 M A = −152 kN ⋅ m

Shear Diag:

V jumps to 8 kN at A, and increases with uniform slope dV = − w = 24 kN/m to 56 kN at C. V is constant at 56 kN to D, then dx drops by 8 kN to 8 kN at D, is then constant at 8 kN to B. Moment Diag:

M starts at M A = −152 kN ⋅ m, with slope 8 kN, which increases to 56 kN  8 + 56  at C, where M = −152 kN ⋅ m +  kN  ( 2 m ) = − 88 kN ⋅ m. Then  2  M increases with uniform slope 56 kN to − 88 kN ⋅ m + ( 56 kN )(1.5 m ) = − 4 kN ⋅ m at D, and finally, with slope 8 kN, to zero at B. From the diagrams,

(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.

V M

max

max

= 56 kN along CD = 152.0 kN ⋅ m at A

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 68.

Note: Cx is omitted to avoid clutter. (a)

By symmetry C y = G y ΣFy = 0:

2C y − 2 (12 lb/in.)(10 in.) − 2 (100 lb ) − 150 lb = 0 C y = 295 lb

G y = 295 lb

Shear Diag:

dV = − w = −12 lb/in. which is uniform to dx C, where V = − (12 lb/in.)(10 in.) = −120 lb ⋅ in. V jumps 295 lb to At A, V = 0 and has slope

+ 175 lb at C, is constant to D where it drops 100 lb to 75 lb, is constant to E where it drops 150 lb to − 75 lb. The diagram can be completed using symmetry.

Moment Diag:

M is zero at A, with zero slope, which decreases linearly to −120 lb at C, 1 where M = − (120 lb )(10 in.) = − 600 lb ⋅ in. M then increases, with 2 uniform slope 175 lb, to − 600 lb ⋅ in. + (175 lb )( 6 in.) = 450 lb ⋅ in. at D. M then increases, at uniform slope 75 lb, to 450 lb + (75 lb)(6 in.) = 900 lb ⋅ in. at E. The diagram can be completed using symmetry.

From the diagrams,

(b)

V

max

= 175.0 lb along CD and FG

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.

max

= 900 lb ⋅ in. at center E

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 69.

(a)

(18 ft )(1 kip/ft )( 6 ft ) − (12 ft ) Dy

ΣM F = 0:

+ ( 4.5 ft )( 2 kips/ft )( 9 ft ) + ( 3 ft )( 33 kips ) = 0 D y = 24 kips

ΣFy = 0:

24 kips + Fy + 33 kips − (1 kip/ft )( 6 ft ) − ( 2 kips/ft )( 9 ft ) = 0

Fy = − 33 kips

Fy = 33 kips Shear Diag:

dV = −1 kip/ft is uniform to C, where V = − 6 kips . dx Then V is constant to D where it jumps up 24 kips to + 18 kips, and dV = − 2 kips/ft and V decreases by remains constant to E. From E to F, dx 18 kips to zero at F, where it drops 33 kips, is constant to B, and jumps V = 0 at A and

33 kips to zero.

Moment Diag:

At A, M = 0 and

dM starts at zero, decreasing to − 6 kips at C, where dx

1 ( 6 kips )( 6 ft ) = −18 kip ⋅ ft. M then decreases linearly by 2 ( 6 kips )( 3 ft ) to − 36 kip ⋅ ft at D, and increases linearly by M =

(18 kip )( 3 ft )

dM decreases from dx

to + 18 kip ⋅ ft at E. From E to F,

1 (18 kips )( 9 ft ) to 99 kip ⋅ ft, at F. 2 M then decreases linearly to zero at B. 18 kips to zero as M increases by

From the diagrams,

(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.

V

max

= 33.0 kips along FB

M

max

= 99.0 kips at F

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 70.

(a)

ΣFy = 0:

Ay + ( 200 N/m )( 0.8 m ) − (120 N/m )( 0.3 m ) = 0 Ay = −124 N

ΣM A = 0:

A y = 124 N

M A − 60 N ⋅ m − ( 0.4 m )( 200 N/m )( 0.8 m ) + (1.35 m )(120 N/m )( 0.3 m ) = 0

M A = 75.4 N ⋅ m

Shear Diag:

dV = 200 N/m from A to C and V increases by dx ( 200 N/m )( 0.8 m ) = 160 N to + 36 N at C. It remains at 36 N to D,

V drops to 124 N at A.

then decreases linearly to zero at B. Note V = 0 where

x 124 N = , 0.8 m 160 N

or x = 0.62 m.

Moment Diag:

dM = −124 N. The slope increases to zero dx 1 at x = 0.62 m, where M = 75.4 N ⋅ m − (124 N )( 0.62 m ) 2 = 36.96 N ⋅ m. The slope then increases as M increases by 1 ( 36 N )( 0.18 m ) = 3.24 N ⋅ m to 40.2 N ⋅ m at C, where it drops 2 60 N ⋅ m to −19.8 N ⋅ m. M increases linearly by

M jumps to 75.4 at A where

( 36 N )( 0.4 m ) = 14.4 N ⋅ m quadratically by

to − 5.4 N ⋅ m, and finally M increases

1 dM is ( 36 N )( 0.3 m ) = 5.4 N ⋅ m to zero at B where 2 dx

also zero.

From the diagrams,

V

max

M

(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.

max

= 124.0 N at A = 75.4 N at A

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 71.

(a)

− ( 3 m )( 9 kN ) − 27 kN ⋅ m − ( 9 m )(12 kN ) + (12 m ) F

ΣM A = 0:

+ (16.5 m )( 3 kN ) + 22.5 kN ⋅ m = 0

F = 7.5 kN ΣFy = 0:

Ay − 9 kN − 12 kN + 7.5 kN + 3 kN = 0 A y = 10.5 kN

Shear Diag:

V is piecewise constant, with jumps at A, C, E, F, and B, equal to the forces there.

Moment Diag:

M is piecewise linear with jumps at D and B equal to the couples there.

M C = (10.5 kN )( 3 m ) = 31.5 kN ⋅ m M D − = 31.5 kN ⋅ m + (1.5 kN )( 3 m ) = 36.0 kN ⋅ m

M D+ = 36 kN ⋅ m + 27 kN ⋅ m = 63 kN ⋅ m M E = 63 kN ⋅ m + (1.5 kN )( 3 m ) = 67.5 kN ⋅ m M F = 67.5 kN ⋅ m − (10.5 kN )( 3 m ) = 36 kN ⋅ m M B − = 36 kN ⋅ m − ( 3 kN )( 4.5 m ) = 22.5 kN ⋅ m

Finally M drops 22.5 kN ⋅ m to zero at B

(b)

From the diagrams,

V

max

= 10.50 kN along AC and EF 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.

max

= 67.5 kN ⋅ m at E

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 72.

(a)

( 3 m ) By − ( 2.1 m )( 2.5 kN/m )( 4.2 m ) = 0

ΣM A = 0:

B y = 7.35 kN

ΣFy = 0:

Ay − ( 2.5 kN/m )( 4.2 m ) + 7.35 kN = 0 A y = 3.15 kN

Shear Diag:

dV = − 2.5 kN/m throughout, and jumps at A and B equal to dx the forces there.

V has slope

VB − = 3.15 kN − ( 2.5 kN/m )( 3 m ) = − 4.35

VB + = − 4.35 kN + 7.35 kN = 3 kN VC = 3 kN − ( 2.5 kN/m )(1.2 m ) = 0 Note, V = 0 where 3.15 kN − ( 2.5 kN/m ) x = 0, x = 1.26 m.

Moment Diag:

dM = 3.15 kN. The slope decreases to zero at dx x = 1.26 m and to − 4.35 kN at B, jumps to 3.0 kN and decreases to 0 at C. 1 M D = ( 3.15 kN )(1.26 m ) = 1.9845 kN ⋅ m 2 1 M B = 1.9845 kN ⋅ m − ( 4.35 kN )( 3 m − 1.26 m ) = −1.80 kN ⋅ m 2 1 M C = −1.80 kN ⋅ m + ( 3 kN )(1.2 m ) = 0 2 At A, M = 0 and

From the diagrams,

(b)

V M

max

max

= 4.35 kN at B

= 1.985 kN at D (1.26 m from 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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 73.

(a)

Note: C x omitted to avoid clutter. ΣM C = 0: − ( 0.5 m )( 2 kN/m )( 2 m ) + (1.5 m )(1 kN )

− ( 3 m )( 4.5 kN/m )(1 m ) + ( 3.5 m ) By = 0

B y = 4.0 kN ΣFy = 0:

C y − ( 2 kN/m )( 2 m ) + 1 kN − ( 4.5 kN/m )(1 m ) + 4.0 kN = 0

C y = 3.5 kN

Shear Diag:

dV = − 2 kN/m, so V decreases to −1 at C, then dm jumps 3.5 kN to + 2.5 kN, and continues to decrease, with the same slope, to − 0.5 kN at D, jumps 1 kN to + 0.5 kN from D to E. Then V decreases at rate 4.5 kN/m, to − 4.0 kN at B. Note that V = 0 where 8 ( − 4.5 kN/m )( x ) = − 4 kN, x = m, and where ( 2 kN/m )( y ) = 2.5, 9 5 y = m.

4 Moment Diag: dM At A, M and = 0, with the slope decreasing to −1 kN at C, where dx 1 M = − (1 kN )( 0.5 m ) = − 0.25 kN ⋅ m. The slope jumps to 2.5 kN 2 and decreases to zero at F and to − 0.5 kN at D. At A, V = 0 and

M F = − 0.25 kN ⋅ m +

1 5 ( 2.5 kN )  m  = 1.3125 kN ⋅ m 2 4 

M D = 1.3125 kN ⋅ m −

1 1 ( 0.5 kN )  m  = 1.25 kN ⋅ m.

2 4 

From D, M increases by ( 0.5 kN )(1 m ) to 1.75 kN ⋅ m at G. M continues to increase to 1.75 kN ⋅ m + then decreases by

(b)

1 1 ( 0.5 kN )  m  = 1.7778 kN ⋅ m at G and 2 9 

1 8 ( 4 kN )  m  = 1.7778 kN ⋅ m, to zero at B. 2 9  V

From the diagrams,

M

max

max

= 4.00 kN at B

4  = 1.778 kN ⋅ m at G  m from B  9  

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 7, Solution 74.

(a)

ΣFy = 0:

Ay − ( 2 kips/ft )( 8 ft ) − 3 kips + 7 kips = 0

A y = 12 kips M A + ( 4 ft )( 2 kips/ft )( 8 ft ) − (14 ft )( 3 kips )

ΣM A = 0:

− ( 20 ft )( 7 kips ) = 0

M A = 34 kip ⋅ ft

Shear Diag: V jumps to 12 kips at A, then decreases at 2 kips/ft to − 4 kips at C to D. V drops 3 kips to − 7 kips from D to B and jumps 7 kips to zero. Note: V = 0 where 12 kips − ( 2 kips/ft ) x = 0, x = 6 ft.

Moment Diag: M jumps to 34 kip ⋅ ft at A and then increases with decreasing slope to 1 34 kip ⋅ ft + (12 kips )( 6 ft ) = 70 kip ⋅ ft at E, and decreases by 2 1 ( 4 kips )( 2 ft ) = 4 kip ⋅ ft, to 66 kip ⋅ ft at C. M then decreases by 2 ( 4 kips )( 6 ft ) to 42 kip ⋅ ft at D, and by ( 7 kips )( 6 ft ) to zero at B.

M

ΣFy = 0:

max

= 70 kip ⋅ ft at E

Ay − ( 2 kips/ft )( 8 ft ) − 3 kips + 10 kips = 0

A y = 9 kips

(b)

ΣM A = 0:

M A + ( 4 ft )( 2 kips/ft )( 8 ft ) + (14 ft )( 3 kips ) − ( 20 ft )(10 kips ) = 0

M A = 94 kip ⋅ ft

Shear Diag: V jumps to 9 kips at A, then decreases, at 2 kips/ft , to − 7 kips at C to D, drops 3 kips to −10 kips from D to B and jumps 10 kips to 0.

Note: V = 0 where 9 kips − ( 2 kips/ft ) x = 0, x = 4.5 ft.

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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Moment Diag: M jumps to 94 kip ⋅ ft at A and increases to 94 kip ⋅ ft +

1 ( 9 kips )( 4.5 ft ) = 114.25 kip ⋅ ft, then decreases by 2

1 ( 7 kips )( 3.5 ft ) to 102 kip ⋅ ft at C. M decreases linearly by 2 ( 7 kips )( 6 ft ) to 60 kip ⋅ ft at D, then by (10 kips )( 6 ft ) to zero at

B.

M

max

= 114.3 kip ⋅ ft at E ( 4.5 ft from 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.

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Chapter 7, Solution 75.

(15 ft ) C y + ( 9 ft )( 600 lb ) − ( 4.5 ft )(800 lb/ft )( 9 ft ) = 0

ΣM A = 0:

C y = 1800 lb = 1.8 kips ΣFy = 0:

Ay + 1800 lb − ( 800 lb )( 9 ft ) + 600 lb = 0

A y = 4800 lb = 4.8 kips Shear Diag: V jumps to 4.8 kips at A then decreases linearly, at 0.8 kips/ft, to − 2.4 kips at B, jumps 0.6 kips to −1.8 kips, is constant to C, and jumps 1.8 kips to zero.

Note: V = 0 at D, where 4.8 kips − ( 800 kip/ft ) x = 0, x = 6.0 ft.

Moment Diag: M starts at zero and increases with decreasing slope to 1 ( 4.8 kips )( 6 ft ) = 14.4 kip ⋅ ft at D, then decreases by 2 1 ( 2.4 kips )( 3 ft ) to 10.8 kip ⋅ ft at B. M then decreases with slope 2 −1.8 kips to zero at C.

M

max

= 14.40 kips at D (6 ft from 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.

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Chapter 7, Solution 76.

(a)

ΣM A = 0:

( 6 m ) By − (1 m )( 20 kN/m )( 2 m ) − 24 kN ⋅ m − ( 5 m )( 20 kN/m )( 2 m ) = 0

B y = 44 kN

ΣFy = 0: Ay − 2 ( 20 kN/m )( 2 m ) + 44 kN = 0 A y = 36 kN

Shear Diag:

V jumps to 36 kN at A, then decreases with slope –20 kN/m to − 4 kN at C, is constant to E, then decreases with slope –20 kN/m to − 44 kN at B.

Note: V = 0 at F where 36 kN − ( 20 kN/m ) x = 0, x = 1.8 m.

Moment Diag:

Starting at zero M increases with decreasing slope to

1 (36 kN )(1.8 m) 2

1 ( 4 kN )( 0.2 m ) to 32 kN ⋅ m at C , 2 then with slope − 4 kN to 28 kN ⋅ m at D, where it jumps to 52 kN ⋅ m, M decreases with slope − 4 kN to 48 kN ⋅ m at E, then with increasingly = 32.4 kN ⋅ m at F , decreases by

(b)

 4 + 44  negative slope by  kN  ( 2 m ) to zero at B. 2   M

max

= 52 kN ⋅ m ( at D ) 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 7, Solution 77.

(a)

ΣM A = 0:

( 6 m ) By − (1 m )( 20 kN/m )( 2 m ) + 24 kN ⋅ m − ( 5 m )( 20 kN/m )( 2 m ) = 0

ΣFy = 0:

B y = 36 kN

Ay − 2 ( 20 kN/m )( 2 m ) + 36 kN = 0

A y = 44 kN Shear Diag: V jumps to 44 kN at A, then decreases with slope − 20 kN/m to 4 kN at C, is constant to E, then decreases with slope − 20 kN/m to −36 kN at B. V = 0 at F where − 36 kN + ( 20 kN/m ) x = 0, x = 1.8 m.

Moment Diag:

 44 + 4  kN ( 2 m ) M starts at zero, increases with decreasing slope to   2  = 48 kN ⋅ m at C , increases with slope 4 kN to 52 kN ⋅ m at D, drops 24 kN ⋅ m to 28 kN ⋅ m then increases with slope 4 kN to 32 kN ⋅ m at E. 1 Then M increases with decreasing slope, by ( 4 kN )( 0.2 m ) to 2 32.4 kN ⋅ m at F and decreases with increasingly negative slope to zero at B. (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.

M

max

= 52.0 kN ⋅ m ( at D )

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Chapter 7, Solution 78.

(a)

Note: The 2 kip force at E has been replaced by the equivalent force and couple at C.

ΣM A = 0: − ( 6 ft )(1 kip/ft )(12 ft ) + 8 kip ⋅ ft − (12 ft )( 2 kips ) + ( 24 ft ) Dy − ( 32 ft )(1 kip ) = 0 ΣFy = 0:

D y = 5 kips

Ay − (1 kip/ft )(12 ft ) − 2 kips + 5 kips − 1 kip = 0 A y = 10 kips

Shear Diag: From 10 kips at A, V decreases with slope −1 kip/ft to − 2 kips at C, drops 2 kips, is constant at − 4 kips to D, jumps 5 kips, and is constant at 1 kip to B. V = 0 at E, where 10 kips − (1 kip/ft ) x = 0, x = 10 ft.

Moment Diag: From zero at A, M increases with decreasing slope to

1 (10 kips)(10 ft ) 2

1 ( 2 kips )( 2 ft ) to 48 kip ⋅ ft at C, drops 2 8 kip ⋅ ft to 40 kip ⋅ ft, then decreases with slope − 4 kips to − 8 kip ⋅ ft at D. Finally M increases with slope 1 kip to zero at B. = 50 kip ⋅ ft at F , decreases by

(b)

M

max

= 50 kip ⋅ ft at F (10.00 ft from A ) 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 7, Solution 79.

(a)

Note: The 2 kip force at E has been replaced by the equivalent force and couple at C. ΣM A = 0: − ( 6 ft )(1 kip/ft )(12 ft ) + (12 ft )( 2 kips ) − 8 kip ⋅ ft

+ ( 24 ft ) Dy − ( 32 ft )(1 kip ) = 0 ΣFy = 0:

Ay − (1 kip/ft )(12 ft ) + 2 kips −

Ay =

Dy =

11 kips 3

11 kips − 1 kip = 0 3

22 kips 3

Shear Diag: 22 14 Starting at kips at A, V decreases with slope −1 kip/ft to − kips 3 3 8 11 kips at C, jumps 2 kips and remains constant at − kips to D, jumps 3 3 and remains constant at 1 kip to B, drops to zero.

V = 0 at F, where

22 kip − (1 kip/ft ) x = 0, 3

x=

22 ft.

3

Moment Diag: Starting from zero, M increases with decreasing slope to 1  22  22  kips  ft  = 26.889 kip ⋅ ft at F . M then decreases by  2 3  3  1  14  14  kips  ft  to 16 kip ⋅ ft at C , jumps to 24 kip ⋅ ft, decreases  2 3  3  8 with slope − kips to − 8 kip ⋅ ft at D, and finally increases with slope 3 1 kip to zero at B. (b)

M

max

= 26.9 kip ⋅ ft at F ( 7.33 ft from A) 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 7, Solution 80. (b)

 x  w = w0  4   − 1  L 

(a) Distributed load

dV = −w, and V ( 0 ) = 0, so dx

Shear:

V = V =

x

∫0

− wdx = − ∫

x/L

∫0

 wo L 1 − 

x/L 0

x Lwd   L

2  x   x  x   x  4    d   = w L    − 2   0  L   L   L  L   

Notes: At x = L, V = −w0 L

x x   = 2  L L

And V = 0 at

2

x 1 = L 2

or

1 x Also V is max where w = 0  =  4 L Vmax =

1 w0 L 8

M ( 0 ) = 0,

Moment:

M =

x

x/L

∫ 0 vdx = L∫ 0 2

M = w0 L

x/L

∫0

dM =V dx

x x V  d   L L

2  x  x  x   − 2    d    L    L   L 

2  x  x  V = w0 L   − 2     L    L 

 1  x  2 2  x 3  M = w0 L    −    3  L    2  L  2

M max =

1 L w0 L2 at x = 24 2

1 M min = − w0 L2 at x = L 6 M max = (c)

w0 L2 L at x = 24 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.

max

= −M min =

w0 L2 at B 6

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Chapter 7, Solution 81.

(b)

1 1 ( 2w0 ) ( 2a ) + w0a = 0 2 2

ΣFy = 0: Ay −

Ay =

3 w0a 2

2 1 7 1 ( 2a ) ( 2w0 ) ( 2a ) −  a  ( w0 ) ( a ) = 0 3 2 3 2

ΣM A = 0: M A +

3 M A = − w0a 2 2 (a) for 0 ≤ x ≤ 2a:

w=

w0 x a V =

3 M = − w0a 2 + 2

∫

x 0

(

3 w0a − 2

∫

x 0

(

)

w0 w xdx = 0 3a 2 − x 2 W a 2a

)

w0 3a 2 − x 2 dx 2a M =

(

)

(

)

w0 − 9a 3 + 9a 2 x − x3 W 6a

Note: V = 0 at x = 3a, where M = 0.232 w0a 2 At x = 2a, M =

1 w0a 2 6

for 2a ≤ x ≤ 3a : 1 V = − w0a + 2

∫

x 2a

w = − 3w0 +

w0 x a

w0 ( 3a − x ) dx a V =

M =

1 w0a + 6

∫

x 2a

(

)

w0 − 9a 2 + 6ax − x 2 dx 2a M =

(c)

w0 − 9a 2 + 6ax − x 2 W 2a

(

)

w0 27a3 − 27a 2 x + 9ax 2 − x3 W 6a 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.

max

=

3 w0a 2 at A ( x = 0 ) W 2

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Chapter 7, Solution 82.

(a)

FBD Beam:

ΣM B = 0:

1  w0 ( 3a )  − 5aAy = 0 2 

( 3a ) 

ΣFy = 0: 0.9w0a −

A y = 0.9w0a

1 w0 ( 3a ) + B = 0 2 B = 0.6w0a

Shear Diag:

V = Ay = 0.9w0a from A to C and V = B = − 0.6w0a from B to D. x1 . If x1 is measured right to left, 3a dV x w dM = + w and = − V . So, from D, V = − 0.6w0a + ∫ 0 1 0 x1dx1, dx1 dx1 3a

Then from D to C, w = w0

2  1  x1   V = w0a  − 0.6 +    6  a   

2

x  Note: V = 0 at  1  = 3.6, x1 = a

3.6 a

Moment Diag:

 dM  M = 0 at A, increasing linearly  = 0.9w0a  to M C = 0.9w0a 2.  dx1   dM  Similarly M = 0 at B increasing linearly  = 0.6w0a  to dx  

M D = 0.6w0a 2. Between C and D 2

M = 0.6w0a + w0a ∫

x1

0

2  0.6 − 1  x1   dx1,  6 a    

3  x  1 x   M = w0a 2 0.6 + 0.6  1  −  1    a  18  a   

(b)

At

x1 = a

3.6, M = M max = 1.359w0a 2 x1 = 1.897a left of D

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 7, Solution 83.

ΣM A = 0: LBy − V =−

w0 L + 20

L1   w0 L  = 0 54 

∫

x1 0

By =

(

w0 L 20

w0 3 w x dx1 = 0 3 5x14 − L4 3 1 20L L V =

or

)

w0  4 5 ( L − x ) − L4  W 3   20 L 

Note: V = 0 at x1 = 5−1/ 4 L = 0.6687 L or

x = 0.3313 L M =0+ =

∫

x1 0

(

)

w0 5x14 − L4 dx1 20L3

(

)

or

M =

w0 x15 − L4 x1 20L3

w0  5 L − x ) − L4 ( L − x )  W ( 3   20L 

M max = M at x1 = 0.6687 L,

x = 0.331 L W

M max = 0.0267 w0 L2 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 7, Solution 84.

(a)

ΣM C = 0: ( 395 − 215 ) N ⋅ m + ( 0.1 m )(1000 N/m )( 0.2 m )

FBD section AC:

− ( 0.2 m )VA = 0,

VA = 1000 N

ΣM B = 0: 395 N ⋅ m + ( 0.45 m )(1 kN/m )( 0.4 m ) − ( 0.65 m )(1 kN ) + ( 0.25 m ) P = 0 FBD whole:

(b)

P = 0.3 kN,

P = 300 N W

ΣFy = 0: 1000 N − 400 N − 300 N − Q = 0 Q = 300 N W

Shear Diag:

Starting at 1000 N, V decreases with slope − 1000 N/m to 600 N at D, drops 300 N and is constant to B where it drops 300 N to zero.

Moment Diag:

Starting at − 395 N ⋅ m, M increases with decreasing slope to −395 N ⋅ m  1000 + 600  N ( 0.4 m ) = − 75 N ⋅ m at D, then increases with slope 300 N +   2 to zero at 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 7, Solution 85.

(a)

ΣM C = 0:

( 395 − 215) N ⋅ m − ( 0.2 m ) Ay

FBD section AC:

+ ( 0.1 m )(1000 N/m )( 0.2 m ) = 0 A y = 1000 N

ΣM B = 0: 395 N ⋅ m + ( 0.325 m )(1000 N/m )( 0.65 m ) + ( 0.25 m ) P = 0 − ( 0.65 m )(1000 N ) = 0

(b) FBD Whole:

P = 175.0 N W

ΣFy = 0: 1000 N − 175 N − Q − (1000 N/m )( 0.65 m ) = 0 Q = 175.0 N W Shear Diag:

V starts at 1000 N, and has slope − 1000 N/m throughout, but with drops of 175 N at D and B. Moment Diag:

M starts at − 395 N ⋅ m and increases with decreasing slope to − 295 N ⋅ m  1000 + 600  + N ( 0.4 m ) = − 75 N ⋅ m at D. There is a discontinuity in slope,   2  425 + 175  N  ( 0.25 m ) to zero at B. and M increases by  2  

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 7, Solution 86.

(a) FBD AC:

ΣM C = 0: 2.7 kN ⋅ m − ( 0.2 m ) Ay = 0, ΣM D = 0:

( 0.2 m ) Dy

− 2.5 kN ⋅ m = 0,

A y = 1.35 kN D y = 1.25 kN

ΣM C = 0: − ( 2 m )(1.35 kN ) − ( 2 m )( 0.4 kN/m )( 4 m ) − ( 4 m ) Q + ( 6 m )(1.25 kN ) = 0,

FBD AB:

Q = 0.4 kN Q = 400 N W

ΣFy = 0:

1.35 kN − P − ( 0.4 kN/m )( 4 m ) − 0.4 kN + 1.25 kN = 0 P = 0.6 kN P = 600 N W

(b)

FBD Whole:

Shear Diag:

V is constant at 1.35 kN from A to C, drops 0.6 kN, then decreases with slope –0.4 kN/m ( −1.6 kN ) to − 0.85 kN at D, drops 0.4 kN to −1.25 kN, and is constant to B. V = 0 where 0.75 kN – (0.4 kN/m)x = 0, x = 1.875 m.

Moment Diag:

From zero at A, M increases with slope 1.35 kN to 2.70 kN ⋅ m at C, the slope drops to 0.75 kN and then decreases to zero at E, where 1 M = 2.7 kN ⋅ m + ( 0.75 kN )(1.875 m ) = 3.403 kN ⋅ m. This curve 2 1 continues to D where M = 3.403 kN ⋅ m − ( 0.85 kN )( 2.125 m ) 2 = 2.50 kN ⋅ m, then M decreases with slope −1.25 kN to zero at 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 7, Solution 87.

(a) FBD AC:

ΣM C = 0: 2.7 kN ⋅ m − (1.35 m ) Ay = 0,

A y = 2 kN

ΣM D = 0: 2.5 kN ⋅ m − ( 2 m ) By = 0,

B y = 1.25 kN

ΣM C = 0: ( 6.65 m )(1.25 kN ) − ( 4.65 m ) Q − ( 2.325 m )( 0.4 kN/m )( 4.65 m ) − (1.35 m )( 2 kN ) = 0 Q = 0.27699 kN

FBD DB:

Q = 277 N

W

ΣFy = 0: 2 kN − P − 0.277 kN − ( 0.4 kN/m )( 4.65 m ) + 1.25 kN = 0 P = 1.113 kN

(b) FBD Whole:

W

Shear Diag:

V is constant at 2 kN from A to C, drops 1.113 kN to 0.887 kN, then decreases with slope − 0.4 kN/m to 0.887 kN − ( 0.4 kN/m ) ( 4.65 m )

= − 0.973 kN, drops 0.277 kN to −1.25 kN and is constant to B. V = 0 where 0.887 kN − ( 0.4 kN/m ) x = 2.2175 m at E .

Moment Diag:

Starting from zero, M increases with slope 2 kN to 2.7 kN ⋅ m at C. The slope drops to 0.887 kN and decreases to zero at E where M = 2.7 kN 1 + ( 0.887 kN )( 2.2175 m ) = 3.68 kN ⋅ m. This curve continues to D 2 where M = 2.5 kN ⋅ m, then M decreases linearly to zero at 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 7, Solution 88.

FBD Cable:

ΣM A = 0:

( 5 m ) Dy − ( 4 m )( 2.25 kN ) − ( 2 m )( 3 kN ) = 0 D y = 3 kN

ΣFy = 0:

Ay − 3 kN − 2.25 kN + 3 kN = 0 A y = 2.25 kN

Point D:

ΣFx = 0: − Ax + Dx = 0,

Ax = Dx

(1)

Since Ax = Dx and Dy > Ay , TCD is Tmax ΣFy = 0: 3 kN − ΣFx = 0: − Point A:

3 TCD = 0 5

TCD = 5 kN

4 ( 5 kN ) + Dy = 0 5 From(1),

dB 2.25 kN = 2m 4 kN

D x = 4 kN

A x = 4 kN (a )

d B = 1.125 m W

(b) A x = 4.00 kN

W

A y = 2.25 kN W (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.

Tmax = 5.00 kN W

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Chapter 7, Solution 89.

FBD Cable:

ΣM A = 0:

( 5 m ) Dy − ( 4 m )( 2.25 kN ) − ( 2 m )( 3 kN ) = 0 D y = 3 kN

ΣFy = 0:

Ay − 3 kN − 2.25 kN + 3 kN = 0 A y = 2.25 kN

ΣFx = 0: − Ax + Dx = 0

Point D:

Since Ax = Dx , and Dy > Ay ,

Ax = Dx Tmax is TCD = 3.6 kN

1 + dC2 dC 1m = = 3 kN Dx 3.6 kN 1.2 dC = 1 + dC2 , 1.44 dC2 = 1 + dC2

Point A:

dC = 1.50756 m 1 m  Also Dx = 3 kN   = 1.98997 kN = Ax  dC 

dB 2m = 2.25 kN 1.9900 kN

(a) d B = 2.26 m W (b) dC = 1.508 m 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 7, Solution 90.

ΣM A = 0:

4aE y − 3a (1.2 kips ) − 2a ( 0.8 kips ) − a ( 0.4 kips ) = 0

FBD Cable:

E y = 1.4 kips ΣFy = 0:

AY − ( 0.4 + 0.8 + 1.2 kips ) + 1.4 kips = 0

A y = 1.0 kips ΣFx = 0:

− Ax + Ex = 0,

Since Ax = Ex and E y > Ay ,

FBD CDE:

ΣM C = 0:

Ax = Ex Tmax = TDE

( 30 ft )(1.4 kips ) − (12 ft ) Ex − (15 ft )(1.2 kips ) = 0 (a)

E x = 2.00 kips W E y = 1.400 kips W

(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.

Tmax = TDE = 2.44 kips W

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Chapter 7, Solution 91.

FBD Cable:

ΣM A = 0:

4aE y − 3a (1.2 kips ) − 2a ( 0.8 kips ) − a ( 0.4 kips ) = 0 E y = 1.4 kips

ΣFy = 0:

Ay − ( 0.4 + 0.8 + 1.2 kips ) + 1.4 kips = 0 A y = 1.0 kips

ΣFx = 0: Point E:

− Ax + Ex = 0,

Since Ax = Ex , and E y > Ay , Ex =

Ax = Ex Tmax = TDE

( 5 kips )2 + (1.4 kips )2

= 4.8 kips

FBD CDE:

ΣM C = 0:

( 30 ft )(1.4 kips ) − dC ( 4.8 kips ) − (15 ft )(1.2 kips ) = 0 dC = 5.00 ft

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 7, Solution 92.

(a) FBD Whole:

ΣM A = 0:

(1.2 m ) Ex + ( 4 m ) E y − (1 m )(1.8 kN ) − ( 2 m )( 3.6 kN ) − ( 3 m )(1.2 m ) = 0 1.2Ex + 4E y = 12.6 kN

FBD CDE:

ΣM C = 0:

(1)

( 2 m ) E y − ( 0.6 m ) Ex − (1 m )(1.2 kN ) = 0 − 0.6Ex + 2E y = 1.2 kN

Solving (1) and (2)

(2)

E x = 4.25 kN E y = 1.875 kN E = 4.65 kN

(a)

(b) From FBD whole:

AB:

ΣFx = 0 ΣFy = 0

− Ax + 4.25 kN = 0

23.8°

A x = 4.25 kN

Ay − 1.8 kN − 3.6 kN − 1.2 kN + 1.875 kN = 0 A y = 4.725 kN

ΣM B = 0:

d B ( 4.25 kN ) − (1 m )( 4.725 kN ) = 0 d B = 1.112 m

DE: ΣM D = 0:

(1 m )(1.875 kN ) − ( d D − 1.2 m )( 4.25 kN ) = 0 d D = 1.641 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 7, Solution 93.

FBD DE:

ΣFy = 0: ΣM D = 0:

FBD Whole:

ΣM A = 0:

E y − 1.2 kN = 0

E y = 1.2 kN

(1 m )(1.2 kN ) − ( dC

− 1.2 m ) Ex = 0

(1)

(1.2 m ) Ex + ( 4 m )(1.2 kN ) − ( 3 m )(1.2 kN ) − ( 2 m )( 3.6 kN ) − (1 m )(1.8 kN ) = 0 E x = 6.5 kN

dC = 1.385 m

(a) then, from (1) ΣFx = 0: ΣFy = 0:

− Ax + 6.5 kN = 0

A x = 6.5 kN

Ay − 1.8 kN − 3.6 kN − 1.2 kN + 1.2 kN = 0

A y = 5.4 kN So

(b)

A = 8.45 kN E = 6.61 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.

39.7° 10.46°

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Chapter 7, Solution 94.

FBD Cable:

Hanger forces at A and F act on the supports, so A y and Fy act on the cable. ΣM F = 0:

( 6 ft + 12 ft + 18 ft + 24 ft )( 400 lb ) − ( 30 ft ) Ay − ( 5 ft ) Ax = 0

Ax + 6 Ay = 4800 lb (1)

FBD ABC: ΣM C = 0:

( 7 ft ) Ax − (12 ft ) Ay + ( 6 ft )( 400 lb ) = 0

(2)

Solving (1) and (2) A x = 800 lb Ay =

2000 lb 3

ΣFx = 0:

From FBD Cable:

− 800 lb + Fx = 0

FBD DEF:

Fx = 800 lb ΣFy = 0:

2000 lb − 4 ( 400 lb ) + Fy = 0 3 Fy =

Since Ax = Fx and Fy > Ay ,

Tmax = TEF =

(a) Tmax = 1229.27 lb,

ΣM D = 0:

2800 lb 3

(800 lb )

2

 2800  lb  +  3 

2

Tmax = 1.229 kips

(12 ft ) 

2800  lb  − d D ( 800 lb ) − ( 6 ft )( 400 lb ) = 0  3 

(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 D = 11.00 ft

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Chapter 7, Solution 95.

FBD CDEF:

ΣM C = 0:

(18 ft ) Fy − ( 9 ft ) Fy − ( 6 ft + 12 ft )( 400 lb ) = 0 Fx − 2Fy = − 800 lb

FBD Cable:

ΣM A = 0:

(1)

( 30 ft ) Fy − ( 5 ft ) Fx − ( 6 ft )(1 + 2 + 3 + 4 )( 400 lb ) = 0 Fx − 6Fy = − 4800 lb

Solving (1) and (2), Fx = 1200 lb

ΣFx = 0: Point F:

ΣFy = 0:

(2)

, Fy = 1000 lb

− Ax + 1200 lb = 0 ,

A x = 1200 lb

Ay + 1000 lb − 4 ( 400 lb ) = 0 ,

A y = 600 lb

Since Ax = Ay and Fy > Ay , Tmax = TEF

Tmax = FBD DEF:

(1 kip )2 + (1.2 kips )2 Tmax = 1.562 kips

(a)

ΣM D = 0:

(12 ft )(1000 lb ) − d D (1200 lb ) − ( 6 ft )( 400 lb ) = 0

(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 D = 8.00 ft

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Chapter 7, Solution 96.

ΣM A = 0:

( 9 ft ) P − ( 6 ft )( 30 lb ) − a ( 20 lb ) = 0

FBD BC:

20a   P =  20 +  lb with a in ft.

9   ΣFx = 0: ΣFy = 0:

− TAB x + P = 0

TAB x = P

TAB y − 20 lb − 30 lb = 0

But Solving (1) and (2),

TAB x a = TAB y 7

(1)

so

a = 4.0645 ft,

TAB y = 50 lb P=

50a 7

(2)

P = 29.032 lb (a) P = 29.0 lb (b) a = 4.06 ft

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 7, Solution 97.

FBD C:

ΣM B = 0:

( 2 ft )( 25 lb ) − ( 6 ft − a )( 30 lb ) = 0 a=

13 ft 3

a = 4.33 ft

FBD BC:

ΣM A = 0:

( b + 2 ft )( 25 lb ) − ( 6 ft )( 30 lb ) − 

13  ft  ( 20 lb ) = 0  3 

b=

32 ft 3 b = 10.67 ft

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 7, Solution 98.

ΣFx = 0: −

FBD B:

ΣFy = 0:

12 3 TAB + TBC + 1.32 kN = 0 13 5

5 4 TAB − TBC = 0 13 5

Solving: TAB = 2.08 kN, TBC = 1 kN By inspection of A, A = 2.08 kN

(a) FBD C:

ΣFx = 0: ΣFy = 0:

12 3 TCD − (1 kN ) = 0 , 13 5

22.6°

TCD = 0.65 kN

4 5 (1 kN ) + ( 0.65 kN ) − w = 0 5 13 w = 1.05 kN

(b) m = (c)

w 1050 N = = 107.03 kg g 9.81 m/s 2 From above

m = 107.0 kg TAB = 2.08 kN TBC = 1.000 kN TCD = 650 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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 99.

FBD C:

(

)

W = mg = (150 kg ) 9.81 m/s 2 = 1471.5 N = 1.4715 kN ΣFx = 0:

12 3 TCD − TBC = 0 13 5 5 4 TCD + TBC − 1.4715 kN = 0 13 5

ΣFy = 0:

Solving: TCD = 0.91093 kN = 911 N, TBC = 1.40143 kN FBD B:

By inspection of D, ΣFx = 0: ΣFy = 0:

P−

(a)

D = 911 N

22.6°

12 3 TAB + (1.40143 kN ) = 0 13 5

5 4 TAB − (1.40143 kN ) = 0, 13 5 (b) From above

(c)

TAB = 2.91497 kN P = 1.850 kN

TAB = 2.91 kN TBC = 1.401 kN TCD = 911 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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 100.

FBD E:

Ey Ex T = = DE 1 1 2

FBD CDE:

ΣM C = 0:

( 6.5 m ) E y − ( 4.2 m ) Ex − ( 3.5 m ) wD Ex = E y =

2.3Ex = 3.5wD ,

ΣM B = 0:

Ex = E y

35 wD 23

35  wD  − ( 5.1 m ) wD − (1.6 m ) wC = 0  23  mC = 1.66304 mD

FBD B:

(a)

ΣFx = 0:

4 − TBC + Ex = 0, 5

ΣFx = 0:

45  1 24 TAB − TBF = 0  Ex  − 54  25 2

Solving:

=0

(8.1 m − 3 m ) 

wC = 1.66304wD ,

ΣFy = 0:

(1)

TBC =

mC = 56.5 kg

5 Ex 4

1 7 3 5  TAB − TBF −  Ex  = 0 25 5 4  2 Ex 31 = TBF , 4 25

TBF =

TBF =

25 Ex 124

25  35    wD = 0.30680wD 124  23 

wD = ( 34 kg )( 9.81 N/kg )

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)

TBF = 102.3 N

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Chapter 7, Solution 101.

FBD E:

Ey Ex = 1 1

ΣM C = 0:

( 6.5 m ) E y − ( 4.2 m ) Ex − ( 3.5 m ) wD 2.3Ex = 3.5wD ,

FBD CDE:

ΣM B = 0:

Ex = E y

(8.1 m − 3 m )

Ex = E y =

=0

35 wD 23

35 wD − ( 5.1 m ) wD − (1.6 m ) wC = 0 23

wD = 0.60131 wC ,

mD = 0.60131 mC = 33.072 kg mD = 33.1 kg

(a) 4 ΣFx = 0: − TBC = Ex , 5 Point B:

ΣFx = 0: ΣFy = 0: Solving:

TBC =

5 Ex 4

45  1 24 TAB − TBF = 0  Ex  − 54  25 2 1 7 3 5  TAB − TBF +  Ex  = 0 25 5 4  2

31 1 TBF = Ex , 25 4 TBF =

TBF =

25 Ex 124

(

25  35  25 35 ( 33.072 kg ) 9.81 m/s2  wD  = 124  23  124 23 (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.

)

TBF = 99.5 N

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Chapter 7, Solution 102.

(a) FBD half wire:

Since h = 1.5 m L = 30 m we can approximate the load as evenly distributed horizontally, and the length S = L. kg   N  W =  0.6  ( 30 m ) = 176.58 N   9.81 m  kg   ΣM B = 0:

(15 m )(176.58 N ) − (1.5 m ) TC

=0

TC = 1765.8 N Tmax = TB = TC2 + W 2 Tmax =

(b)

(1765.8 N )2 + (176.58 N )2 ,

Tmax = 177.5 kN

2 4   2  yB  2  yB   S B = xB 1 +   −   + ... 3  xB  5  xB      2 4   2  1.5  2  1.5  = ( 30 m ) 1 +  −    + ... = 30.05 m 3  30  5  30   

Note: the third term in the brackets is unnecessary

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.

S tot = 60.1 m

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Chapter 7, Solution 103.

Half-cable FBDs:

T1x = T2 x to create zero horizontal force on tower → thus T01 = T02 FBD I:

(15 m )  w ( 30 m ) − h1T0

ΣM B = 0:

=0

( 450 m ) w = 2

h1 FBD II:

T0

( 2 m ) T0 − (10 m )  w ( 20 m )

ΣM B = 0:

=0

T0 = (100 m ) w

( 450 m ) w = 4.50 m = 2

h1

(a)

ΣFx = 0:

FBD I:

T1x − T0 = 0

T1x = (100 m ) w ΣFy = 0:

T1y − ( 30 m ) w = 0

T1y = ( 30 m ) w

T1 =

(100 m )2 + ( 30 m )2 w

(

= (104.4 m )( 0.4 kg/m ) 9.81 m/s 2

)

= 409.7 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.

(100 m ) w

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ΣFy = 0:

FBD II:

T2y − ( 20 m ) w = 0 T2y = ( 20 m ) w

T2 x = T1x = (100 m ) w T2 =

(100 m )2 + ( 20 m )2 w = 400.17 N (b)

T1 = 410 N T2 = 400 N

*Since h L it is reasonable to approximate the cable weight as being distributed uniformly along the horizontal. The methods of section 7.10 are more accurate for cables sagging under their own weight.

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 7, Solution 104.

FBD half-span:

(a)

 2075 ft    ( 23032.5 kips ) − ( 464 ft ) T0 = 0  2 

ΣM B = 0:

T0 = 47, 246 kips

Tmax = T02 + W 2 =

( 47, 246 kips )2 + ( 23, 033 kips )2

2  2 y 2 s = x 1 +   −  3 x  5 

(b)

= 56, 400 kips

4  y  + " x 

2 4   2  464 ft  2  464 ft  sB = ( 2075 ft ) 1 +  −    + " 3  2075 ft  5  2075 ft   

sB = 2142 ft

l = 2sB

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.

l = 4284 ft

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Chapter 7, Solution 105.

FBD half-span:

W = ( 9.75 kips/ft )(1750 ft ) = 17, 062.5 kips ΣM B = 0: ( 875 ft )(17, 065 kips ) − ( 316 ft ) T0 = 0 T0 = 47, 246 kips Tmax = T02 + W 2 =

( 47, 246 kips )2 + (17, 063 kips )2 (a)

Tmax = 50, 200 kips

2 4   2 y 2 y s = x 1 +   −   + " 3 x  5 x   2 4   2  316 ft  2  316 ft  − sB = (1750 ft ) 1 +     + " 3  1750 ft  5  1750 ft   

= 1787.3 ft

(b) *

To get 3-digit accuracy, only two terms are needed.

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.

l = 2sB = 3575 ft

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Chapter 7, Solution 106.

FBD pipe:

neglecting friction ΣM B = 0:

FBD half-cord:

r (Tmax − WB ) = 0,

Tmax = WB = 60 lb

Assuming the weight to be evenly distributed horizontally, and S=L

W = ( 0.02 lb/ft )( 75 ft ) = 1.5 lb T0 =

( 60 lb )2 − (1.5 lb )2

= 59.981 lb

ΣM B = ( 37.5 ft )(1.5 lb ) − h ( 59.981 lb ) = 0 (a)

(b)

h = 0.93780 ft, h = 11.25 in.

θ B = sin −1

1.5 lb = 1.43254°, 60 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.

θ B = 1.433°

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Chapter 7, Solution 107.

(a) FBD ship:

ΣFx = 0:

T0 − 300 N = 0,

T0 = 300 N

FBD half-span:*

Tmax = T02 + W 2 = (b)

ΣM A = 0: hTx −

L W = 0, 4

2   2 4 s = x 1 +   + L 3 x  

( 2.5 m ) =

L 2

h=

( 300 N )2

2

= ( 54 N ) = 305 N

LW 4Tx

but

yA = h =

LW 4Tx

so

yA W = xA 2Tx

2   2  53.955 N  1 +   − L → L = 4.9732 m 3  600 N   

So h =

LW = 0.2236 m 4Tx

*See note Prob. 7.103

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 = 224 mm

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Chapter 7, Solution 108.

2  2 y 2 s = x 1 +   −  x 3 5    

Knowing

l = 2sTOT

4  y  + L x 

2 2   2 h  2 h  = L 1 +   −   + L 3  L/2  5  L/2     

Winter: 2 4   2  386 ft  2  386 ft  L lw = ( 4260 ft ) 1 +  − +  = 4351.43 ft    3  2130 ft  5  2130 ft   

Summer: 2 4   2  394 ft  2  394 ft  ls = ( 4260 ft )  1 +   −   + L = 4355.18 ft 3  2130 ft  5  2130 ft   

∆l = ls − lw = 3.75 ft

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 7, Solution 109.

FBD whole:

ΣM D = 0:

( 2m ) TAx − ( 5 m )( 98.1 + 147.15) N = 0 TAx = 1103.6 N

ΣFx = 0:

FBD half-cable:

ΣFy = 0:

T0 − TAx = 0,

TAy = 49.05 N = 0,

ΣM A = 0:

T0 = 1103.6 N TAy = 49.05 N

h (1103.6 N ) − ( 2.5 m )( 49.05 N ) = 0 h = 0.11111 m h = 111.1 mm

(a) (b)

θ = tan −1

TAy TAx

= tan −1

49.05 = 2.5449°, 1103.6

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.

θ = 2.54°

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Chapter 7, Solution 110.

FBD AB:

ΣM A = 0:

(1100 ft ) TBy − ( 496 ft ) TBx − ( 550 ft )W 11TBy − 4.96TBx = 5.5W

=0 (1)

FBD CB:

ΣM C = 0:

( 550 ft ) TBy − ( 278 ft ) TBx − ( 275 ft ) 11TBy − 5.56TBx = 2.75W

Solving (1) and (2)

TBy = 28,798 kips

Solving (1) and (2)

TBx = 51, 425 kips tan θ B =

Tmax = TB = TB2x + TB2y ,

So that

(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.

W =0 2 (2)

TB y TBx

Tmax = 58,900 kips

θ B = 29.2°

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Chapter 7, Solution 111.

Cable profile:

Eqn:

y=

w 2 x 2T0

xB − x A = 45 in. at A:

4 in. =

w ( xB − 45 in.)2 2T0

(1)

at B:

1 in. =

w 2 xB 2T0

(2)

∴ or

( xB − 45 in.)2 xB2

=4

xB2 + 30 xB − 675 = 0 xB = ( −15 ± 30 ) in.

xB = 15 in.

x A = xB − 45 in.

x A = − 30 in.

(a)

lowest point (x = 0) is 30 in. from A T0 =

From (2),

wxB2 1  0.18 lb  2 =   (15 in.)

2 2  12 in. 

T0 = 1.6875 lb Tmax occurs at A where slope is maximum Tmax = T02 + ( wx A ) = 2

(1.6875 lb )2 + 

0.18 lb  2  ( − 30 in.) = 1.74647 lb  12 in. 

(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.

Tmax = 1.747 lb

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Chapter 7, Solution 112.

ΣM A = 0:

y AT0 −

a wa = 0 2

ΣM B = 0: − yBT0 + yA =

d = ( yB − yB ) = But

∴

or Using

L = 6 m, yields

wa 2 2T0

yB =

(

w 2 b − a2 2T0

2

(

= w2 b 2 − a 2

)

2

2

(

= L2 w2 4b 2 − 4 Lb + L2

 T2 4 L2 + d 2 b 2 − 4 L3b +  L4 − 4d 2 max w2 

(

d = 0.9 m,

)

Tmax = 8 kN,

b = ( 2.934 ± 1.353) m,

wb 2 2T0

)

2 T0 = TB2 − ( wb ) = Tmax − ( wb )

2 2 − ( wb )  ( 2d )2 Tmax 

b wb = 0 2

)

  = 0 

(

)

w = ( 85 kg/m ) 9.81 m/s2 = 833.85 N/m

b = 4.287 m

( since b > 3 m )

(a)

a = 6 m − b = 1.713 m

2

2 T0 = Tmax − ( wb ) = 7156.9 N

yA wa = = 0.09979, 2T0 xA

yB wb = = 0.24974 2T0 xB

2 2     2  yA  2  yB     l = s A + sB = a 1 +  + L + b 1 + + L    3  xA  3  xB         

2 2 2 2   = (1.713 m ) 1 + ( 0.09979 )  + ( 4.287 m ) 1 + ( 0.24974 )  = 6.19 m 3 3     (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.

l = 6.19 m

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Chapter 7, Solution 113.

x wx − yT0 = 0 2

ΣM P = 0:

Geometry:

y =

wx 2 2T0

y wx = x 2T0

so

and d = yB − y A =

(

w 2 b − a2 2T0

)

Also 2 2   2  yA   2  yB   l = s A + sB a 1 +    + b 1 +    3  a   3  b    

FBD Segment:

l−L

2 2 2  y A  w2 3 y   a + b3   +  B   =  b   6T02 3  a  

(

1 4d 2 = 6 b2 − a 2

(

)

2

(a

3

+b

3

)

)

(

2 3 3 2d a +b = 3 b2 − a 2 2

(

)

)

Using l = 6.4 m, L = 6 m, d = 0.9 m, b = 6 m − a, and solving for a, knowing that a < 3 ft

a = 2.2196 m

a = 2.22 m

(a) T0 =

Then

(

w 2 b − a2 2d

(

)

)

w = ( 85 kg/m ) 9.81 m/s 2 = 833.85 N/m

And with

And b = 6 m − a = 3.7804 m Tmax = TB = T02 + ( wb ) =

T0 = 4338 N

2

( 4338 N )2 + (833.85 N/m )2 ( 3.7804 m )2

Tmax = 5362 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.

(b)

Tmax = 5.36 kN

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Chapter 7, Solution 114.

FBD Cable:

ΣM B = 0: LACy + aT0 − ΣM B loads = 0

(1)

(Where ΣM B loads includes all applied loads)

x  ΣM C = 0: xACy −  h − a  T0 − ΣM C left = 0 L 

FBD AC:

(2)

(Where ΣM C left includes all loads left of C)

x (1) − ( 2 ): L

hT0 −

x ΣM B loads + ΣM C left = 0 L

(3)

ΣM B = 0: LABy − ΣM B loads = 0

(4)

FBD Beam: ΣM C = 0: xABy − ΣM C left − M C = 0

FBD AC:

x ( 4 ) − ( 5): L

−

(5)

x ΣM B loads + ΣM C left + M C = 0 L

Comparing (3) and (6)

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.

M C = hT0

Q.E.D.

(6)

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Chapter 7, Solution 115.

FBD Beam:

ΣM D = 0:

(1 m )( 2.25 kN ) + ( 3 m )( 3 kN ) − ( 5 m ) ABy = 0 A By = 2.25 kN

ΣFy = 0: 2.25 kN − 3 kN − 2.25 kN + DBy = 0 D By = 3 kN FBD AB:

ΣM B = 0:

M B − ( 2 m )( 2.25 kN ) = 0,

M B = 4.5 kN ⋅ m

Note, since A and D are in line horizontally, ACy = A By and DCy = D By .

Also, since Dy > Ay ,

Tmax = TCD = 3.6 kN

2 2 T0 = TCD − DCy =

Cable:

( 3.6 kN )2 − ( 3 kN )2

T0 = 3.96 kN

dB =

M B 4.5 kN ⋅ m = = 2.2613 m T0 3.96 kN d B = 2.26 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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 116.

FBD Beam:

ΣM E = 0:

(1 m )(1.2 kN ) + ( 2 m )( 3.6 kN ) + ( 3 m )(1.8 kN ) − ( 4 m ) ABy = 0

A By = 3.45 kN

ΣFx = 0: ABx = 0 FBD AB:

ΣM B = 0: M B − (1 m )( 3.45 kN ) = 0,

M B = 3.45 kN ⋅ m

FBD AC:

ΣM C = 0: M C + (1 m )(1.8 kN ) − ( 2 m )( 3.45 kN ) = 0 M C = 5.1 kN ⋅ m ΣM D = 0: M D + (1 m )( 3.6 kN ) + ( 2 m )(1.8 kN ) − ( 3 m )( 3.45 kN ) = 0 FBD AD:

M D = 3.15 kN ⋅ m hC = dC − 0.6 m = 1.8 m − 0.6 m = 1.2 m

Cable:

T0 =

M C 5.1 kN ⋅ m = = 4.25 kN hC 1.2 m

hB =

M B 3.45 kN ⋅ m = = 0.81176 m T0 4.25 kN d B = hB + 0.3 m = 1.11176 m,

hD =

d B = 1.112 m

M D 3.15 kN ⋅ m = = 0.74118 m T0 4.25 kN d D = hD + 0.9 m = 1.64118 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.

d D = 1.641 m

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Chapter 7, Solution 117.

FBD Beam:

ΣFx = 0:

ABx = 0

By symmetry A By = FBy = 800 lb

FBD AC:

ΣM C = 0: M C + ( 6 ft )( 400 lb ) − (12 ft )( 800 lb ) = 0 M C = 7200 lb ⋅ ft By symmetry, M D = M C = 7200 lb ⋅ ft ∴

hC = hD

Cable:

hC = dC − 3 ft = 12 ft − 3 ft = 9 ft d D = hD + 2 ft = 9 ft + 2 ft = 11 ft d D = 11.00 ft

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 7, Solution 118.

FBD Beam:

ΣFx = 0:

ABx = 0

By symmetry A By = FBy = 800 lb

FBD AC:

ΣM C = 0:

M C + ( 6 ft )( 400 lb ) − (12 ft )( 800 lb ) = 0 M C = 7200 lb ⋅ ft

By symmetry, M D = M C = 7200 lb ⋅ ft ∴

Cable:

hC = hD

hC = dC − 3 ft = 9 ft − 3 ft = 6 ft d D = hD + 2 ft = 6 ft + 2 ft = 8 ft d D = 8.00 ft

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 7, Solution 119.

FBD Elemental segment:

Ty ( x + ∆x ) − Ty ( x ) − w ( x ) ∆ x = 0

ΣFy = 0:

Ty ( x + ∆x )

So

T0

Ty

But

T0 dy dx

So

− x + ∆x

∆x

In

lim :

∆x → 0

dy dx

x

=

dy dx

=

w( x) T0

w( x) d2y = 2 T0 dx

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.

−

Ty ( x ) T0

=

Q.E.D.

w( x) ∆x T0

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Chapter 7, Solution 120.

w ( x ) = w0 cos

πx L

From Problem 7.119

w ( x ) w0 πx d2y = = cos 2 T0 T0 L dx  dy  using dx 

dy W0 L πx = sin dx T0π L

So

y =

But

And

w0 L2  πx 1 − cos   using y ( 0 ) = 0  2  L   T0π 

w L2  π L y   = h = 0 2 1 − cos  2 T0π  2 T0 = Tmin Tmax = TA = TB :

TBy =

T0

=

T0 =

so

Tmin =

so TBy

dy dx

= x = L2

w0 L T0π

w0 L

π

2 TB = TBy + T02 =

w0 L

π

0

 = 0 

 L  1+   πh 

2

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.

w0 L2 π 2h

w0 L2 π 2h

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Chapter 7, Solution 121.

Elemental Segment:

dx = cosθ ds,

But

w0 ds cos 2 θ w0 w( x) = cos3 θ

w ( x ) dx =

Load on segment*

so

From Problem 7.119

d2y w( x) w0 = = 2 T0 dx T0 cos3 θ

In general

d2y d  dy  d dθ = ( tan θ ) = sec2 θ  = 2 dx  dx  dx dx dx

So

dθ w0 w0 = = 3 2 dx T0 cosθ T0 cos θ sec θ

or

T0 cosθ dθ = dx = rdθ cosθ w0

Giving r =

T0 = constant. So curve is circular arc w0

*For large sag, it is not appropriate to approximate ds by dx.

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.

Q.E.D.

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Chapter 7, Solution 122.

(

)

w = ( 0.07 kg/m ) 9.81 m/s 2 = 0.6867 N/m

FBD half-span:

L = 10 m SCB =

12 m =6m 2

SCB = c sinh

xB , c

6 m = c sinh

5m c

Solving numerically, c = 4.6954 m yB = c cosh

 5m  xB = ( 4.6954 m ) cosh   = 7.6188 m c  4.6954 m 

hB = yB − c = 7.6188 m − 4.6954 m = 2.9234 m (a)

hB = 2.92 m

TB = wyB = ( 0.6867 N/m )( 7.6188 m ) = 5.2318 N (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.

TB = 5.23 N

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Chapter 7, Solution 123.

sB = 30 ft,

w=

hB = 24 ft,

120 lb = 2 lb/ft 60 ft L 2

xB =

yB2 = c 2 + sB2 = ( hB + c )

2

= hB2 + 2chB + c 2

( 30 ft ) − ( 24 ft ) sB2 − hB2 = 2hB 2 ( 24 ft ) 2

c=

2

c = 6.75 ft Then

sB = c sinh

xB s → xB = c sinh −1 B c c

 30 ft  xB = ( 6.75 ft ) sinh −1   = 14.83 ft  6.75 ft  (a)

L = 2 xB = 29.7 ft

Tmax = TB = wyB = w ( c + hB ) = ( 2 lb/ft )( 6.75 ft + 24 ft ) = 61.5 lb (b)

Tmax = 61.5 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.

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Chapter 7, Solution 124.

S B = 250 ft

FBD half-span:

w = 2.8 lb/ft hB = 125 ft yB = hB + c = 125 ft + c yB2 − sB2 = c 2

(125 ft − c )2 − ( 250 ft )2 = c 2 c = 187.5 ft

sB = c sinh

xB , c

xB 4 = sinh −1 = 1.0986, c 3

250 ft = (187.5 ft ) sinh

xB c

xc = 205.99 ft

(a)

span L = 2 xB = 411.98 ft

L = 412 ft

(b)

Tmax = wyB = ( 2.8 lb/ft )(125 ft + 187.5 ft ) = 875 lb Tmax = 875 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 7, Solution 125.

FBD half-span:

sB = 65 m,

(

hB = 30 m

)

w = ( 3.4 kg/m ) 9.81 m/s 2 = 33.35 N/m

yB2 = c 2 + sB2

( c + hB )2

= c 2 + sB2 2

c=

( 65 m ) − ( 30 m ) sB2 − hB2 = 2hB 2 ( 30 m )

2

= 55.417 m

Now

sB = c sinh

xB s  65 m  → xB = c sinh −1 B = ( 55.417 m ) sinh −1   c c  55.417 m  = 55.335 m

L = 2 xB = 2 ( 55.335 m ) = 110.7 m Tmax = wyB = w ( c + hB ) = ( 33.35 N/m )( 55.417 m + 30 m ) = 2846 N Tmax = 2.85 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 7, Solution 126.

FBD Cable:

s = 30 m

30 m   = 15 m   so sB = 2  

(

)

w = ( 0.3 kg/m ) 9.81 m/s 2 = 2.943 N/m hB = 12 m yB2 = ( c + hB ) = c 2 + s 2B 2

c=

So

c= Now

sB = c sinh

sB2 − hB2 2hB

(15 m )2 − (12 m )2 2 (12 m )

= 3.375 m

xB s  15 m  → xB = c sinh −1 B = ( 3.375 m ) sinh −1   c c  3.375 m  xB = 7.4156 m

P = T0 = wc = ( 2.943 N/m )( 3.375 m ) L = 2 xB = 2 ( 7.4156 m )

(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.

P = 9.93 N L = 14.83 m

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Chapter 7, Solution 127.

FBD Cable:

sT = 30 m,

(

)

w = ( 0.3 kg/m ) 9.81 m/s 2 = 2.943 N/m P = T0 = wc, c=

c=

P w

30 N = 10.1937 m 2.943 N/m

yB2 = ( hB + c ) = c 2 + sB2 2

h 2 + 2ch − sB2 = 0,

sB =

30 m = 15 m 2

h 2 + 2 (10.1937 m ) h − 225 m 2 = 0 h = 7.9422 m sB = c sinh

(a)

xA s  15 m  → xB = c sinh −1 B = (10.1937 m ) sinh −1   c c  10.1937 m 

= 12.017 m L = 2 xB = 2 (12.017 m ) (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.

L = 24.0 m

h = 7.94 m

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Chapter 7, Solution 128.

FBD half-span:

xD =

3.6 w = 1.8 m 2

kg  m  w =  3.72  9.81 2  = 36.4932 N/m m s    sD =

length 3.8 m = = 1.9 m 2 2

sD = c sinh

xD c

1.9 m = c sinh

1.8 m c

Solving numerically, c = 3.1433 m yD2 − sD2 = c 2

yD2 = (1.9 m ) + ( 3.1433 m ) 2

2

yD = 3.6729 h = yD − c = 3.6729 m − 3.1433 m = 0.5296 m h = 0.530 m (a)

h = 530 mm

(b)

TDx = 114.7 N

N  TDx = wc =  36.4932  ( 3.1433 m ) = 114.71 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 7, Solution 129.

FBD half-span:

sB =

90 m = 45 m 2

xB =

L 60 m = = 30 m 2 2

TB = Tmax = 300 N sB = c sinh

xB , c

Solving numerically, yB = c cosh

45 m = c sinh

30 m c

c = 18.495 m

 30 m  xB = (18.495 m ) cosh   c  18.495 m 

= 48.651 m hB = yB − c = 48.651 m − 18.495 m = 30.156 m (a) Tmax = wyB

h = 30.2 m

300 N = w ( 48.651 m ) w = 6.1664

N m

N  W = w ( length ) =  6.1664  ( 90 m ) = 554.97 N m  m=

W 554.97 N = = 56.57 kg g 9.81 m/s 2 (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.

m = 56.6 kg

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Chapter 7, Solution 130.

sB =

45 ft = 22.5 ft 2 xB =

L = 20 ft

L = 10 ft 2

sB = c sinh

xB c

22.5 ft = c sinh

10 ft c

Solving numerically:

yB = c cosh

c = 4.2023 ft

xB c

= ( 4.2023 ft ) cosh

10 ft = 22.889 ft 4.2023 ft

hB = yB − c = 22.889 ft − 4.202 ft hB = 18.69 ft

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 7, Solution 131.

l = total length xB , c

yB = c cosh

L + c = c cosh 1 = cosh

or

So

xB c

c=

L L − 2c c L = 4.933 c

Solving numerically, sB = c sinh

L/2 c

l L = c sinh 2 2c l

 L 2sinh    2c 

=

10 ft = 0.8550 ft  4.933  2sinh    2 

L = 4.933 c = ( 4.933)( 0.8550 ft ) = 4.218 ft (a) from y = c cosh

x c

tan θ B =

L = 4.22 ft

dy x = sinh dx c dy dx

= sinh B

L  4.933  = sinh   = 5.848 2c  2 

(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.

θ B = 80.3°

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Chapter 7, Solution 132.

(

)

w = ( 3 kg/m ) 9.81 m/s 2 = 29.43 N/m L = 48 m,

Tmax ≤ 1800 N

Tmax = wyB → yB =

yB ≤

Tmax w

1800 N = 61.162 m 29.43 N/m

x yB = c cosh B , c Solving numerically,

48 m 61.162 m = c cosh 2 * c

c = 55.935 m

h = yB − c = 61.162 m − 55.935 m

h = 5.23 m *Note: There is another value of c which will satisfy this equation. It is much smaller, thus corresponding to a much larger h.

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 7, Solution 133.

Pulley B:

TB = wa but TB = wyB , now So

yB = c cosh a = c cosh

xB =a c 3 ft c

sB = c sinh

also

yB = a

so

(1) xB c

24 ft − a 3 ft = c sinh 2 c or

12 ft = c sinh = c sinh

3 ft a + c 2

(2)

3 ft c 3 ft + cosh c 2 c

c = 1.13194 ft

Solving numerically

or 17.7167 ft a = c cosh

from (1)

3 ft 3 ft = (1.13194 ft ) cosh c 1.13194 ft a = 8.05 ft

or

  3 ft a = (17.7167 ft ) cosh    17.7167 ft  a = 17.97 ft

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 7, Solution 134.

FBD Cable:

s A = length = 10 ft ΣFx = 0:

− F + T0 = 0,

y A = c + h = c cosh So

1+

xA a h = c cos h = c cosh c c c

h h = cosh c c

Solving numerically, s A = c sinh

F = T0

h = 1.61614 c

h = c sinh (1.61614 ) = 2.41748c = 10 ft c

So c = 4.1365 ft,

h = 6.6852 ft

y A = c + h = 10.8217 ft  20 lb  F = T0 = wc =   ( 4.1365 ft ) = 8.2730 lb  10 ft  (a)

F = 8.27 lb

(b)

h = a = 6.69 ft

(c)

Tmax = 21.6 lb

 lb  Tmax = wy A =  2  (10.8217 ft ) = 21.643 lb  ft 

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 7, Solution 135.

FBD Cable:

Pulley

TA = mg = Tmax m  mg = ( 40 kg )  9.81 2  = 392.4 N s   Also L = 15 m, So x A = −

15 m = − 7.5 m, 2

y A = c cosh So

h=5m

c cosh

xB = + 7.5 m

xA =c+h c

− 7.5 m =c+5m c

Solving numerically, sB = c sinh

c = 6.3175 m

 7.5 m  xB = ( 6.3175 m ) sinh   = 9.390 m c  6.3175 m  cable length = 2sB = 18.78 m

(a)

Tmax = wyB = w ( c + h ) = w ( 6.3175 m + 5 m ) w= (b)

392.4 N = 34.672 N/m 11.3175 m mass/length =

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 34.672 N/m kg = = 3.53 g 9.81 N/kg m

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Chapter 7, Solution 136.

yD = c cosh

xD c

h + c = c cosh

a c

10.8 ft   12 ft = c  cosh − 1 c   Solving numerically, c = 6.2136 ft Then

yB = ( 6.2136 ft ) cosh

10.8 ft = 18.2136 ft 6.2136 ft

F = Tmax = wyB = (1.5 lb/ft )(18.2136 ft ) F = 27.3 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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 7, Solution 137.

yD = c cosh

xD c

c + h = c cosh

a c

a   h = c  cosh − 1 c  

18 ft   12 ft = c  cosh − 1 c   Solving numerically

c = 15.162 ft

yB = h + c = 12 ft + 15.162 ft = 27.162 ft

F = TD = wyD = (1.5 lb/ft )( 27.162 ft ) = 40.74 lb F = 40.7 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.

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Chapter 7, Solution 138.

(

)

w = 4 kg/m 9.81 m/s 2 = 39.24 N/m P = T0 = wc,

c=

P 800 N = w 39.24 N/m

c = 20.387 m y = c cosh

x c

dy x = sinh dx c tan θ = −

dy dx

= − sinh −a

−a a = sinh c c

a = c sinh −1 ( tan θ ) = ( 20.387 m ) sinh −1 ( tan 60° ) a = 26.849 m y A = c cosh

a 26.849 m = ( 20.387 m ) cosh = 40.774 m c 20.387 m

b = y A − c = 40.774 m − 20.387 m = 20.387 m So

s = c sinh

(a)

B is 26.8 m right and 20.4 m down from A

a 26.849 m = ( 20.387 m ) sinh = 35.31 m c 20.387 m (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.

s = 35.3 m

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Chapter 7, Solution 139.

(

)

w = ( 4 kg/m ) 9.81 m/s 2 = 39.24 N/m P = T0 = wc

c=

P 600 N = w 39.24 N/m

c = 15.2905 m x y = c cosh , c At A: So

tan θ = −

dy dx

dy x = sinh dx c = − sinh −a

−a a = sinh c c

a = c sinh −1 ( tan θ ) = (15.2905 m ) sinh −1 ( tan 60° ) = 20.137 m

yB = h + c = c cosh

a c

a   h = c  cosh − 1 c     20.137 m = (15.2905 m )  cosh − 1 15.2905 m   = 15.291 m So

s = c sinh

(a)

B is 20.1 m right and 15.29 m down from A

a 20.137 m = (15.291 m ) sinh = 26.49 m c 15.291 m (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.

s = 26.5 m

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Chapter 7, Solution 140.

Cable:

Since y A = c cosh

xA , c

yB = c cosh

xB c

( xA < 0 )

and xB − x A = 8 m

( c + 0.5 m ) = c cosh

xA c xB    c 

( c + 1.2 m ) = c cosh  1.2 m  0.5 m  8 m  −1  cosh −1 1 +  + cosh 1 + = c  c  c   Solving numerically, So

c = 9.9987 m   0.5 m   x A = cosh −1  1 +   ( 9.9987 m ) 9.9987 m     = 3.15 m C is 3.15 m from house

(a)

Tmax = TB = wyB

kg   N  =  2.1  (1.2 m + 9.9987 m )   9.81 m  kg   = 230.7 N (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.

Tmax = 231 N

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Chapter 7, Solution 141.

y A = c cosh

−a = c + 6 ft c

6 ft   a = c cosh −1 1 +  c   yB = c cosh

b = c + 11.4 ft c

11.4 ft   b = c cosh −1 1 +  c   So Solving numerically,

 6 ft  11.4 ft    −1  a + b = c cosh −1 1 +  + cosh 1 +   = 36 ft c  c     c = 20.446 ft

11.4 ft   b = ( 20.446 ft ) cosh −1 1 +  = 20.696 ft 20.446 ft   (a)

C is 20.7 ft left of and 11.4 ft below B

 20.696 ft  Tmax = wyB = ( 0.3 lb/ft )( 20.446 ft ) cosh   = 9.554 lb  20.446 ft  (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.

Tmax = 9.55 lb

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Chapter 7, Solution 142.

(a)

tan θ =

dy x = sinh dx c

s = c sinh

x = c tan θ, Q.E.D. c

( cosh x = sinh ( tan θ + 1) = c sec θ

y 2 = s2 + c2 ,

2

So

y2 = c2

2

And

y = c secθ , Q.E.D.

(b)

Also

2

2

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.

2

)

x +1

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Chapter 7, Solution 143.

TB = Tmax = wyB = wc cosh L 2c

ξ =

Let

so

xB L  2c  L = w   cosh c 2 L  2c

Tmax =

wL cosh ξ 2ξ

 dTmax wL  1 =  sinh ξ − cosh ξ  dξ 2ξ  ξ  tanh ξ −

min Tmax ,

For

Solving numerically

(Tmax )min

=

=0

ξ

ξ = 1.1997

wL cosh (1.1997 ) = 0.75444wL 2 (1.1997 ) Lmax =

(a) If

Tmax T = 1.3255 max 0.75444w w

(

)

Tmax = 32 kN and w = ( 0.34 kg/m ) 9.81 m/s 2 = 3.3354 N/m

Lmax = 1.3255 (b)

1

32.000 N = 12 717 m 3.3354 N/m Lmax = 12.72 km

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 7, Solution 144.

ymax = c cosh Tmax = wymax , ymax =

L =h+c 2c ymax =

80 lb = 40 ft 2 lb/ft c cosh

Solving numerically,

Tmax w

9 ft = 40 ft c

c1 = 2.6388 ft c2 = 38.958 ft h = ymax − c h1 = 40 ft − 2.6388 ft

h1 = 37.4 ft

h2 = 40 ft − 38.958 ft

h2 = 1.042 ft

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 7, Solution 145.

Tmax = wyB = 2wsB y B = 2sB c cosh

L L = 2c sinh 2c 2c

tanh

L 1 = 2c 2

L 1 = tanh −1 = 0.549306 2c 2 hB y −c L = B = cosh −1 c c 2c

= 0.154701 hB hB /c = L 2( L/2c) =

0.5 ( 0.154701) = 0.14081 0.549306 hB = 0.1408 L

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 7, Solution 146.

Tmax = wyB = wc cosh

(a)

L 2c

dTmax L L L  = w  cosh − sinh  dc 2c 2c 2c   For

min Tmax ,

tanh

L 2c = 2c L

dTmax =0 dc

L = 1.1997 2c

yB L = cosh = 1.8102 2c c h y = B − 1 = 0.8102 c c h h  1 h  2c   0.8102 = = 0.3375 = 0.338   = L  2 c  L   2 (1.1997 ) L

T0 = wc,

(b)

L , 2c

T0 = Tmax cosθ B ,

But So

Tmax = wc cosh

Tmax L y = cosh = B 2c T0 c Tmax = secθ B T0

y 

θ B = sec−1  B  = sec−1 (1.8102 )  c  = 56.46°

Tmax = wyB = w

θ B = 56.5°

yB  2c  L  L    = w (1.8102 ) c  L  2  2 (1.1997 ) Tmax = 0.755wL

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 7, Solution 147.

FBD AB:

4  3  ΣM A = 0: r  C  + r  C  − 2r ( 70 lb ) = 0 5   5  C = 100 lb

ΣFx = 0: − Ax +

4 (100 lb ) = 0 5

A x = 80 lb ΣFy = 0: Ay +

3 (100 lb ) − 70 lb = 0 5

A y = 10 lb

FBD AJ:

ΣFx′ = 0: F − ( 80 lb ) sin 30° − (10 lb ) cos30° = 0 F = 48.66 lb F = 48.7 lb

60°

ΣFy′ = 0: V − ( 80 lb ) cos30° + (10 lb ) sin 30° = 0 V = 64.28 lb V = 64.3 lb

30°

ΣM 0 = 0: ( 8 in.)( 48.66 lb ) − (8 in.)(10 lb ) − M = 0

M = 309.28 lb ⋅ in.

M = 309 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 7, Solution 148.

FBD Whole: ΣM A = 0: ( 5.4 ft ) Bx − ( 7.8 ft )( 90 lb ) = 0

B x = 130 lb

FBD BE with pulleys and cord:

ΣM E = 0: ( 5.4 ft )(130 lb ) − ( 7.2 ft ) By + ( 4.8 ft )( 90 lb ) − ( 0.6 ft )( 90 lb ) = 0

B y = 150 lb ΣFx = 0: Ex − 130 lb = 0

E x = 130 lb ΣFy = 0: E y + 150 lb − 90 lb − 90 lb = 0

E y = 30 lb ΣFx′ = 0: − F − 90 lb +

FBD JE and pulley:

4 3 (130 lb ) + ( 90 lb − 30 lb ) = 0 5 5 F = 50.0 lb

3 4 (130 lb ) + ( 30 lb − 90 lb ) = 0 5 5 V = −30 lb V = 30.0 lb

ΣFy′ = 0: V +

ΣM J = 0: − M + (1.8 ft )(130 lb ) + ( 2.4 ft )( 30 lb ) + ( 0.6 ft )( 90 lb ) − ( 3.0 ft )( 90 lb ) = 0

M = 90.0 lb ⋅ ft

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 7, Solution 149.

FBD Rod: ΣFx = 0: ΣM B = 0:

Ax = 0 2r

π

W − rAy = 0

α = 15°, weight of segment = W FBD AJ: r =

r

α

ΣFy′ = 0:

sin α = 2W

π F=

Ay =

2W

π

30° W = 90° 3

r sin15° = 0.9886r π /12 cos 30° −

W cos30° − F = 0 3

W 3  2 1  −  2 π 3

2W  W  ΣM 0 = M + r  F − =0  + r cos15° 3 π   M = 0.0557 Wr

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 7, Solution 150.

(a)

Along AC:

ΣFy = 0: ΣM J = 0:

− 3 kip − V = 0

V = −3 kips

M + x ( 3 kips ) = 0

M = ( 3 kips ) x

M = −9 kip ⋅ ft at C Along CD:

ΣFy = 0:

− 3 kips − 5 kips − V = 0

ΣM K = 0:

V = −8 kips

M + ( x − 3 ft )( 5 kips ) + x ( 3 kips ) = 0 M = +15 kip ⋅ ft − ( 8 kips ) x

M = −16.2 kip ⋅ ft at D ( x = 3.9 ft ) Along DE:

ΣFy = 0:

− 3 kips − 5 kips + 6 kips − V = 0 V = −2 kips

ΣM L = 0:

M − x1 ( 6 kips ) + (.9 ft + x1 )( 5 kips ) + ( 3.9 ft + x1 )( 3 kips ) = 0 M = −16.2 kip ⋅ ft − ( 2 kips ) x1

M = −18.6 kip ⋅ ft at E

( x1 = 1.2 ft ) 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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Along EB:

ΣFy = 0: −3 kips − 5 kips + 6 kips − 4 kips − V = 0

ΣM N = 0:

V = −6 kips

M + ( 4 kips ) x2 + ( 2.1 ft + x2 )( 5 kips ) + ( 5.1 ft + x2 )( 3 kips ) − (1.2 ft + x2 )( 6 kips ) = 0 M = −18.6 kip ⋅ ft − ( 6 kips ) x2 M = −33 kip ⋅ ft at B

(b)

From diagrams:

( x2

= 2.4 ft ) V

max

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.

max

= 8.00 kips on CD = 33.0 kip ⋅ ft at B

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Chapter 7, Solution 151.

(a)

By symmetry:

Ay = B = 8 kN +

1 ( 4 kN/m )( 5 m ) 2

A y = B = 18 kN

Along AC:

ΣFy = 0: ΣM J = 0:

18 kN − V = 0 M − x (18 kN )

V = 18 kN M = (18 kN ) x

M = 36 kN ⋅ m at C ( x = 2 m ) Along CD:

ΣFy = 0:

18 kN − 8 kN − ( 4 kN/m ) x1 − V = 0 V = 10 kN − ( 4 kN/m ) x1 V = 0 at x1 = 2.5 m ( at center )

ΣM K = 0: M +

x1 ( 4 kN/m ) x1 + (8 kN ) x1 − ( 2 m + x1 )(18 kN ) = 0 2

M = 36 kN ⋅ m + (10 kN/m ) x1 − ( 2 kN/m ) x12

M = 48.5 kN ⋅ m at x1 = 2.5 m Complete diagram by symmetry (b)

From diagrams:

V

max

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.

= 18.00 kN on AC and DB max

= 48.5 kN ⋅ m at center

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Chapter 7, Solution 152.

Ay = B = 60 kN − P

By symmetry: Along AC:

ΣM J = 0: M − x ( 60 kN − P ) = 0

M = ( 60 kN − P ) x

M = 120 kN ⋅ m − ( 2 m ) P at x = 2 m Along CD:

ΣM K = 0:

M + ( x − 2 m )( 60 kN ) − x ( 60 kN − P ) = 0 M = 120 kN ⋅ m − Px M = 120 kN ⋅ m − ( 4 m ) P at x = 4 m

Along DE:

ΣM L = 0:

M − ( x − 4 m ) P + ( x − 2 m )( 60 kN ) − x ( 60 kN − P ) = 0 M = 120 kN ⋅ m − ( 4 m ) P

(const)

Complete diagram by symmetry For minimum M

max

, set M max = −M min

120 kN ⋅ m − ( 2 m ) P = − 120 kN ⋅ m − ( 4 m ) P 

(a) (b)

P = 40.0 kN

M min = 120 kN ⋅ m − ( 4 m) P

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.

M

max

= 40.0 kN ⋅ m

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Chapter 7, Solution 153.

FBD Beam:

ΣM A = 0:

( 4 m ) B − (1 m )( 20 kN/m )( 2 m ) − M B = 10 kN +

(a)

(a)

B = 10 kN

(b)

B = 13 kN

ΣFy = 0:

=0

M 4m

Ay − ( 20 kN/m )( 2 m ) + B = 0 Ay = 40 kN − B

(a)

A y = 30 kN

(b)

A y = 27 kN

Shear Diags:

 dV  VA = Ay , then V is linear  = −20 kN/m  to C.  dx  VC = Ay − ( 20 kN/m )( 2 m ) = Ay − 40 kN

(b)

(a)

VC = −10 kN

(b)

VC = −13 kN V = 0 = Ay − ( 20 kN/m ) x1 at x1 =

(a)

x1 = 1.5 m

(b)

x1 = 1.35 m

Ay m 20 kN

V is constant from C to B.

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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Moment Diags:

 dM  M A = applied M . Then M is parabolic  decreases with V  dx   M is max where V = 0. M max = M + (a)

M

max

1 Ay x1.

2 1 ( 30 kN )(1.5 m ) = 22.5 kN ⋅ m 2

=

1.500 m from A (b)

M max = 12 kN ⋅ m +

1 ( 27 kN )(1.35 m ) = 30.225 kN ⋅ m 2

M M C = M max −

max

= 30.2 kN, 1.350 m from A

1 VC ( 2 m − x1 ) 2

(a)

M C = 20 kN ⋅ m

(b)

M C = 26 kN ⋅ m  dM  Finally, M is linear  = VC  to zero at B.  dx 

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 7, Solution 154.

(b)

(a)

x  Distributed load w = w0 1 −  L   ΣM A = 0: ΣFy = 0:

1    total = w0 L  2  

L 1   w0 L  − LB = 0 3 2  Ay −

B=

1 wL w0 L + 0 = 0 2 6

w0 L 6

Ay =

w0 L 3

Shear:

VA = Ay =

w0 L , 3

dV = − w → V = VA − dx

Then



x

x

∫0 w0 1 − L  dx

 1 x 1  x 2  1 w0 2 w L V =  0  − w0 x + x = w0 L  − +    2 L  3   3 L 2  L   Note: At x = L, V = −

w0 L ; 6

2

x x x 2 V = 0 at   − 2   + = 0 → =1− 3 L L L    

1 3

Moment: M A = 0,

Then

 dM   =V → M =  dx 

M = w0 L2 ∫

x

x / L 1

0

x/L

∫0 Vdx = L∫0

x 1 x   − +   3 2 L L 

x x V  d   L L

2

x d    L 

 1  x  1  x  2 1  x 3  M = w0 L2    −   +    6  L    3  L  2  L  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

 x =1− M max  at  L

1 2  = 0.06415w0 L 3 

(a)

 1 x 1  x 2  V = w0 L  − +     3 L 2  L  

 1  x  1  x  2 1  x 3  M = w0 L2    −   +    6  L    3  L  2  L  (c)

M max = 0.0642 w0 L2 at x = 0.423L

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 7, Solution 155.

ΣFy = 0: wg L −

(a)

wg L = Define ξ =

L 4w0

∫0

L2

( Lx − x ) dx = 0 2

4w0  1 2 1 3  2  LL − L  = w0 L 3  3 L2  2

2w0 3

wg =

 x  x 2  2  −    − w0  L  L   3

x dx so dξ = → net load w = 4w0 L L  1  w = 4w0  − + ξ − ξ 2  6  

or

V = V(0) −

ξ

 1

∫ 0 4w0 L  − 6 + ξ − ξ 1 6

1 2

1 3

2

 dξ 

 

= 0 + 4w0 L  ξ + ξ 2 − ξ 3  V = M = M0 + = (b)

x

2

ξ

∫ 0 Vdx = 0 + 3 w0 L ∫ 0 2

(

2 w0 L ξ − 3ξ 2 + 2ξ 3 3

(ξ − 3ξ

2

)

)

+ 2ξ 3 dξ

(

2 1  1 1 w0 L2  ξ 2 − ξ 3 + ξ 4  = w0 L2 ξ 2 − 2ξ 3 + ξ 4 3 2  3 2

Max M occurs where V = 0 → 1 − 3ξ + 2ξ 2 = 0 → ξ =

)

1 2

1 1 1  w L2  1 2 M  ξ =  = w0 L2  − +  = 0 2 3 48   4 8 16 

M max =

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.

w0 L2 at center of beam 48

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Chapter 7, Solution 156.

(a) FBD cable:

ΣM E = 0:

( 4 m )(1.2 kN ) + (8 m )( 0.8 kN ) + (12 m )(1.2 kN ) − ( 3 m ) Ax − (16 m ) Ay = 0 3 Ax + 16 Ay = 25.6 kN

(1)

ΣM C = 0: ( 4 m )(1.2 kN ) + (1 m ) Ax − ( 8 m ) Ay = 0

FBD ABC:

Ax − 8 Ay = −4.8 kN Solving (1) and (2)

Ax = 3.2 kN

Ay = 1 kN So A = 3.35 kN

(b) cable:

(2)

17.35°

ΣFx = 0: − Ax + Ex = 0 Ex = Ax = 3.2 kN ΣFy = 0: Ay − (1.2 + 0.8 + 1.2 ) kN + E y = 0 E y = 3.2 kN − Ay = ( 3.2 − 1) kN = 2.2 kN So E = 3.88 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.

34.5°

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Chapter 7, Solution 157.

FBD AC:

FBD CB:

ΣM A = 0: (13.5 ft ) T0 −

a ( 57.5 lb/ft ) a = 0 2

(

)

T0 = 2.12963 lb/ft 2 a 2 ΣM B = 0:

60 ft − a ( 57.5 lb/ft )( 60 ft − a ) − ( 6 ft ) T0 = 0 2

(

)

6T0 = 28.75 lb/ft 2 3600 ft 2 − (120 ft ) a + a 2  Using (1) in (2), Solving: So (a)

0.55a 2 − (120 ft ) a + 3600 ft 2 = 0

a = (108 ± 72 ) ft,

a = 36 ft (180 ft out of range)

C is 36 ft from A C is 6 ft below and 24 ft left of B T0 = 2.1296 lb/ft 2 ( 36 ft ) = 2760 lb 2

W1 = ( 57.5 lb/ft )( 36 ft ) = 2070 lb (b)

Tmax = TA = T02 + W12 =

( 2760 lb )2 + ( 2070 lb )2

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.

= 3450 lb

(2)

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Chapter 7, Solution 158.

sB = 100 ft,

w=

4 lb = 0.02 lb/ft 200 ft

Tmax = 16 lb Tmax = TB = wyB yB =

TB 16 lb = = 800 ft w 0.02 lb/ft c 2 = yB2 − sB2

c= But

(800 ft )2 − (100 ft )2

yB = xB cosh

= 793.73 ft

xB y → xB = c cosh −1 B c c

 800 ft  = ( 793.73 ft ) cosh −1   = 99.74 ft  793.73 ft  L = 2 xB = 2 ( 99.74 ft ) = 199.5 ft

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.