Cap 03

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Chapter 3, Solution 1.

Resolve 90 N force into vector components P and Q where Q = ( 90 N ) sin 40° = 57.851 N Then M B = − rA/BQ

= − (0.225 m )(57.851 N ) = −13.0165 N ⋅ m M B = 13.02 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 3, Solution 2.

Fx = ( 90 N ) cos 25°

= 81.568 N Fy = ( 90 N ) sin 25° = 38.036 N x = ( 0.225 m ) cos 65° = 0.095089 m

y = (0.225 m ) sin 65° = 0.20392 m

M B = xFy − yFx = ( 0.095089 m )( 38.036 N ) − ( 0.20392 m )( 81.568 N ) = −13.0165 N ⋅ m

M B = 13.02 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 3, Solution 3.

Px = ( 3 lb ) sin 30° = 1.5 lb

Py = ( 3 lb ) cos 30° = 2.5981 lb

M A = xB/ A Py + yB/ A Px = ( 3.4 in.)( 2.5981 lb ) + ( 4.8 in.)(1.5 lb ) = 16.0335 lb ⋅ in.

M A = 16.03 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 3, Solution 4.

For P to be a minimum, it must be perpendicular to the line joining points A and B with rAB =

( 3.4 in.)2 + ( 4.8 in.)2

= 5.8822 in.  y  

α = θ = tan −1   x  4.8 in.  = tan −1    3.4 in. 

= 54.689° Then

M A = rAB Pmin

or

Pmin =

M A 19.5 lb ⋅ in. = rAB 5.8822 in.

= 3.3151 lb ∴ Pmin = 3.32 lb

54.7° or

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

Pmin = 3.32 lb

35.3°

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Chapter 3, Solution 5.

M A = rB/ A P sin θ

By definition where

θ = φ + ( 90° − α )

and

φ = tan −1 

 4.8 in.    3.4 in. 

= 54.689° Also

rB/ A =

( 3.4 in.)2 + ( 4.8 in.)2

= 5.8822 in.

Then

(17 lb ⋅ in.) = ( 5.8822 in.)( 2.9 lb ) sin ( 54.689° + 90° − α )

or

sin (144.689° − α ) = 0.99658

or

144.689° − α = 85.260°; 94.740° ∴ α = 49.9°, 59.4°

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 3, Solution 6.

(a)

(a) M A = rB/ A × TBF

M A = xTBFy + yTBFx = ( 2 m )( 200 N ) sin 60° + ( 0.4 m )( 200 N ) cos 60° = 386.41 N ⋅ m or M A = 386 N ⋅ m (b)

(b) For FC to be a minimum, it must be perpendicular to the line joining A and C.

∴ M A = d ( FC )min d =

with

( 2 m )2 + (1.35 m )2

= 2.4130 m Then 386.41 N ⋅ m = ( 2.4130 m ) ( FC )min

( FC )min and

= 160.137 N  1.35 m   = 34.019°  2m 

φ = tan −1 

θ = 90 − φ = 90° − 34.019° = 55.981° ∴ ( FC )min = 160.1 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.

56.0°

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Chapter 3, Solution 7.

(a)

M A = xTBFy + yTBFx

= ( 2 m )( 200 N ) sin 60° + ( 0.4 m )( 200 N ) cos 60° = 386.41 N ⋅ m or M A = 386 N ⋅ m

(b)

Have or

M A = xFC

FC =

MA 386.41 N ⋅ m = 2m x

= 193.205 N

(c)

∴ FC = 193.2 N

For FB to be minimum, it must be perpendicular to the line joining A and B

∴ M A = d ( FB )min with Then

d =

= 2.0396 m

386.41 N ⋅ m = ( 2.0396 m ) ( FC )min

( FC )min and

( 2 m )2 + ( 0.40 m )2

= 189.454 N

 2m   = 78.690°  0.4 m 

θ = tan −1 

or

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

( FC )min

= 189.5 N

78.7°

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Chapter 3, Solution 8.

(a)

(

)

M B = rA/B cos15° W

= (14 in.)( cos15° )( 5 lb ) = 67.615 lb ⋅ in. or

M B = 67.6 lb ⋅ in.

(b)

M B = rD/B P sin 85° 67.615 lb ⋅ in. = ( 3.2 in.) P sin 85° or

(c)

P = 21.2 lb

For ( F )min, F must be perpendicular to BC. Then,

M B = rC/B F 67.615 lb ⋅ in. = (18 in.) F or

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

F = 3.76 lb

75.0°

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Chapter 3, Solution 9.

Slope of line EC =

(a)

Then

and Then

TABx =

35 in.

5 = 76 in. + 8 in. 12

12 (TAB ) 13

=

12 ( 260 lb ) = 240 lb 13

TABy =

5 ( 260 lb ) = 100 lb 13

M D = TABx ( 35 in.) − TABy ( 8 in.)

= ( 240 lb )( 35 in.) − (100 lb )( 8 in.) = 7600 lb ⋅ in. or M D = 7600 lb ⋅ in. (b) Have

M D = TABx ( y ) + TABy ( x ) = ( 240 lb )( 0 ) + (100 lb )( 76 in.) = 7600 lb ⋅ in. or M D = 7600 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 3, Solution 10.

Slope of line EC =

35 in.

7 = 112 in. + 8 in. 24

Then

TABx =

24 TAB 25

and

TABy =

7 TAB 25

M D = TABx ( y ) + TABy ( x )

Have

∴ 7840 lb ⋅ in. =

24 7 TAB ( 0 ) + TAB (112 in.)

25 25 TAB = 250 lb or TAB = 250 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 3, Solution 11.

The minimum value of d can be found based on the equation relating the moment of the force TAB about D: M D = (TAB max ) y ( d ) M D = 1152 N ⋅ m

where

(TAB max ) y Now

sin θ =

= TAB max sin θ = ( 2880 N ) sin θ

1.05 m

(d

 ∴ 1152 N ⋅ m = 2880 N   

+ 0.24 ) + (1.05 ) m 2

2

1.05

(d

+ 0.24 ) + (1.05 ) 2

( d + 0.24 )2 + (1.05)2

or or

2

  (d )  

(d

or

= 2.625d

+ 0.24 ) + (1.05 ) = 6.8906d 2 2

2

5.8906d 2 − 0.48d − 1.1601 = 0

Using the quadratic equation, the minimum values of d are 0.48639 m and −0.40490 m.

Since only the positive value applies here, d = 0.48639 m or d = 486 mm

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 3, Solution 12.

with d AB =

( 42 mm )2 + (144 mm )2

= 150 mm sin θ =

42 mm 150 mm

cosθ =

144 mm 150 mm

and FAB = − FAB sin θ i − FAB cosθ j =

2.5 kN ( − 42 mm ) i − (144 mm ) j 150 mm 

= − ( 700 N ) i − ( 2400 N ) j Also rB/C = − ( 0.042 m ) i + ( 0.056 m ) j Now M C = rB/C × FAB = ( − 0.042 i + 0.056 j) × ( − 700 i − 2400 j) N ⋅ m = (140.0 N ⋅ m ) k or

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

M C = 140.0 N ⋅ m

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Chapter 3, Solution 13.

( 42 mm )2 + (144 mm )2

with d AB =

= 150 mm sin θ =

42 mm 150 mm

cosθ =

144 mm 150 mm

FAB = − FAB sin θ i − FAB cosθ j =

2.5 kN ( − 42 mm ) i − (144 mm ) j 150 mm 

= − ( 700 N ) i − ( 2400 N ) j Also rB/C = − ( 0.042 m ) i − ( 0.056 m ) j Now M C = rB/C × FAB = ( − 0.042 i − 0.056 j) × ( − 700i − 2400 j) N ⋅ m = ( 61.6 N ⋅ m ) k or

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

M C = 61.6 N ⋅ m

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Chapter 3, Solution 14.

ΣM D :

 88   105  M D = ( 0.090 m )  × 80 N  − ( 0.280 m )  × 80 N  137 137     = −12.5431 N ⋅ m or

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

M D = 12.54 N ⋅ m

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Chapter 3, Solution 15.

Note: B = B ( cos β i + sin β j)

B′ = B ( cos β i − sin β j) C = C ( cos α i + sin α j) By definition:

B × C = BC sin (α − β )

(1)

B′ × C = BC sin (α + β )

(2)

Now ... B × C = B ( cos β i + sin β j) × C ( cos α i + sin α j) = BC ( cos β sin α − sin β cos α ) k and

(3)

B′ × C = B ( cos β i − sin β j) × C ( cos α i + sin α j) = BC ( cos β sin α + sin β cos α ) k

(4)

Equating the magnitudes of B × C from equations (1) and (3) yields: BC sin (α − β ) = BC ( cos β sin α − sin β cos α )

(5)

Similarly, equating the magnitudes of B′ × C from equations (2) and (4) yields: BC sin (α + β ) = BC ( cos β sin α + sin β cos α )

(6)

Adding equations (5) and (6) gives: sin (α − β ) + sin (α + β ) = 2cos β sin α or

sin α cos β =

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

1 1 sin (α + β ) + sin (α − β ) 2 2

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Chapter 3, Solution 16.

Have d = λ AB × rO/ A where λ AB =

rB/ A rB/ A

and rB/ A = ( −210 mm − 630 mm ) i + ( 270 mm − ( −225 mm ) ) j

= − ( 840 mm ) i + ( 495 mm ) j rB/ A =

( −840 mm )2 + ( 495 mm )2 = 975 mm

Then λ AB = =

− ( 840 mm ) i + ( 495 mm ) j 975 mm

1 ( −56i + 33j) 65

Also rO/ A = ( 0 − 630 ) i + ( 0 − (−225) ) j

= − ( 630 mm ) i + ( 225 mm ) j

∴d =

1 ( −56i + 33j) × − ( 630 mm ) i + ( 225 mm ) j 65

= 126.0 mm d = 126.0 mm 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 3, Solution 17.

(a) where

λ =

A×B A×B

A = 12i − 6 j + 9k B = − 3i + 9 j − 7.5k

Then i j k A × B = 12 − 6 9 − 3 9 − 7.5

= ( 45 − 81) i + ( −27 + 90 ) j + (108 − 18 ) k = 9 ( − 4i + 7 j + 10k ) And A × B = 9 (− 4) 2 + (7)2 + (10)2 = 9 165

∴λ =

9 ( − 4i + 7 j + 10k ) 9 165 or λ =

(b) where

λ =

A×B A×B

A = −14i − 2 j + 8k B = 3i + 1.5j − k

Then

j k i A × B = −14 − 2 8 3 1.5 −1 = ( 2 − 12 ) i + ( 24 − 14 ) j + ( −21 + 6 ) k

= 5 ( −2i + 2 j − 3k ) and

A × B = 5 (−2)2 + (2)2 + (−3)2 = 5 17

∴λ = or λ =

5 ( −2i + 2 j − 3k ) 5 17

1 ( − 2i + 2 j − 3k ) 17

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

1 ( − 4i + 7 j + 10k ) 165

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Chapter 3, Solution 18.

(a)

Have A = P × Q i j k P × Q = 3 7 −2 in.2 −5 1 3 = [ (21 + 2)i + (10 − 9) j + (3 + 35)k ] in.2

(

) (

) (

)

= 23 in.2 i + 1 in.2 j + 38 in.2 k ∴A=

(23)2 + (1)2 + (38) 2 = 44.430 in.2 or A = 44.4 in.2 W

(b)

A = P×Q i j k P × Q = 2 − 4 3 in.2 6 −1 5 = [ (−20 − 3)i + (−18 − 10) j + (−2 + 24)k ] in.2

(

) (

) (

)

= − 23 in.2 i − 28 in.2 j + 22 in.2 k ∴A=

(− 23)2 + (−28)2 + (22) 2 = 42.391 in.2 or A = 42.4 in.2 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 3, Solution 19.

(a)

Have

MO = r × F

i

j

k

= − 6 3 1.5 N ⋅ m 7.5 3 − 4.5 = [ (−13.5 − 4.5)i + (11.25 − 27) j + (−18 − 22.5)k ] N ⋅ m = ( −18.00i − 15.75 j − 40.5k ) N ⋅ m or M O = − (18.00 N ⋅ m ) i − (15.75 Ν ⋅ m ) j − ( 40.5 N ⋅ m ) k W (b)

Have

MO = r × F i j k = 2 − 0.75 −1 N ⋅ m 7.5 3 − 4.5 = [ (3.375 + 3)i + (−7.5 + 9) j + (6 + 5.625)k ] N ⋅ m = ( 6.375i + 1.500 j + 11.625k ) N ⋅ m or M O = ( 6.38 N ⋅ m ) i + (1.500 Ν ⋅ m ) j + (11.63 Ν ⋅ m ) k W

(c)

Have

MO = r × F i j k = − 2.5 −1 1.5 N ⋅ m 7.5 3 4.5 = [ (4.5 − 4.5)i + (11.25 − 11.25) j + (−7.5 + 7.5)k ] N ⋅ m or M O = 0 W

This answer is expected since r and F are proportional ( F = −3r ) . Therefore, vector F has a line of action passing through the origin at O.

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 3, Solution 20.

(a)

Have

MO = r × F i j k = − 7.5 3 − 6 lb ⋅ ft 3 −6 4 = [ (12 − 36)i + (−18 + 30) j + (45 − 9)k ] lb ⋅ ft or M O = − ( 24.0 lb ⋅ ft ) i + (12.00 lb ⋅ ft ) j + ( 36.0 lb ⋅ ft ) k W

(b)

Have

MO = r × F i j k = − 7.5 1.5 −1 lb ⋅ ft 3 −6 4 = [ (6 − 6)i + (−3 + 3) j + (4.5 − 4.5)k ] lb ⋅ ft or M O = 0 W

(c)

Have

MO = r × F i j k = − 8 2 −14 lb ⋅ ft 3 −6 4 = [ (8 − 84)i + (−42 + 32) j + (48 − 6)k ] lb ⋅ ft or M O = − ( 76.0 lb ⋅ ft ) i − (10.00 lb ⋅ ft ) j + ( 42.0 lb ⋅ ft ) k W

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

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Chapter 3, Solution 21.

With

TAB = − ( 369 N ) j TAB = TAD

JJJG AD = ( 369 N ) AD

( 2.4 m ) i − ( 3.1 m ) j − (1.2 m ) k ( 2.4 m )2 + ( −3.1 m )2 + ( −1.2 m )2

TAD = ( 216 N ) i − ( 279 N ) j − (108 N ) k Then

R A = 2 TAB + TAD = ( 216 N ) i − (1017 N ) j − (108 N ) k

Also

rA/C = ( 3.1 m ) i + (1.2 m ) k

Have

M C = rA/C × R A i j k = 0 3.1 1.2 N ⋅ m 216 −1017 −108 = ( 885.6 N ⋅ m ) i + ( 259.2 N ⋅ m ) j − ( 669.6 N ⋅ m ) k M C = ( 886 N ⋅ m ) i + ( 259 N ⋅ m ) j − ( 670 N ⋅ m ) k W

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

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Chapter 3, Solution 22.

Have

M A = rC/ A × F

where

rC/ A = ( 215 mm ) i − ( 50 mm ) j + (140 mm ) k Fx = − ( 36 N ) cos 45° sin12°

Fy = − ( 36 N ) sin 45° Fz = − ( 36 N ) cos 45° cos12°

∴ F = − ( 5.2926 N ) i − ( 25.456 N ) j − ( 24.900 N ) k and

i j k M A = 0.215 − 0.050 0.140 N ⋅ m − 5.2926 − 25.456 − 24.900 = ( 4.8088 N ⋅ m ) i + ( 4.6125 N ⋅ m ) j − ( 5.7377 N ⋅ m ) k M A = ( 4.81 N ⋅ m ) i + ( 4.61 N ⋅ m ) j − ( 5.74 N ⋅ m ) k W

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

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Chapter 3, Solution 23.

Have

M O = rA/O × R

where

rA/D = ( 30 ft ) j + ( 3 ft ) k T1 = − ( 62 lb ) cos10°  i − ( 62 lb ) sin10°  j

= − ( 61.058 lb ) i − (10.766 lb ) j JJJG AB T2 = T2 AB = ( 62 lb )

( 5 ft ) i − ( 30 ft ) j + ( 6 ft ) k ( 5 ft )2 + ( − 30 ft )2 + ( 6 ft )2

= (10 lb ) i − ( 60 lb ) j + (12 lb ) k ∴ R = − ( 51.058 lb ) i − ( 70.766 lb ) j + (12 lb ) k MO

i j k = 0 30 3 lb ⋅ ft − 51.058 −70.766 12 = ( 572.30 lb ⋅ ft ) i − (153.17 lb ⋅ ft ) j + (1531.74 lb ⋅ ft ) k M O = ( 572 lb ⋅ ft ) i − (153.2 lb ⋅ ft ) j + (1532 lb ⋅ ft ) k W

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

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Chapter 3, Solution 24.

(a)

Have

M O = rB/O × TBD

where

rB/O = ( 2.5 m ) i + ( 2 m ) j TBD = TBD

JJJG BD BD

 − (1 m ) i − ( 2 m ) j + ( 2 m ) k  = ( 900 N )  ( −1 m ) 2 + ( − 2 m ) 2 + ( 2 m ) 2

= − ( 300 N ) i − ( 600 N ) j + ( 600 N ) k Then

MO

i j k = 2.5 2 0 N⋅m − 300 − 600 600 M O = (1200 N ⋅ m ) i − (1500 N ⋅ m ) j − ( 900 N ⋅ m ) k W 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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(b)

Have

M O = rB/O × TBE

where

rB/O = ( 2.5 m ) i + ( 2 m ) j TBE = TBE

JJJG BE BE

 − ( 0.5 m ) i − ( 2 m ) j − ( 4 m ) k  = ( 675 N )  ( 0.5 m )2 + ( −2 m )2 + ( − 4 m )2

= − ( 75 N ) i − ( 300 N ) j − ( 600 N ) k Then

MO

i j k = 2.5 2 0 N⋅m − 75 − 300 − 600 M O = − (1200 N ⋅ m ) i + (1500 N ⋅ m ) j − ( 600 N ⋅ m ) k W

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

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Chapter 3, Solution 25.

Have M C = rA/C × P where

rA/C = rB/C + rA/B = (16 in.)( − cos80° cos15°i − sin 80° j − cos80° sin15°k )

+ (15.2 in.)( − sin 20° cos15°i + cos 20° j − sin 20° sin15°k ) = − ( 7.7053 in.) i − (1.47360 in.) j − ( 2.0646 in.) k and

P = (150 lb )( cos 5° cos 70°i + sin 5° j − cos 5° sin 70°k ) = ( 51.108 lb ) i + (13.0734 lb ) j − (140.418 lb ) k

Then

i j k M C = −7.7053 −1.47360 −2.0646 lb ⋅ in.

51.108 13.0734 −140.418 = ( 233.91 lb ⋅ in.) i − (1187.48 lb ⋅ in.) j − ( 25.422 lb ⋅ in.) k or M C = (19.49 lb ⋅ ft ) i − ( 99.0 lb ⋅ ft ) j − ( 2.12 lb ⋅ ft ) k W

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

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Chapter 3, Solution 26.

Have

M C = rA/C × FBA

where

rA/C = ( 0.96 m ) i − ( 0.12 m ) j + ( 0.72 m ) k

and

FBA = λ BA FBA   − ( 0.1 m ) i + (1.8 m ) j − ( 0.6 m ) k   = 228 N )  ( ( 0.1)2 + (1.8)2 + ( 0.6 )2 m  

= − (12.0 N ) i + ( 216 N ) j − ( 72 N ) k i j k ∴ M C = 0.96 −0.12 0.72 N ⋅ m −12.0 216 −72 = − (146.88 N ⋅ m ) i + ( 60.480 N ⋅ m ) j + ( 205.92 N ⋅ m ) k or M C = − (146.9 N ⋅ m ) i + ( 60.5 N ⋅ m ) j + ( 206 N ⋅ m ) k W

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

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Chapter 3, Solution 27.

Have where

M C = TAD d

JJJG d = Perpendicular distance from C to line AD

with

M C = rA/C TAD

and

rA/C = ( 3.1 m ) j + (1.2 m ) k JJJG AD TAD = TAD AD ( 2.4 m ) i − ( 3.1 m ) j − (1.2 m ) k  TAD = ( 369 N )  ( 2.4 m )2 + ( − 3.1 m )2 + ( −1.2 m )2

= ( 216 N ) i − ( 279 N ) j − (108 N ) k

Then

i j k 3.1 1.2 N ⋅ m MC = 0 216 − 279 −108 = ( 259.2 N ⋅ m ) j − ( 669.6 N ⋅ m ) k

and

MC =

( 259.2 N ⋅ m )2 + ( −669.6 N ⋅ m )2

= 718.02 N ⋅ m ∴ 718.02 N ⋅ m = ( 369 N ) d

or d = 1.946 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 3, Solution 28.

Have where

M O = TAC d

d = Perpendicular distance from O to rope AC

with

M O = rA/O × TAC

and

rA/O = ( 30 ft ) j + ( 3 ft ) k TAC = − ( 62 lb ) cos10° i − ( 62 lb ) sin10° j

= − ( 61.058 lb ) i − (10.766 lb ) j

Then

MO

i j k = 0 30 3 lb ⋅ ft − 61.058 −10.766 0 = ( 32.298 lb ⋅ ft ) i − (183.174 lb ⋅ ft ) j + (1831.74 lb ⋅ ft ) k

and

MO =

( 32.298 lb ⋅ ft )2 + ( −183.174 lb ⋅ ft )2 + (1831.74 lb ⋅ ft )2

= 1841.16 lb ⋅ ft ∴ 1841.16 lb ⋅ ft = ( 62 lb ) d

or d = 29.7 ft 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 3, Solution 29.

M O = TAB d

Have where

d = Perpendicular distance from O to rope AB

with

M O = rA/O × TAB

and

rA/O = ( 30 ft ) j + ( 3 ft ) k JJJG AB TAB = TAB AB ( 5 ft ) i − ( 30 ft ) j + ( 6 ft ) k  = ( 62 lb )  ( 5 ft )2 + ( − 30 ft )2 + ( 6 ft )2

= (10 lb ) i − ( 60 lb ) j + (12 lb ) k

Then

MO

i j k = 0 30 3 lb ⋅ ft 10 − 60 12 = ( 540 lb ⋅ ft ) i + ( 30 lb ⋅ ft ) j − ( 300 lb ⋅ ft ) k

and

MO =

( 540 lb ⋅ ft )2 + ( 30 lb ⋅ ft )2 + ( −300 lb ⋅ ft )2

= 618.47 lb ⋅ ft ∴ 618.47 lb ⋅ ft = ( 62 lb ) d

or d = 9.98 ft 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 3, Solution 30.

Have where

M C = TBD d d = Perpendicular distance from C to cable BD

with

M C = rB/C × TB/D

and

rB/C = ( 2 m ) j JJJG BD TBD = TBD BD  − (1 m ) i − ( 2 m ) j + ( 2 m ) k  = ( 900 N )  ( −1 m )2 + ( − 2 m )2 + ( 2 m )2

= − ( 300 N ) i − ( 600 N ) j + ( 600 N ) k

Then i j k 2 0 N⋅m MC = 0 −300 − 600 600 = (1200 N ⋅ m ) i + ( 600 N ⋅ m ) k

and

MC =

(1200 N ⋅ m )2 + ( 600 N ⋅ m )2

= 1341.64 N ⋅ m ∴ 1341.64 = ( 900 N ) d

or d = 1.491 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 3, Solution 31.

Have M C = Pd From the solution of problem 3.25 M C = ( 233.91 lb ⋅ in.) i − (1187.48 lb ⋅ in.) j − ( 25.422 lb ⋅ in.) k Then MC =

( 233.91)2 + ( −1187.48)2 + ( −25.422 )2

= 1210.57 lb ⋅ in. and

d =

MC 1210.57 lb.in. = 150 lb P or d = 8.07 in. 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 3, Solution 32.

| M D | = FBAd

Have

d = perpendicular distance from D to line AB.

where

M D = rA/D × FBA rA/D = − ( 0.12 m ) j + ( 0.72 m ) k FBA = λ BA FBA =

( − ( 0.1 m ) i + (1.8 m ) j − ( 0.6 m ) k ) ( 228 N ) ( 0.1)2 + (1.8)2 + ( 0.6 )2

m

= − (12.0 N ) i + ( 216 N ) j − ( 72 N ) k ∴ MD

i j k = 0 −0.12 0.72 N ⋅ m −12.0 216 −72 = − (146.88 N ⋅ m ) i − ( 8.64 N ⋅ m ) j − (1.44 N ⋅ m ) k

and

|MD | =

(146.88)2 + (8.64 )2 + (1.44 )2

= 147.141 N ⋅ m

∴ 147.141 N ⋅ m = ( 228 N ) d

d = 0.64536 m or d = 0.645 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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 3, Solution 33.

Have

| M C | = FBAd

d = perpendicular distance from C to line AB.

where

M C = rA/C × FBA

rA/C = ( 0.96 m ) i − ( 0.12 m ) j + ( 0.72 m ) k FBA = λ BA FBA =

( − ( 0.1 m ) i + (1.8 m ) j − ( 0.6 ) k ) ( 228 N ) ( 0.1)2 + (1.8)2 + ( 0.6 )2

m

= − (12.0 N ) i + ( 216 N ) j − ( 72 N ) k i

j

k

∴ M C = 0.96 −0.12 0.72 N ⋅ m −12.0 216 −72 = − (146.88 N ⋅ m ) i − ( 60.48 N ⋅ m ) j + ( 205.92 N ⋅ m ) k and

| MC | =

(146.88)2 + ( 60.48)2 + ( 205.92 )2

= 260.07 N ⋅ m

∴ 260.07 N ⋅ m = ( 228 N ) d

d = 1.14064 m or d = 1.141 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 3, Solution 34.

(a) Have d = rC/ A sin θ = λ AB × rC/ A where d = Perpendicular distance from C to pipe AB with λ AB =

AB = AB

=

7i + 4 j − 32k

( 7 )2 + ( 4 )2 + ( −32 )2 1 ( 7i + 4 j − 32k ) 33

and rC/ A = − (14 ft ) i + ( 5 ft ) j + ( L − 22 ) ft  k Then

λ AB × rC/ A

i j k 1 7 4 − 32 ft = 33 −14 5 L − 22

{

}

1  4 ( L − 22 ) + 32 ( 5 )  i + 32 (14 ) − 7 ( L − 22 )  j +  7 ( 5 ) + 4 (14 )  k ft 33  1 ( 4L + 72 ) i + ( −7 L + 602 ) j + 91k  ft = 33 

=

and

d =

1 33

( 4L + 72 )2 + ( −7 L + 602 )2 + ( 91)2

or or

dd 2 1 = 2  2 ( 4 )( 4L + 72 ) + 2 ( −7 )( −7 L + 602 )  = 0 dL 33 65L − 3926 = 0 L = 60.400 ft

But

L > Lgreenhouse

For

(b) with

(d ) min ,

L = 30 ft,

so d =

1 33

L = 30.0 ft W

( 4 × 30 + 72 )2 + ( −7 × 30 + 602 )2 + ( 91)2 or d = 13.51 ft W

Note:

with

L = 60.4 ft,

d =

1 33

( 4 × 60.4 + 72 )2 + ( −7 × 60.4 + 602 )2 + ( 91)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.

= 11.29 ft

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Chapter 3, Solution 35.

P ⋅ Q = ( − 4i + 8j − 3k ) ⋅ ( 9i − j − 7k )

= ( − 4 )( 9 ) + ( 8 )( −1) + ( − 3)( − 7 ) = − 23 or P ⋅ Q = −23 W P ⋅ S = ( − 4i + 8 j − 3k ) ⋅ ( 5i − 6 j + 2k )

= ( − 4 )( 5 ) + ( 8 )( − 6 ) + ( − 3)( 2 ) = − 74 or P ⋅ S = −74 W Q ⋅ S = ( 9i − j − 7k ) ⋅ ( 5i − 6 j + 2k )

= ( 9 )( 5 ) + ( −1)( − 6 ) + ( − 7 )( 2 )

= 37 or Q ⋅ S = 37 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 3, Solution 36.

By definition B ⋅ C = BC cos (α − β )

where

B = B ( cos β ) i + ( sin β ) j

C = C ( cos α ) i + ( sin α ) j

∴ ( B cos β )( C cos α ) + ( B sin β )( C sin α ) = BC cos (α − β ) cos β cos α + sin β sin α = cos (α − β )

or

(1)

By definition B′⋅ C = BC cos (α + β )

where

B′ = ( cos β ) i − ( sin β ) j

∴ ( B cos β )( C cos α ) + ( − B sin β )( C sin α ) = BC cos (α + β ) or

cos β cos α − sin β sin α = cos (α + β )

(2)

Adding Equations (1) and (2), 2 cos β cos α = cos (α − β ) + cos (α + β ) or cos α cos β =

1 1 cos (α + β ) + cos (α − β ) W 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.

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Chapter 3, Solution 37.

First note: rB/ A = ( 0.56 m ) i + ( 0.9 m ) j rC/ A = ( 0.9 m ) j − ( 0.48 m ) k rD/ A = − ( 0.52 m ) i + ( 0.9 m ) j + ( 0.36 m ) k

By definition

rB/ A =

( 0.56 m )2 + ( 0.9 m )2

rC/ A =

( 0.9 m )2 + ( − 0.48 m )2

rD/ A =

( − 0.52 m )2 + ( 0.9 m )2 + ( 0.36 m )2

= 1.06 m = 1.02 m = 1.10 m

rB/ A ⋅ rD/ A = rB/ A rD/ A cosθ

or ( 0.56i + 0.9 j) ⋅ ( − 0.52i + 0.9 j + 0.36k ) = (1.06 )(1.10 ) cosθ

( 0.56 )( − 0.52 ) + ( 0.9 )( 0.9 ) + ( 0 )( 0.36 ) = 1.166 cosθ cosθ = 0.44494

θ = 63.6° 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 3, Solution 38.

From the solution to problem 3.37 rC/ A = 1.02 m with rC/ A = ( 0.9 m ) i − ( 0.48 m ) j rD/ A = 1.10 m with rD/ A = − ( 0.52 m ) i + ( 0.9 m ) j + ( 0.36 m ) k

Now by definition rC/ A ⋅ rD/ A = rC/ A rD/ A cosθ

or ( 0.9 j − 0.48k ) ⋅ ( − 0.52i + 0.9 j + 0.36k ) = (1.02 )(1.10 ) cosθ 0 ( − 0.52 ) + ( 0.9 )( 0.9 ) + ( − 0.48)( 0.36 ) = 1.122cosθ

cosθ = 0.56791 or θ = 55.4° 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 3, Solution 39.

(a) By definition λ BC + λ EF = (1) (1) cosθ where λ BC =

=

λ EF =

=

Therefore

( 32 ft ) i − ( 9 ft ) j − ( 24 ft ) k ( 32 )2 + ( − 9 )2 + ( − 24 )2 ft 1 ( 32i − 9 j − 24k ) 41 − (14 ft ) i − (12 ft ) j + (12 ft ) k

( −14 )2 + ( −12 )2 + ( 12 )2 ft 1 ( −7i − 6 j + 6k ) 11

( 32i − 9 j − 24k ) ⋅ ( −7i − 6 j + 6k ) 41

11

= cosθ

( 32 )( −7 ) + ( −9 )( −6 ) + ( −24 )( 6 ) = ( 41)(11) cosθ cosθ = − 0.69623 or (b) By definition

(TEG ) BC

θ = 134.1° W

= (TEF ) cosθ = (110 lb )( −0.69623) = −76.585 lb or (TEF ) BC = −76.6 lb W

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

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Chapter 3, Solution 40.

(a) By definition λ BC ⋅ λ EG = (1) (1) cosθ where λ BC =

=

λ EG =

=

Therefore

( 32 ft ) i − ( 9 ft ) j − ( 24 ft ) k ( 32 )2 + ( − 9 )2 + ( − 24 )2 ft 1 ( 32i − 9 j − 24k ) 41

(16 ft ) i − (12 ft ) j + ( 9.75) k (16 )2 + ( −12 )2 + ( 9.75)2 ft 1 (16i − 12 j + 9.75k ) 22.25

( 32i − 9 j − 24k ) ⋅ (16i − 12 j + 9.75k ) 41

22.25

= cosθ

( 32 )(16 ) + ( −9 )( −12 ) + ( −24 )( 9.75) = ( 41)( 22.25) cosθ cosθ = 0.42313 or (b) By definition

(TEG )BC

θ = 65.0° W

= (TEG ) cosθ = (178 lb )( 0.42313) = 75.317 lb or (TEG ) BC = 75.3 lb W

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

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Chapter 3, Solution 41.

First locate point B: d 3.5 = 22 14 d = 5.5 m

or

d BA =

(a)

( 5.5 + 0.5)2 + ( −22 )2 + ( −3)2

= 23 m

Locate point D: ( −3.5 − 7.5sin 45° cos15° ) , (14 + 7.5cos 45° ) ,

( 0 + 7.5sin 45° sin15° ) m or ( −8.6226 m, 19.3033 m, 1.37260 m ) Then d BD =

( −8.6226 + 5.5)2 + (19.3033 − 22 )2 + (1.37260 − 0 )2

m

= 4.3482 m and

cosθ ABD =

( 6i − 22 j − 3k ) ⋅ ( −3.1226i − 2.6967 j + 1.37260k ) d BA ⋅ d BD = d BA d BD ( 23)( 4.3482 ) = 0.36471 or

(b)

(TBA )BD

θ ABD = 68.6° W

= TBA cosθ ABD = ( 230 N )( 0.36471) or (TBA ) BD = 83.9 N W

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

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Chapter 3, Solution 42.

First locate point B: d 3.5 = 22 14 d = 5.5 m

or

(a) Locate point D: ( −3.5 − 7.5sin 45° cos10° ) , (14 + 7.5cos 45° ) ,

( 0 + 7.5sin 45° sin10° ) m or ( −8.7227 m, 19.3033 m, 0.92091 m ) Then

d DC = ( 5.2227 m ) i − ( 5.3033 m ) j − ( 0.92091 m ) k and d DB =

( −5.5 + 8.7227 )2 + ( 22 − 19.3033)2 + ( 0 − 0.92091)2

m

= 4.3019 m and

cos θ BDC =

( 3.2227i + 2.6967 j − 0.92091k ) ⋅ ( 5.2227i − 5.3033j − 0.92091k ) d DB ⋅ d DC = d DB d DC ( 4.3019 )( 7.5)

= 0.104694 or (b)

(TBD )DC

θ BDC = 84.0° W

= TBD cosθ BDC = ( 250 N )( 0.104694 ) or

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

(TBD )DC

= 26.2 N W

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Chapter 3, Solution 43.

Volume of parallelopiped is found using the mixed triple product (a)

Vol = P ⋅ ( Q × S ) 3 −4 1 = − 7 6 − 8 in.3 9 −2 −3 = ( −54 + 288 + 14 − 48 + 84 − 54 ) in.3 = 230 in.3 or Volume = 230 in.3 W

(b)

Vol = P ⋅ ( Q × S ) −5 −7 4 = 6 − 2 5 in.3 −4 8 −9 = ( −90 + 140 + 192 + 200 − 378 − 32 ) in.3 = 32 in.3 or Volume = 32 in.3 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 3, Solution 44.

For the vectors to all be in the same plane, the mixed triple product is zero.

P ⋅(Q × S ) = 0 −3 −7 5 ∴ 0 = −2 1 −4 8 Sy −6

0 = 18 + 224 − 10S y − 12S y + 84 − 40 So that

22 S y = 286 S y = 13

or S y = 13.00 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 3, Solution 45.

Have

rC = ( 2.25 m ) k JJJG CE TCE = TCE CE ( 0.90 m ) i + (1.50 m ) j − ( 2.25 m ) k  TCE = (1349 N )  ( 0.90 )2 + (1.50 )2 + ( −2.25 )2 m

= ( 426 N ) i + ( 710 N ) j − (1065 N ) k

Now

M O = rC × TCE i j k = 0 0 2.25 N ⋅ m 426 710 −1065 = − (1597.5 N ⋅ m ) i + ( 958.5 N ⋅ m ) j ∴ M x = −1598 N ⋅ m, M y = 959 N ⋅ m, M z = 0 W

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

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Chapter 3, Solution 46.

Have

rE = ( 0.90 m ) i + (1.50 m ) j TDE = TDE

JJJG DE DE

 − ( 2.30 m ) i + (1.50 m ) j − ( 2.25 m ) k  = (1349 N )  ( − 2.30 )2 + (1.50 )2 + ( − 2.25)2 m

= − ( 874 N ) i + ( 570 N ) j − ( 855 N ) k Now

M O = rE × TDE i j k = 0.90 1.50 0 N ⋅ m − 874 570 − 855 = − (1282.5 N ⋅ m ) i + ( 769.5 N ⋅ m ) j + (1824 N ⋅ m ) k ∴ M x = −1283 N ⋅ m, M y = 770 N ⋅ m, M z = 1824 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 3, Solution 47.

Have

M z = k ⋅ ( rB ) y × TBA  + k ( rC ) y × TCD     

where

M z = − ( 48 lb ⋅ ft ) k

( rB ) y TBA

= ( rC ) y = ( 3 ft ) j JJJG ( 4.5 ft ) i − ( 3 ft ) j + ( 9 ft ) k  BA = TBA = (14 lb )  BA ( 4.5)2 + ( − 3)2 + ( 9 )2 ft

= ( 6 lb ) i − ( 4 lb ) j + (12 lb ) k TCD

JJJG ( 6 ft ) i − ( 3 ft ) j − ( 6 ft ) k  CD = TCD = TCD  CD ( 6 )2 + ( − 3)2 + ( − 6 )2 ft

= Then

TCD (2i − j − 2k ) 3

{

}

− ( 48 lb ⋅ ft ) = k ⋅ ( 3 ft ) j × ( 6 lb ) i − ( 4 lb ) j + (12 lb ) k   T  + k ⋅ ( 3 ft ) j ×  CD ( 2 i − j − 2 k )   3   

or

− 48 = −18 − 2TCD

TCD = 15.00 lb

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

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Chapter 3, Solution 48.

Have

M y = j ⋅ ( rB ) z × TBA  × j⋅ ( rC ) z × TCD 

where

M y = 156 lb ⋅ ft

( rB ) z TBA

= ( 24 ft ) k;

( rC ) z

= ( 6 ft ) k

JJJG ( 4.5 ft ) i − ( 3 ft ) j + ( 9 ft ) k  BA = TBA = TBA  BA ( 4.5)2 + ( − 3)2 + ( 9 )2 ft

TBA ( 4.5i − 3j + 9k ) 10.5 JJJG ( 6 ft ) i − ( 3 ft ) j + ( 9 ft ) k  CD = TCD = ( 7.5 lb )  CD ( 6 )2 + ( − 3)2 + ( 9 )2 ft

=

TCD

= ( 5 lb ) i − ( 2.5 lb ) j − ( 5 lb ) k T   Then (156 lb ⋅ ft ) = j ⋅ ( 24 ft ) k × BA ( 4.5i − 3j + 9k )  10.5  

{

}

+ j ⋅ ( 6 ft ) k × ( 5 lb ) i − ( 2.5 lb ) j − ( 5 lb ) k  or

156 =

108 TBA + 30 10.5 TBA = 12.25 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 3, Solution 49.

Based on M x = ( P cos φ ) ( 0.225 m ) sin θ  − ( P sin φ ) ( 0.225 m ) cosθ 

By

(1)

M y = − ( P cos φ )( 0.125 m )

(2)

M z = − ( P sin φ )( 0.125 m )

(3)

Equation ( 3) M z − ( P sin φ )( 0.125 ) : = Equation ( 2 ) M y − ( P cos φ )( 0.125 ) or

−4 = tan φ ∴ − 23

φ = 9.8658° or

φ = 9.87°

From Equation (2) − 23 N ⋅ m = − ( P cos 9.8658° )( 0.125 m ) P = 186.762 N or P = 186.8 N From Equation (1) 26 N ⋅ m = (186.726 N ) cos 9.8658° ( 0.225 m ) sin θ  − (186.726 N ) sin 9.8658°  ( 0.225 m ) cosθ 

or 0.98521sin θ − 0.171341cosθ = 0.61885 Solving numerically,

θ = 48.1°

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 3, Solution 50.

Based on M x = ( P cos φ ) ( 0.225 m ) sin θ  − ( P sin φ ) ( 0.225 m ) cosθ 

(1)

M y = − ( P cos φ )( 0.125 m )

(2)

M z = − ( P sin φ )( 0.125 m ) By

Equation ( 3) M z − ( P sin φ )( 0.125 ) : = Equation ( 2 ) M y − ( P cos φ )( 0.125 ) or

− 3.5 = tan φ ; φ = 9.9262° − 20

From Equation (3): − 3.5 N ⋅ m = − ( P sin 9.9262° )( 0.125 m ) P = 162.432 N From Equation (1): M x = (162.432 N )( 0.225 m )( cos 9.9262° sin 60° − sin 9.9262° cos 60° ) = 28.027 N ⋅ m or M x = 28.0 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.

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Chapter 3, Solution 51.

First note:

TBA

JJJG BA = TBA BA

= ( 70 lb )

= ( 70 lb )

( 4 ) i + 1.5 − ( LBC

+ 1)  j + ( − 6 ) k 2

( 4 )2 + 1.5 − ( LBC

+ 1)  + ( − 6 )

2

4 i + ( 0.5 − LBC ) j − 6 k 52 + ( 0.5 − LBC )

2

rA = ( 4 ft ) i + (1.5 ft ) j − (12 ft ) k Have M O = rA × TBA =

For the i components: − 763 lb ⋅ ft =

70 52 + ( 0.5 − LBC )

2

70 lb 52 + ( 0.5 − LBC )

j k 1.5 ft −12 ft ( 0.5 − LBC ) − 6

1.5 ( − 6 ) + 12 ( 0.5 − LBC )  lb ⋅ ft

or

10.9 52 + ( 0.5 − LBC ) = 3 + 12 LBC

or

(10.9 )2 52 + ( 0.5 − LBC )2  = 9 + 72LBC

or

25.19L2BC + 190.81LBC − 6198.8225 = 0

Then

2

i 4 ft 4

2

LBC =

−190.81 ±

+ 144 L2BC

(190.81)2 − 4 ( 25.19 )( − 6198.8225) 2 ( 25.19 )

Taking the positive root

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

LBC = 12.35 ft

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Chapter 3, Solution 52.

First note:

JJJG BA = TBA BA

TBA

= ( 70 lb )

= ( 70 lb )

( 4 ) i + 1.5 − ( LBC

+ 1)  j + ( − 6 ) k 2

( 4 )2 + 1.5 − ( LBC

+ 1)  + ( − 6 )

2

4 i + ( 0.5 − LBC ) j − 6 k 52 + ( 0.5 − LBC )

2

rA = ( 4 ft ) i + (1.5 ft ) j − (12 ft ) k Have M O = rA × TBA =

For the i components: − 900 lb ⋅ ft = 300 =

or

315 =

Then, or

(1) ( 2)

⇒

52 + ( 0.5 − LBC ) TBA

52 + ( 0.5 − LBC )

For the k components: − 315 lb ⋅ ft =

or

TBA

2

2

52 + ( 0.5 − LBC )

52 + ( 0.5 − LBC )

2

52 + ( 0.5 − LBC )

2

i 4 ft 4

j k −12 ft 1.5 ft ( 0.5 − LBC ) − 6

1.5 ( − 6 ) + 12 ( 0.5 − LBC )  lb ⋅ ft

(1 + 4LBC )

TBA 4TBA

TBA

2

(1)

 4 ( 0.5 − LBC ) − 1.5 ( 4 )  lb ⋅ ft

(1 + LBC )

(2)

300 1 + 4LBC = 315 4 (1 + LBC ) LBC =

59 ft 4 LBC = 14.75 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 3, Solution 53.

(

Have

M AD = λ AD ⋅ rB/ A × TBH

where

λ AD =

)

( 0.8 m ) i − ( 0.6 m ) k ( 0.8 m )2 + ( − 0.6 m )2

= 0.8 i − 0.6 k

rB/ A = ( 0.4 m ) i TBH = TBH

JJJJG ( 0.3 m ) i + ( 0.6 m ) j − ( 0.6 m ) k  BH = (1125 N )  BH ( 0.3)2 + ( 0.6 )2 + ( − 0.6 )2 m

Then M AD

0.8 0 − 0.6 = 0.4 0 0 = −180 N ⋅ m 375 750 − 750 or M AD = −180.0 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.

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Chapter 3, Solution 54.

(

)

Have

M AD = λ AD ⋅ rB/ A × TBG

where

λ AD = ( 0.8 m ) i − ( 0.6 m ) k rB/ A = ( 0.4 m ) i TBG = TBG

JJJG  − ( 0.4 m ) i + ( 0.74 ) j − ( 0.32 m ) k  BG = (1125 N )  BG ( − 0.4 m )2 + ( 0.74 m )2 + ( − 0.32 m )2

= − ( 500 N ) i + ( 925 N ) j − ( 400 N ) k Then M AD

0.8 0 − 0.6 = 0.4 0 0 − 500 925 − 400 or M AD = − 222 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 3, Solution 55.

Have

(

M AD = λ AD ⋅ rE/ A × FEF

where

λ AD

)

JJJG AD = AD

λ AD =

( 7.2 m ) i + ( 0.9 m ) j ( 7.2 m )2 + ( 0.9 m )2

= 0.99228 i + 0.124035 j

rE/ A = ( 2.1 m ) i − ( 0.9 m ) j FEF = FEF

JJJG ( 0.3 m ) i + (1.2 m ) j + ( 2.4 m ) k  EF = ( 24.3 kN )  EF ( 0.3 m )2 + (1.2 m )2 + ( 2.4 m )2

= ( 2.7 kN ) i + (10.8 kN ) j + ( 21.6 kN ) k Then

0.99228 0.124035 M AD =

2.1 2.7

− 0.9 10.8

0 0 kN ⋅ m 21.6

= −19.2899 − 5.6262 = − 24.916 kN ⋅ m or M AD = − 24.9 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.

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Chapter 3, Solution 56.

(

Have

M AD = λ AD ⋅ rG/ A × E EF

Where

λ AD =

)

( 7.2 m ) i + ( 0.9 m ) j ( 7.2 m )2 + ( 0.9 m )2

= 0.99228 i + 0.124035 j rG/ A = ( 6 m ) i − (1.8 m ) j FGH = FGH

JJJJG ( −1.2 m ) i + ( 2.4 m ) j + ( 2.4 m ) k  GH = ( 21.3 kN )  GH ( −1.2 m )2 + ( 2.4 m )2 + ( 2.4 m )2

= − ( 7.1 kN ) i + (14.2 kN ) j + (14.2 kN ) k Then

M AD

0.99228 0.124035 0 6 0 kN ⋅ m = −1.8 14.2 14.2 − 7.1 = − 25.363 − 10.5678 = − 35.931 kN ⋅ m or

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

M AD = − 35.9 kN ⋅ m

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Chapter 3, Solution 57.

(

M OA = λ OA ⋅ rC/O × P

Have

)

where From triangle OBC

Since or

a 2

( OA) x

=

( OA) z

= ( OA ) x tan 30° =

( OA)2

= ( OA ) x + ( OA ) y + ( OAz )

2

a 1  a =   2 3  2 3

2

2

 a  2 a a =   + ( OA ) y +   2 2 3 2

∴

( OA) y

=

2 2

a2 −

a2 a2 2 − =a 4 12 3

Then

rA/O =

a 2 a i +a j+ k 2 3 2 3

and

λ OA =

1 i+ 2

2 1 j+ k 3 2 3

P = λ BC P

= =

( a sin 30°) i − ( a cos30° ) k a

P i − 3k 2

(

)

rC/O = ai

∴ M OA

1 2 = 1 1 =

=

2 1 3 2 3 P ( a )   0 0 2 0 − 3

aP  2  −  (1) − 3 2  3 

(

aP aP M OA = 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.

)

( P)

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Chapter 3, Solution 58.

(a) For edge OA to be perpendicular to edge BC, uuur uuur OA ⋅ BC = 0 where From triangle OBC

a 2

( OA) x

=

( OA) z

= ( OA ) x tan 30° =

a 1  a = 2  3  2 3

uuur  a   a  ∴ OA =   i + ( OA )y j +  k 2 2 3

and

Then

or

so that

uuur BC = ( a sin 30° ) i − ( a cos 30° ) k =

a a 3 i− k 2 2

=

a i − 3k 2

(

)

a a  a   =0  i + ( OA) y j +   k  ⋅ i − 3k 2 2 2 3  

(

)

a2 a2 + ( OA )y ( 0 ) − =0 4 4 uuur uuur ∴ OA ⋅ BC = 0 uuur uuur OA is perpendicular to BC.

(b) Have M OA = Pd , with P acting along BC and d the uuur uuur perpendicular distance from OA to BC. From the results of Problem 3.57,

M OA = ∴

Pa 2

Pa = Pd 2

or

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

d =

a

2

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Chapter 3, Solution 59.

(

M DI = λ DI ⋅ rF /I × TEF

Have where

λ DI

JJJG DI = = DI

)

( 4.8 ft ) i − (1.2 ft ) j ( 4.8 ft )2 + ( −1.2 ft )2

= 0.97014 i − 0.24254 j rF /I = (16.2 ft ) k

TEF = TEF

JJJG ( 3.6 ft ) i − (10.8 ft ) j + (16.2 ft ) k  EF = ( 29.7 lb )  EF ( 3.6 ft )2 + ( −10.8 ft )2 + (16.2 ft )2

= ( 5.4 lb ) i − (16.2 lb ) j + ( 24.3 lb ) k Then M DI

0.97014 − 0.24254 0 0 0 16.2 lb ⋅ ft = 5.4 −16.2 24.3 = − 21.217 + 254.60 = 233.39 lb ⋅ ft or

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

M DI = 233 lb ⋅ ft

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Chapter 3, Solution 60.

Have where

(

M DI = λ DI ⋅ rG/I × TEG λ DI

JJJG DI = = DI

)

( 4.8 ft ) i − (1.2 ft ) j ( 4.8 ft )2 + ( −1.2 ft )2

= 0.97014 i − 0.24254 j rG/I = − ( 35.1 ft ) k

TEG = TEG

JJJG ( 3.6 ft ) i − (10.8 ft ) j − ( 35.1 ft ) k  EG = ( 24.6 lb )  EG ( 3.6 ft )2 + ( −10.8 ft )2 + ( − 35.1 ft )2

= ( 2.4 lb ) i − ( 7.2 lb ) j − ( 23.4 lb ) k Then M DI

0.97014 − 0.24254 0 0 0 = − 35.1 lb ⋅ ft 2.4 − 7.2 − 23.4 = 20.432 − 245.17 = − 224.74 lb ⋅ ft

or

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

M DI = − 225 lb ⋅ ft

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Chapter 3, Solution 61.

F1 = F1λ1

First note that

F2 = F2λ 2

and

Let M1 = moment of F2 about the line of action of M1

and M 2 = moment of F1 about the line of action of M 2

Now, by definition

(

)

(

)

(

)

(

)

M1 = λ1 ⋅ rB/ A × F2 = λ1 ⋅ rB/ A × λ 2 F2 M 2 = λ 2 ⋅ rA/B × F1 = λ 2 ⋅ rA/B × λ1 F1

Since

F1 = F2 = F

rA/B = −rB/ A

and

(

)

M1 = λ1 ⋅ rB/ A × λ 2 F

(

)

M 2 = λ 2 ⋅ −rB/ A × λ1 F

Using Equation (3.39)

(

)

(

λ1 ⋅ rB/ A × λ 2 = λ 2 ⋅ −rB/ A × λ1

so that

(

)

)

M 2 = λ1 ⋅ rB/ A × λ 2 F ∴ M12 = M 21

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 3, Solution 62.

From the solution of Problem 3.53: λ AD = 0.8i − 0.6k TBH = ( 375 N ) i + ( 750 N ) j − ( 750 N ) k;

TBH = 1125 N

M AD = −180 N ⋅ m

Only the perpendicular component of TBH contributes to the moment of TBH about line AD. The parallel component of TBH will be used to find the perpendicular component.

Have

( TBH )Parallel

= λ AD ⋅ TBH = [ 0.8i − 0.6k ] ⋅ ( 375 N ) i + ( 750 N ) j − ( 750 N ) k 

= ( 300 + 450 ) N = 750 N

Since

TBH = ( TBH )Perpendicular + ( TBH )Parallel

Then

(TBH )Perpendicular

=

( TBH )2 − ( TBH )2Parallel

=

(1125 N )2 − ( 750 N )2

= 838.53 N

and

M AD = ( TBH )Perpendicular d

180 N ⋅ m = ( 838.53 N ) d d = 0.21466 m

or

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

d = 215 mm

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Chapter 3, Solution 63.

From the solution of Problem 3.54: λ AD = 0.8i − 0.6k TBG = − ( 500 N ) i + ( 925 N ) j − ( 400 N ) k TBG = 1125 N

M AD = − 222 N ⋅ m

Only the perpendicular component of TBG contributes to the moment of TBG about line AD. The parallel component of TBG will be used to find the perpendicular component.

Have

( TBG )Parallel

= λ AD ⋅ TBG = [ 0.8i − 0.6k ] ⋅  − ( 500 N ) i + ( 925 N ) j − ( 400 N ) k 

= ( − 400 + 240 ) N = −160 N

Since

TBG = ( TBG )Perpendicular + ( TBG )Parallel

Then

(TBG )Perpendicular

=

(TBG )2 − (TBG )2Parallel

=

(1125 N )2 − ( −160 N )2

= 1113.56 N

and

M AD = (TBG )Perpendicular d

222 N ⋅ m = (1113.56 N ) d d = 0.199361 m

or d = 199.4 mm

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 3, Solution 64.

From the solution of Problem 3.59: λ DI = 0.97014 i − 0.24254 j TEF = ( 5.4 lb ) i − (16.2 lb ) j + ( 24.3 lb ) k TEF = 29.7 lb

M DI = 233.39 lb ⋅ ft

Only the perpendicular component of TEF contributes to the moment of TEF about line DI. The parallel component of TEF will be used to find the perpendicular component.

Have

( TEF )Parallel

= λ DI ⋅ TEF = [ 0.97014 i − 0.24254 j] ⋅ ( 5.4 lb ) i − (16.2 lb ) j + ( 24.3 lb ) k 

= ( 5.2388 + 3.9291) = 9.1679 lb

Since

TEF = ( TEF )Perpendicular + ( TEF )Parallel

Then

(TEF )Perpendicular

=

( TEF )2 − ( TEF )2Parallel

=

( 29.7 )2 − ( 9.1679 )2

= 28.250 lb

and

M DI = ( TEF )Perpendicular d

233.39 lb ⋅ ft = ( 28.250 lb ) d d = 8.2616 ft

or d = 8.26 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 3, Solution 65.

From the solution of Problem 3.60: λ DI = 0.97014 i − 0.24254 j TEG = ( 2.4 lb ) i − ( 7.2 lb ) j − ( 23.4 lb ) k TEG = 24.6 lb

M DI = − 224.74 lb ⋅ ft

Only the perpendicular component of TEG contributes to the moment of TEG about line DI. The parallel component of TEG will be used to find the perpendicular component.

Have

( TEG )Parallel

= λ DI ⋅ TEG = [ 0.97014 i − 0.24254 j] ⋅ ( 2.4 lb ) i − ( 7.2 lb ) j − ( 23.4 lb ) k 

= ( 2.3283 + 1.74629 ) = 4.0746 lb

Since

TEG = ( TEG )Perpendicular + (TEG )Parallel

Then

(TEG )Perpendicular

=

(TEG )2 − (TEG )2Parallel

=

( 24.6 )2 − ( 4.0746 )2

= 24.260 lb

and

M DI = (TEG )Perpendicular d

224.74 lb ⋅ ft = ( 24.260 lb ) d d = 9.2638 ft

or d = 9.26 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 3, Solution 66.

From the solution of Prob. 3.55: λ AD = 0.99228 i + 0.124035 j FEF = ( 2.7 kN ) i + (10.8 kN ) j + ( 21.6 kN ) k FEF = 24.3 kN M AD = − 24.916 kN ⋅ m

Only the perpendicular component of FEF contributes to the moment of FEF about edge AD. The parallel component of FEF will be used to find the perpendicular component.

Have

( FEF )Parallel

= λ AD ⋅ FEF = [ 0.99228 i + 0.124035 j] ⋅ ( 2.7 kN ) i + (10.8 kN ) j + ( 21.6 kN ) k 

= 4.0187 kN

Since

FEF = ( FEF )Perpendicular + ( FEF )Parallel

Then

( FEF )Perpendicular

=

( FEF )2 − ( FEF )2Parallel

=

( 24.3)2 − ( 4.0187 )2

= 23.965 kN

and

M AD = ( FEF )Perpendicular d

24.916 kN ⋅ m = ( 23.965 kN ) d d =1.039683m

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

or

d = 1.040 m

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Chapter 3, Solution 67.

From the solution of Prob. 3.56: λ AD = 0.99228 i + 0.124035 j FGH = − ( 7.1 kN ) i + (14.2 kN ) j + (14.2 kN ) k FGH = 21.3 kN M AD = − 35.931 kN ⋅ m

Only the perpendicular component of FGH contributes to the moment of FGH about edge AD. The parallel component of FGH will be used to find the perpendicular component.

Have

( FGH )Parallel

= λ AD ⋅FGH = ( 0.99228 i + 0.124035 j) ⋅  − ( 7.1 kN ) i + (14.2 kN ) j + (14.2 kN ) k 

= − 5.2839 kN

Since

FGH = ( FGH )Perpendicular + ( FGH )Parallel

Then

( FGH )Perpendicular

=

( FGH )2 − ( FGH )2Parallel

=

( 21.3)2 − ( 5.2839 )2

= 20.634 kN

and

M AD = ( FGH )Perpendicular d

35.931 kN ⋅ m = ( 20.634 kN ) d d = 1.741349m

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

or

d = 1.741 m

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Chapter 3, Solution 68.

(a)

M1 = d1F1

Have Where

d1 = 0.6 m and F1 = 40 N

∴ M1 = ( 0.6 m )( 40 N )

or M1 = 24.0 N ⋅ m (b)

Have

M Total = M1 + M 2

8 N ⋅ m = 24.0 N ⋅ m − ( 0.820 m )( cos α )( 24 N ) ∴ cos α = 0.81301

(c)

Have

or

α = 35.6°

or

d 2 = 1.000 m

M1 + M 2 = 0

24 N ⋅ m − d 2 ( 24 N ) = 0

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

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Chapter 3, Solution 69.

(a)

M = Fd

12 N ⋅ m = F ( 0.45 m ) or F = 26.7 N (b)

M = Fd

12 N ⋅ m = F ( 0.24 m ) or F = 50.0 N (c)

M = Fd

Where d =

( 0.45 m )2 + ( 0.24 m )2

= 0.51 m

12 N ⋅ m = F ( 0.51 m ) or F = 23.5 N

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

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Chapter 3, Solution 70.

(a)

Note when a = 8 in., rC/F is perpendicular to the inclined 10 lb forces.

Have

M = ΣFd (

)

= − (10 lb )  a + 8 in. + 2 (1 in.)  − (10 lb )  2a 2 + 2 (1 in.)   

a = 8 in.,

For

M = − (10 lb )(18 in. + 24.627 in.)

= − 426.27 lb ⋅ in.

or M = 426 lb ⋅ in. (b)

Have

M = 480 lb ⋅ in.

Also

M = Σ ( M + Fd ) (

)

= Moment of couple due to horizontal forces at A and D + Moment of force-couple systems at C and F about C. Then

− 480 lb ⋅ in. = −10 lb  a + 8 in. + 2 (1 in.)  +  M C + M F + FX ( a + 8 in.) + Fy ( 2a ) 

Where M C = − (10 lb )(1 in.) = −10 lb ⋅ in.

M F = M C = −10 lb ⋅ in.

Fx =

−10 lb 2

Fy =

−10 lb 2

∴ − 480 lb ⋅ in. = −10 lb ( a + 10 in.) − 10 lb ⋅ in. − 10 lb ⋅ in. −

10 lb 10 lb ( a + 8 in.) − ( 2a ) 2 2

303.43 = 31.213 a or a = 9.72in.

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 3, Solution 71.

(a)

Have

M = ΣFd (

)

= ( 9 lb )(13.8 in.) − ( 2.5 lb )(15.2 in.) = ( 86.2 lb ⋅ in.)

M = 86.2 lb ⋅ in.

(b)

Have

M = Td = 86.2 lb ⋅ in.

For T to be a minimum, d must be maximum. ∴ Tmin must be perpendicular to line AC.

tan θ =

15.2 in.

11.4 in.

θ = 53.130° or θ = 53.1° (c)

Have

M = Tmin d max Where M = 86.2 lb ⋅ in. d max =

(15.2 in.)2 + (11.4 in.)2

+ 2 (1.2 in.)

= 21.4 in. ∴ 86.2 lb ⋅ in. = Tmin ( 21.4 in.)

Tmin = 4.0280 lb

or Tmin = 4.03 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 3, Solution 72.

Based on M = M1 + M 2 M1 = (18 N ⋅ m ) k M 2 = ( 7.5 N ⋅ m ) i ∴ M = ( 7.5 N ⋅ m ) i + (18 N ⋅ m ) k

and

M =

( 7.5 N ⋅ m )2 + (18 N ⋅ m )2

= 19.5 N ⋅ m

or M = 19.50 N ⋅ m With

λ = =

Then

( 7.5 N ⋅ m ) i + (18 N ⋅ m ) k M = M 19.5 N ⋅ m 5 12 i+ k 13 13

cos θ x =

5 13

cosθ y = 0 cosθ z =

12 13

∴ θ x = 67.380° ∴ θ y = 90° ∴ θ z = 22.620° or θ x = 67.4°,

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 y = 90.0°,

θ z = 22.6°

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Chapter 3, Solution 73.

Have

M = M1 + M 2

Where M1 = rC/B × PIC rC/B = ( 38.4 in.) i − (16 in.) j PIC = − ( 25 lb ) k i j k ∴ M1 = 38.4 −16 0 lb ⋅ in.

0 0 −25 = ( 400 lb ⋅ in.) i + ( 960 lb ⋅ in.) j and

M 2 = rD/ A × PZE rD/ A = ( 8 in.) j − ( 22 in.) k JJJG  − (19.2 in.) i + ( 22 in.) k  ED PZ E = PZE = ( 36.5 lb )  ED ( −19.2 in.)2 + ( 22 in.)2 = − ( 24 lb ) i + ( 27.5 lb ) k i j k ∴ M 2 = 0 8 −22 lb ⋅ in. −24 0 27.5 M 2 = ( 220 lb ⋅ in.) i + ( 528 lb ⋅ in.) j + (192 lb ⋅ in.) k

and

M = M1 + M 2 = ( 400 lb ⋅ in.) i + ( 960 lb ⋅ in.) j + ( 220 lb ⋅ in.) i + ( 528 lb ⋅ in.) j + (192 lb ⋅ in.) k 

= ( 620 lb ⋅ in.) i + (1488 lb ⋅ in.) j + (192 lb ⋅ in.) k

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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 M = 

( 620 )2 + (1488)2 + (192 )2  lb ⋅ in. 

= 1623.39 lb ⋅ in.

or M = 1.623 kip ⋅ in. λ=

( 620 lb ⋅ in.) i + (1488 lb ⋅ in.) j + (192 lb ⋅ in.) k M = M 1623.39 lb ⋅ in. = 0.38192 i + 0.91660 j + 0.118271k

cosθ x = 0.38192

or θ x = 67.5°

cosθ y = 0.91660

or θ y = 23.6°

cosθ z = 0.118271

or θ z = 83.2°

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 3, Solution 74.

Have

M = M1 + M 2

Where M1 = rE/D × FD = − ( 0.7 m ) k × ( 80 N ) j = ( 56.0 N ⋅ m ) i

And

M 2 = rG/F × FB

Now

d BF =

( −0.300 m )2 + ( 0.540 m )2 + ( 0.350 m )2

= 0.710 m

Then

FB = λBF FB =

( −0.300 m ) i + ( 0.540 m ) j + ( 0.350 m ) k 0.710 m

( 71 N )

= − ( 30 N ) i + ( 54 N ) j + ( 35 N ) k

∴

M 2 = ( 0.54 m ) j ×  − ( 30 N ) i + ( 54 N ) j + ( 35 N ) k 

= (18.90 N ⋅ m ) i + (16.20 N ⋅ m ) k

Finally M = ( 56.0 N ⋅ m ) i + (18.90 N ⋅ m ) i + (16.20 N ⋅ m ) k  = ( 74.9 N ⋅ m ) i + (16.20 N ⋅ m ) k

and

M =

( 74.9 N ⋅ m )2 + (16.20 N ⋅ m )2

= 76.632 N ⋅ m

cosθ x =

74.9 76.632

or M = 76.6 N ⋅ m cosθ y =

0 76.632

cosθ z =

16.20 76.632

or θ x = 12.20°

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 y = 90.0°

θ z = 77.8°

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Chapter 3, Solution 75.

M = ( M1 + M 2 ) + M P

Have

From the solution to Problem 3.74

( M1 + M 2 ) = ( 74.9 N ⋅ m ) i + (16.20 N ⋅ m ) k Now

M P = rD / E × PE = ( 0.54 m ) j + ( 0.70 m ) k  × ( 90 N ) i

= ( 63.0 N ⋅ m ) j − ( 48.6 N ⋅ m ) k

∴

M = ( 74.9 i + 16.20 k ) + ( 63.0 j − 48.6 k )

= ( 74.9 N ⋅ m ) i + ( 63.0 N ⋅ m ) j − ( 32.4 N ⋅ m ) k and

M =

( 74.9 N ⋅ m )2 + ( 63.0 N ⋅ m )2 + ( − 32.4 N ⋅ m )2

= 103.096 N ⋅ m or M = 103.1 N ⋅ m and

cosθ x =

74.9 103.096

cosθ y =

63.0 103.096

cosθ z =

− 32.4 103.096

or θ x = 43.4°

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 y = 52.3°

θ z = 108.3°

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Chapter 3, Solution 76.

Have

M = M1 + M 2 + M P

From Problem 3.73 solution: M1 = ( 400 lb ⋅ in.) i + ( 960 lb ⋅ in.) j M 2 = ( 220 lb ⋅ in.) i + ( 528 lb ⋅ in.) j + (192 lb ⋅ in.) k Now

M P = rE/ A × PE

rE/ A = (19.2 in.) i + ( 8 in.) j − ( 44 in.) k PE = ( 52.5 lb ) j Therefore i j k M P = 19.2 8 − 44 0 52.5 0 = ( 2310 lb. in.) i + (1008 lb. in.) k and

M = M1 + M 2 + M P

= [(400 + 220 + 2310)i + (960 + 528)j + (192 + 1008)k ] lb ⋅ in. = ( 2930 lb ⋅ in.) i + (1488 lb ⋅ in.) j + (1200 lb ⋅ in.) k M =

( 2930 )2 + (1488)2 + (1200 )2

= 3498.4 lb ⋅ in. or M = 3.50 kip ⋅ in. 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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λ =

M 2930i + 1488j + 1200k = M 3498.4

= 0.83753i + 0.42534 j + 0.34301k cosθ x = 0.83753 or cosθ y = 0.42534 or θ y = 64.8° cosθ z = 0.34301 or θ z = 69.9°

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.

θ x = 33.1°

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Chapter 3, Solution 77.

Have

M = M1 + M 2 + M 3

Where

M1 = − (1.2 lb ⋅ ft ) cos 25° j − (1.2 lb ⋅ ft ) sin 25°k M 2 = − (1.3 lb ⋅ ft ) j M 3 = − (1.4 lb ⋅ ft ) cos 20° j + (1.4 lb ⋅ ft ) sin 20°k

∴

M = ( −1.08757 − 1.3 − 1.31557 ) j + ( − 0.507142 + 0.478828 ) k = − ( 3.7031 lb ⋅ ft ) j − ( 0.028314 lb ⋅ ft ) k

and

M =

( −3.7031)2 + ( − 0.028314 )2

= 3.7032 lb ⋅ ft or M = 3.70 lb ⋅ ft

λ =

M −3.7031j − 0.028314k = M 3.7032

= − 0.99997 j − 0.0076458k cosθ x = 0 or

θ x = 90°

or

θ y = 179.6°

or

θ z = 90.4°

cosθ y = − 0.99997 cosθ z = −0.0076458

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 3, Solution 78.

(a)

FB = P :

∴ FB = 160.0 N

50.0°

M B = −rBA P cos10° = − ( 0.355 m )(160 N ) cos10° = −55.937 N ⋅ m or M B = 55.9 N ⋅ m (b)

FC = P :

∴ FC = 160.0 N

50.0°

M C = M B − rCB ( FB )⊥ = M B − rCB FB sin 55° = −55.937 N ⋅ m − ( 0.305 m )(160 N ) sin 55°

or M C = 95.9 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 3, Solution 79.

(a)

ΣF :

FB = 135 N or FB = 135 N

ΣM :

M B = P dB = (135 N )( 0.125 m ) = 16.875 N ⋅ m

or M B = 16.88 N ⋅ m (b)

ΣM B :

M B = FC d

16.875 N ⋅ m = FC ( 0.075 m ) FC = 225 N

or FC = 225 N ΣF :

0 = − FB + FC FB = FC = 225 N

or FB = 225 N

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

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Chapter 3, Solution 80.

ΣF : PC = P = 700 N

(a) Based on

or PC = 700 N

60°

ΣM C : M C = − Px dCy + Py dCx

Px = ( 700 N ) cos60° = 350 N

where

Py = ( 700 N ) sin 60° = 606.22 N dCx = 1.6 m dCy = 1.1 m ∴ M C = − ( 350 N )(1.1 m ) + ( 606.22 N )(1.6 m ) = −385 N ⋅ m + 969.95 N ⋅ m = 584.95 N ⋅ m or M C = 585 N ⋅ m ΣFx : PDx = P cos60°

(b) Based on

= ( 700 N ) cos 60°

= 350 N ΣM D :

( P cos 60°)( d DA ) =

PB ( d DB )

( 700 N ) cos 60° ( 0.6 m ) = PB ( 2.4 m )

PB = 87.5 N or PB = 87.5 N

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

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ΣFy : P sin 60° = PB + PDy

( 700 N ) sin 60° = 87.5 N + PDy PDy = 518.72 N PD = =

( PDx )2 + ( PDy )

2

( 350 )2 + ( 518.72 )2

= 625.76 N

 PDy  −1  518.72   = tan   = 55.991°  350   PDx 

θ = tan −1 

or PD = 626 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.

56.0°

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Chapter 3, Solution 81.

ΣFx :

2.8cos 65° = FA cos θ + FC cos θ = ( FA + FC ) cosθ

ΣFy :

(1)

2.8sin 65° = FA sin θ + FC sin θ = ( FA + FC ) sin θ

Then

(2)

(2) ⇒ tan 65° = tan θ (1) or θ = 65.0° ΣM A :

( 27 m )( 2.8 kN ) sin 65° = ( 72 m ) ( FC ) sin 65° or FC = 1.050 kN

From Equation (1): 2.8 kN = FA + 1.050 kN or FA = 1.750 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.

∴ FA = 1.750 kN

65.0°

FC = 1.050 kN

65.0°

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Chapter 3, Solution 82.

Based on ΣFx : − ( 54 lb ) cos 30° = − FB cos α − FC cos α

( FB + FC ) cosα

= ( 54 lb ) cos 30°

( 54 lb ) sin 30° =

ΣFy :

(1)

FB sin α + FC sin α

or ( FB + FC ) sin α = ( 54 lb ) sin 30° From

(2)

Eq ( 2 ) : tan α = tan 30° Eq (1) ∴ α = 30°

Based on

ΣM C :

( 54 lb ) cos ( 30° − 20° )  (10 in.) = ( FB cos10° )( 24 in.)

∴ FB = 22.5 lb

From Eq. (1),

or FB = 22.5 lb

30°

or FC = 31.5 lb

30°

( 22.5 + FC ) cos 30° = ( 54 ) cos 30° FC = 31.5 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 3, Solution 83.

(a) Based on − ( 54 lb ) cos 30° = − FC cos 30°

ΣFx : ∴

FC = 54 lb or FC = 54.0 lb

30°

( 54 lb ) cos10°  (10 in.) = M C

ΣM C :

∴

M C = 531.80 lb ⋅ in. or

M C = 532 lb ⋅ in.

(b) Based on ΣFy :

( 54 lb ) sin 30° =

FB sin α = 27

or ΣM B :

FB sin α (1)

531.80 lb ⋅ in. − ( 54 lb ) cos10° ( 24 in.) = − FC ( 24 in.) cos 20° 

FC = 33.012 lb or And

ΣFx :

FC = 33.0 lb

− ( 54 lb ) cos 30° = − 33.012 lb − FB cos α FB cos α = 13.7534

From

Eq (1) : Eq ( 2 )

From Eq. (1),

FB =

tan α =

27 13.7534

(2)

∴ α = 63.006°

27 = 30.301 lb sin ( 63.006° ) or FB = 30.3 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.

63.0°

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Chapter 3, Solution 84.

(a) Have ΣFy :

FC + FD + FE = F F = −200 lb + 150 lb − 150 lb F = − 200 lb or F = 200 lb

ΣM G :

Have

FC ( d − 4.5 ft ) − FD ( 6 ft ) = 0

( 200 lb )( d − 4.5 ft ) − (150 lb )( 6 ft ) = 0 d = 9 ft or d = 9.00 ft (b) Changing directions of the two 150-lb forces only changes the sign of the couple.

∴ F = − 200 lb or F = 200 lb And

ΣM G :

FC ( d − 4.5 ft ) + FD ( 6 ft ) = 0

( 200 lb )( d

− 4.5 ft ) + (150 lb )( 6 ft ) = 0 d =0

or d = 0

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

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Chapter 3, Solution 85.

(a)

Based on ΣFz : − 200 N + 200 N + 240 N = FA FA = 240 N or FA = ( 240 N ) k Based on ΣM A :

(b)

( 200 N )( 0.7 m ) − ( 200 N )( 0.2 m ) = M A M A = 100 N ⋅ m or M A = (100.0 N ⋅ m ) j Based on ΣFz : − 200 N + 200 N + 240 N = F F = 240 N or F = ( 240 N ) k Based on ΣM A :

(c)

100 N ⋅ m = ( 240 N )( x ) x = 0.41667 m or x = 0.417 m From A along AB Based on ΣM B : − ( 200 N )( 0.3 m ) + ( 200 N )( 0.8 m ) − P (1 m ) = R ( 0 ) P = 100 N or P = 100.0 N

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

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Chapter 3, Solution 86.

Let R be the single equivalent force... ΣF :

R = FA + FC = ( 260 N )( cos10° i − sin10° k ) + ( 320 N )( − cos8° i − sin 8° k ) = − ( 60.836 N ) i − ( 89.684 N ) k or R = − ( 60.8 N ) i − ( 89.7 N ) k

ΣM A :

rAD Rx = rAC FC cos8° rAD ( 60.836 N ) = ( 0.690 m )( 320 N ) cos8° rAD = 3.5941 m ∴ R Would have to be applied 3.59 m to the right of A on an extension of handle ABC.

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 3, Solution 87.

(a)

Have

ΣF :

FB + FC + FD = FA

Since

FB = − FD ∴

FA = FC = 22 lb

20° or FA = 22.0 lb

Have

ΣM A :

20°

− FBT ( r ) − FCT ( r ) + FDT ( r ) = M A − ( 28 lb ) sin15°  ( 8 in.) − ( 22 lb ) sin 25°  ( 8 in.) + ( 28 lb ) sin 45°  ( 8 in.) = M A

M A = 26.036 lb ⋅ in. or M A = 26.0 lb ⋅ in. (b)

Have

ΣF :

FA = FE or FE = 22.0 lb

ΣM : ∴

20°

M A = [ FE cos 20°] ( a ) 26.036 lb ⋅ in. = ( 22 lb ) cos 20°  ( a )

a = 1.25941 in. or a = 1.259 in. Below 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 3, Solution 88.

(a)

Let R be the single equivalent force. Then

R = (120 N ) k ΣM B :

R = 120 N

− a (120 N ) = − ( 0.165 m )( 90 N ) cos15° + ( 0.201 m )( 90 N ) sin15° a = 0.080516 m

∴ The line of action is y =

201 mm − 80.516 mm = 19.984 mm 2 or y = 19.98 mm

(b)

ΣM B :

Then

− ( 0.201 − 0.040 ) m  (120 N ) = − ( 0.165 m )( 90 N ) cosθ + ( 0.201 m )( 90 N ) sin θ

or

cosθ − 1.21818sin θ = 1.30101

or

cos 2 θ = (1.30101 + 1.21818sin θ )

or

1 − sin 2 θ = 1.69263 + 3.1697sin θ + 1.48396sin 2 θ

or

2.48396sin 2 θ + 3.1697 sin θ + 0.69263 = 0 sin θ =

−3.1697 ±

2

( 3.1697 )2 − 4 ( 2.48396 )( 0.69263) 2 ( 2.48396 ) or θ = −16.26°

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

and

θ = −85.0°

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Chapter 3, Solution 89.

(a)

First note that F = P and that F must be equivalent to (P, MD) at point D, Where

M D = 57.6 N ⋅ m

For F = ( F )min

F must act as far from D as possible ∴ Point of application is at point B

(b) For ( F )min Now

F must be perpendicular to BD d DB =

( 630 mm )2 + ( −160 mm )2

= 650 mm

tan α =

63 16

α = 75.7° Then

M D = d DB F

57.6 N ⋅ m = ( 0.650 m ) F F = 88.6 N

or F = 88.6 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.

75.7°

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Chapter 3, Solution 90.

Have

ΣF :

− ( 250 kN ) j = F or F = − ( 250 kN ) j

Also have

ΣM G :

rP × P = M

i j k − 0.030 0 0.060 kN ⋅ m = M 0 − 250 0 ∴ M = (15 kN ⋅ m ) i + ( 7.5 kN ⋅ m ) k or M = (15.00 kN ⋅ m ) i + ( 7.50 kN ⋅ m ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 3, Solution 91.

Have ΣF :

where TAB

TAB = F JJJG AB = TAB AB

= ( 54 lb )

2.25i − 18 j + 9k

( 2.25)2 + ( −18)2 + ( 9 )2

= ( 6 lb ) i − ( 48 lb ) j + ( 24 lb ) k F = ( 6.00 lb ) i − ( 48.0 lb ) j + ( 24.0 lb ) k

So that Have ΣM E :

rA/E × TAB = M

i j k 0 22.5 0 lb ⋅ ft = M 6 − 48 24 ∴ M = ( 540 lb ⋅ ft ) i − (135 lb ⋅ ft ) k or

M = ( 540 lb ⋅ ft ) i − (135.0 lb ⋅ ft ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 3, Solution 92.

Have ΣF :

where TCD

TCD = F JJJG CD = TCD CD = ( 61 lb )

− 0.9i − 16.8j + 7.2k

( − 0.9 )2 + ( −16.8)2 + ( 7.2 )2

= − ( 3 lb ) i − ( 56 lb ) j + ( 24 lb ) k So that

F = − ( 3.00 lb ) i − ( 56.0 lb ) j + ( 24.0 lb ) k

Have ΣM O = rC/D × TCD = M i j k 0 22.5 0 lb ⋅ ft = M − 3 − 56 24 ∴ M = ( 540 lb ⋅ ft ) i + ( 67.5 lb ⋅ ft ) k M = ( 540 lb ⋅ ft ) i + ( 67.5 lb ⋅ ft ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 3, Solution 93.

Have ΣF :

TAB = F

where TAB = TAB

JJJG AB AB

= (10.5 kN )

− i − 4.75j + 2k

( −1)2 + ( − 4.75)2 + ( 2 )2

= − ( 2 kN ) i − ( 9.5 kN ) j + ( 4 kN ) k F = − ( 2.00 kN ) i − ( 9.50 kN ) j + ( 4.00 kN ) k

So that Have ΣM O :

rA × TAB = M

i j k 3 4.75 0 kN ⋅ m = M − 2 − 9.5 4 ∴ M = (19 kN ⋅ m ) i − (12 kN ⋅ m ) j − (19 kN ⋅ m ) k M = (19.00 kN ⋅ m ) i − (12.00 kN ⋅ m ) j − (19.00 kN ⋅ m ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 3, Solution 94.

Let ( R, M O ) be the equivalent force-couple system Then R = ( 220 N )( − sin 60° j − cos 60° k )

(

= (110 N ) − 3 j − k

) or R = − (190.5 N ) j − (110 N ) k

Now

ΣM O :

Where

rOC = ( 0.2 m ) i + ( 0.1 − 0.4sin 20° ) m  j + ( 0.4 cos 20° m ) k

Then

MO

M O = rOC × R

i j k = − ( 0.1)(110 N ) 2 (1 − 4sin 20° ) 4 cos 20° ( m ) 0

3

1

{

= − (11 N ⋅ m ) (1 − 4sin 20° )(1) − ( 4cos 20° )  or

( 3 ) i − 2 j + 2

3k

}

M O = ( 75.7 N ⋅ m ) i + ( 22.0 N ⋅ m ) j − ( 38.1 N ⋅ m ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 3, Solution 95.

Have ΣF :

F = FD JJG AI where F = F AI = ( 63 lb )

14.4i − 4.8j + 7.2k

(14.4 )2 + ( − 4.8)2 + ( 7.2 )2 F = ( 54.0 lb ) i − (18.00 lb ) j + ( 27.0 lb ) k

So that Have ΣM D :

M + rI /O × F = M D

JJJG AC where M = M AC

= ( 560 lb ⋅ in.)

9.6 i − 7.2 k

( 9.6 )2 + ( − 7.2 )2

= ( 448 lb ⋅ in.) i − ( 336 lb ⋅ in.) k Then M D

i j k = ( 448 lb ⋅ in.) i − ( 336 lb ⋅ in.) k + 0 0 14.4 lb ⋅ in.

54 −18 27

= ( 448 lb ⋅ in.) i − ( 336 lb ⋅ in.) k + ( 259.2 lb ⋅ in.) i + ( 777.6 lb ⋅ in.) j

or

M D = ( 707 lb ⋅ in.) i + ( 778 lb ⋅ in.) j − ( 336 lb ⋅ in.) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 3, Solution 96.

First assume that the given force W and couples M1 and M 2 act at the origin.

Now

W = − Wj

and

M = M1 + M 2 = − ( M 2 cos 25° ) i + ( M1 − M 2 sin 25° ) k

Note that since W and M are perpendicular, it follows that they can be replaced with a single equivalent force.

F =W

(a) Have

or

F = − Wj = − ( 2.4 N ) j or F = − ( 2.40 N ) j W

(b) Assume that the line of action of F passes through point P (x, 0, z).

Then for equivalence M = rP/O × F rP/O = xi + zk

where ∴

− ( M 2 cos 25° ) i + ( M1 − M 2 sin 25° ) k

i

j

k

= x 0 z = (Wz ) i − (Wx ) k 0 −W 0

Equating the i and k coefficients,

z = (b) For

−M z cos 25° W

and

 M − M 2 sin 25°  x = − 1  W  

W = 2.4 N, M1 = 0.068 N ⋅ m, M 2 = 0.065 N ⋅ m x=

0.068 − 0.065sin 25° = − 0.0168874 m − 2.4 or x = −16.89 mm W

z =

− 0.065cos 25° = − 0.024546 m 2.4 or z = − 24.5 mm 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 3, Solution 97.

ΣM Bz : M 2 z = 0

(a) Have

(

)

k ⋅ rH /B × F1 + M1z = 0

(1)

rH /B = ( 31 in.) i − ( 2 in.) j

where

F1 = λ EH F1 = =

( 6 in.) i + ( 6 in.) j − ( 7 in.) k 11.0 in.

( 20 lb )

20 lb ( 6i + 6 j − 7k ) 11.0

M1z = k ⋅ M1 M1 = λ EJ M1 =

−di + ( 3 in.) j − ( 7 in.) k d 2 + 58 in.

( 480 lb ⋅in.)

Then from Equation (1), 0

0

1

31 −2 0 6 6 −7

20 lb ⋅ in. ( −7 )( 480 lb ⋅ in.) + =0 11.0 d 2 + 58 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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Solving for d, Equation (1) reduces to 20 lb ⋅ in.

3360 lb ⋅ in. =0 (186 + 12 ) − 2 11.0 d + 58

d = 5.3955 in.

From which

or d = 5.40 in. (b)

F2 = F1 =

20 lb ( 6i + 6 j − 7k ) 11.0

= (10.9091i + 10.9091j − 12.7273k ) lb or F2 = (10.91 lb ) i + (10.91 lb ) j − (12.73 lb ) k M 2 = rH /B × F1 + M1 i j k 20 lb ⋅ in. ( − 5.3955 ) i + 3j − 7k = 31 − 2 0 + ( 480 lb ⋅ in.)

11.0 9.3333 6 6 −7 = ( 25.455i + 394.55j + 360k ) lb ⋅ in. + ( − 277.48i + 154.285j − 360k ) lb ⋅ in.

M 2 = − ( 252.03 lb ⋅ in.) i + ( 548.84 lb ⋅ in.) j or M 2 = − ( 21.0 lb ⋅ ft ) i + ( 45.7 lb ⋅ ft ) j

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

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Chapter 3, Solution 98.

(a) a : ΣFy :

Ra = − 400 N − 600 N

or R a = 1000 N ΣM B :

M a = ( 2 kN ⋅ m ) + ( 2 kN ⋅ m ) + ( 5 m )( 400 N )

or M a = 6.00 kN ⋅ m b : ΣFy :

Rb = −1200 N + 200 N

or R b = 1000 N ΣM B :

M b = ( 0.6 kN ⋅ m ) + ( 5 m )(1200 N )

or M b = 6.60 kN ⋅ m c : ΣFy :

Rc = 200 N − 1200 N

or R c = 1000 N ΣM B :

M c = − ( 4 kN ⋅ m ) − (1.6 kN ⋅ m ) − ( 5 m )( 200 N )

or M c = 6.60 kN ⋅ m d : ΣFy :

Rd = − 800 N − 200 N

or R d = 1000 N ΣM B :

M d = − (1.6 kN ⋅ m ) + ( 4.2 kN ⋅ m ) + ( 5 m )( 800 N )

or M d = 6.60 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.

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e : ΣFy :

Re = − 500 N − 400 N

or R e = 900 N ΣM B :

M e = ( 3.8 kN ⋅ m ) + ( 0.3 kN ⋅ m ) + ( 5 m )( 500 N )

or M e = 6.60 kN ⋅ m f : ΣFy :

R f = 400 N − 1400 N

or R f = 1000 N ΣM B :

M f = ( 8.6 kN ⋅ m ) − ( 0.8 kN ⋅ m ) − ( 5 m )( 400 N )

or M f = 5.80 kN ⋅ m g : ΣFy :

Rg = −1200 N + 300 N

or R g = 900 N ΣM B :

M g = ( 0.3 kN ⋅ m ) + ( 0.3 kN ⋅ m ) + ( 5 m )(1200 N ) or M g = 6.60 kN ⋅ m

h : ΣFy :

Rh = − 250 N − 750 N or R h = 1000 N

ΣM B :

M h = − ( 0.65 kN ⋅ m ) + ( 6 kN ⋅ m ) + ( 5 m )( 250 N ) or M h = 6.60 kN ⋅ m (b) The equivalent loadings are (b), (d), (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 3, Solution 99.

The equivalent force-couple system at B is... ΣFy :

R = − 650 N − 350 N or R = 1000 N

ΣM B :

M = (1.6 m )( 800 N ) + (1.27 kN ⋅ m ) + ( 5 m )( 650 N ) or M = 5.80 kN ⋅ m

∴ The equivalent loading of Problem 3.98 is (f)

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

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Chapter 3, Solution 100.

Equivalent force system... (a) ΣFy :

R = − 400 N − 200 N or R = 600 N

ΣM A :

− d ( 600 N ) = − ( 200 N ⋅ m ) + (100 N ⋅ m ) − ( 4 m )( 200 N ) or d = 1.500 m

(b) ΣFy :

R = − 400 N + 100 N or R = 300 N

ΣM A :

− d ( 300 N ) = − ( 200 N ⋅ m ) − ( 600 N ⋅ m ) + ( 4 m )(100 N ) or d = 1.333 m

(c) ΣFy :

R = − 400 N − 100 N or R = 500 N

ΣM A :

− d ( 500 N ) = − ( 200 N ⋅ m ) − ( 200 N ⋅ m ) − ( 4 m )(100 N ) or d = 1.600 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 3, Solution 101.

The equivalent force-couple system at A for each of the five force-couple systems will be determined and compared to F = ( 2 lb ) j

M = ( 48 lb ⋅ in.) i + ( 32 lb ⋅ in.) k

To determine if they are equivalent Force-couple system at B: Have ΣF :

F = ( 2 lb ) j

and ΣM A :

M = ΣM B + rB/ A × FB

(

)

M = ( 32 lb ⋅ in.) i + (16 lb ⋅ in.) k + ( 8 in.) i × ( 2 lb ) j = ( 32 lb ⋅ in.) i + ( 32 lb ⋅ in.) k ∴ is not equivalent

Force-couple system at C: Have ΣF : And ΣM A :

F = ( 2 lb ) j

(

M = M C + rC/ A × FC

)

M = ( 68 lb ⋅ in.) i + ( 8 in.) i + (10 in.) k  × ( 2 lb ) j

= ( 48 lb ⋅ in.) i + (16 lb ⋅ in.) k ∴ is not equivalent 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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Force-couple system at E: Have ΣF :

F = ( 2 lb ) j

and ΣM A :

M = M E + rE/ A × FE

(

)

M = ( 48 lb ⋅ in.) i + (16 in.) i − ( 3.2 in.) j × ( 2 lb ) j

= ( 48 lb ⋅ in.) i + ( 32 lb ⋅ in.) k ∴ is equivalent

Force-couple system at G: Have ΣF :

F = ( 2 lb ) i + ( 2 lb ) j

F has two force components

∴ is not equivalent

Force-couple system at I: Have ΣF :

F = ( 2 lb ) j

and ΣM A :

ΣM I + rI / A × FI

(

)

M = ( 80 lb ⋅ in.) i − (16 in.) k + (16 in.) i − ( 8 in.) j + (16 in.) k  × ( 2 lb ) j

M = ( 48 lb ⋅ in.) i + (16 lb ⋅ in.) k

∴ is not equivalent

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 3, Solution 102.

WA = mA g = ( 38 kg ) g

First

WB = mB g = ( 29 kg ) g WC = mC g = ( 27 kg ) g

(a)

For resultant weight to act at C, Then

ΣM C = 0

( 38 kg ) g  ( 2 m ) − ( 27 kg ) g  ( d ) − ( 29 kg ) g  ( 2 m ) = 0

∴ d =

76 − 58 = 0.66667 m 27

or d = 0.667 m WC = mC g = ( 24 kg ) g

(b)

For resultant weight to act at C, Then

ΣM C = 0

( 38 kg ) g  ( 2 m ) − ( 24 kg ) g  ( d ) − ( 29 kg ) g  ( 2 m ) = 0

∴ d =

76 − 58 = 0.75 m 24 or d = 0.750 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 3, Solution 103.

(a) Have ΣF :

− WC − WD − WE = R

∴ R = − 200 lb − 175 lb − 135 lb = − 510 lb or R = 510 lb

Have ΣM A :

− ( 200 lb )( 4.5 ft ) − (175 lb )( 7.8 ft ) − (135 lb )(12.75 ft ) = − R ( d ) ∴ − 3986.3 lb ⋅ ft = ( − 510 lb ) d or d = 7.82 ft (b) For equal reactions at A and B, The resultant R must act at midspan.

From

L ΣM A = − R   2 ∴ − ( 200 lb )( 4.5 ft ) − (175 lb )( 4.5 ft + a ) − (135 lb )( 4.5 ft + 2.5 a ) = − ( 510 lb )( 9 ft ) or 2295 + 512.5 a = 4590 and a = 4.48 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 3, Solution 104.

Have ΣF :

−12 kN − WL − 18 kN = − 40 kN − 40 kN

WL = 50 kN or WL = 50.0 kN ΣM B :

(12 kN )( 5 m ) + ( 50 kN ) d

= ( 40 kN )( 5 m )

d = 2.8 m or heaviest load ( 50 kN ) is located 2.80 m from front axle

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 3, Solution 105.

(a) ΣF :

R = ( 80 N ) i − ( 40 N ) j − ( 60 N ) j + ( 90 N )( − sin 50°i − cos 50° j) = (11.0560 N ) i − (157.851 N ) j

R=

(11.0560 N )2 + ( −157.851 N )2

= 158.2 N

tan θ =

−157.851 11.0560

θ = 86.0° or R = 158.2 N

86.0°

(b)

ΣM F : d − (157.851 N ) = ( 0.32 m )( 80 N ) − ( 0.15 m )( 40 N ) − ( 0.35 m )( 60 N ) − ( 0.61 m )( 90 N ) cos 50° − ( 0.16 m )( 90 N ) sin 50°

or d = 302 mm to the right of F

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

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Chapter 3, Solution 106.

(a) ΣM I :

0 = ( 0.32 m )( 80 N ) + ( 0.1 m )( 40 N ) − ( 0.1 m )( 60 N ) − ( 0.36 m )( 90 N ) cos α − ( 0.16 m )( 90 N ) sin α or 4sin α + 9cos α = 6.5556

( 9cosα )2

= ( 6.5556 − 4sin α )

(

2

)

81 1 − sin 2 α = 42.976 − 52.445sin α + 16sin 2 α 97sin 2 α − 52.445sin α − 38.024 = 0 Solving by the quadratic formula gives for the positive root sin α = 0.95230

α = 72.233° or α = 72.2° Note: The second root (α = − 24.3° ) is rejected since 0 < α < 90°. (b) ΣF :

R = ( 80 N ) i − ( 40 N ) j − ( 60 N ) j + ( 90 N )( − sin 72.233°i − cos 72.233° j) = − ( 5.7075 N ) i − (127.463 N ) j R=

( − 5.7075 N )2 + ( −127.463 N )2

= 127.6 N

tan θ =

−127.463 − 5.7075

θ = 87.4° or

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

R = 127.6 N

87.4°

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Chapter 3, Solution 107.

(a) Have

ΣM D :

0 = M − ( 0.8 in.)( 40 lb ) − ( 2.9 in.)( 20 lb ) cos 30° − ( 3.3 in.)( 20 lb ) sin 30° or M = 115.229 lb ⋅ in. or

M = 115.2 lb ⋅ in.

Now, R is oriented at 45° as shown (since its line of action passes through B and D).

0 = ( 40 lb ) cos 45° − ( 20 lb ) cos15°

Have ΣFx′ :

− ( 90 lb ) cos (α + 45° ) or α = 39.283° or (b) ΣFx :

α = 39.3°

Rx = 40 − 20sin 30° − 90cos 39.283°

= − 39.663 lb Now

R=

2 Rx

or

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

R = 56.1 lb

45.0°

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Chapter 3, Solution 108.

(a) Reduce system to a force and couple at B:

Have R = ΣF = − (10 lb ) j + ( 25 lb ) cos 60°i + ( 25 lb ) sin 60° j − ( 40 lb ) i = − ( 27.5 lb ) i + (11.6506 lb ) j or R =

( − 27.5 lb )2 + (11.6506 lb )2

= 29.866 lb

 11.6506 

θ = tan −1   = 22.960°  27.5  or

R = 29.9 lb

Also M B = ΣM B = ( 80 lb ⋅ in.) k − (12 in.) i × ( −10 lb ) j − ( 8 in.) j × ( − 40 lb ) i = − (120 lb ⋅ in.) k (b)

Have M B = − (120 lb ⋅ in.) k = − ( u ) i × (11.6506 lb ) j − (120 lb ⋅ in.) k = − (11.6506 lb )( u ) k u = 10.2999 in. and x = 12 in. − 10.2999 in.

= 1.7001 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.

23.0°

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Have M B = − (120 lb ⋅ in.) k = − ( v ) j × ( − 27.5 lb ) i − (120 lb ⋅ in.) k = − ( 27.5 lb )( v ) k v = 4.3636 in.

and y = 8 in. − 4.3636 in. = 3.6364 in. or

1.700 in. to the right of A and 3.64 in. above C

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

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Chapter 3, Solution 109. (a)

Position origin along centerline of sheet metal at the intersection with line EF. (a)

ΣF = R

Have

R =  − 0.52 j − 1.05 j − 2.1( sin 45° i + cos 45° j) − 0.64 i  kips

R = − ( 2.1249 kips ) i − ( 3.0549 kips ) j

( − 2.1249 )2 + ( − 3.0549 )2

R=

= 3.7212 kips  − 3.0549   = 55.179°  − 2.1249 

θ = tan −1 

or Have

R = 3.72 kips

55.2°

M EF = ΣM EF

Where M EF = ( 0.52 kip )( 3.6 in.) + (1.05 kips )(1.6 in.)

− ( 2.1 kips )( 0.8 in.) − ( 0.64 kip ) (1.6 in.) sin 45° + 1.6 in. = 0.123923 kip ⋅ in.

To obtain distance d left of EF, Have

M EF = dRy = d ( − 3.0549 kips ) d =

0.123923 kip ⋅ in. = −0.040565 in. − 3.0549 kips

or

d = 0.0406 in. left of EF

(b) Have

M EF = ΣM EF = 0

M EF = ( 0.52 kip )( 3.6 in.) + (1.05 kips )(1.6 in.) − ( 2.1 kips )( 0.8 in.) − ( 0.64 kip ) (1.6 in.) sin α + 1.6 in. ∴

(1.024 kip ⋅ in.) sin α

= 0.848 kip ⋅ in.

or

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

α = 55.9°

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Chapter 3, Solution 110.

(a)

Have

ΣF = R R =  − 0.52 j − 1.05j − 2.1( sin α i + cos α j) − 0.64i  kips = − 0.64 kip + ( 2.1 kips )( sin α ) i  − 1.57 kips + ( 2.1 kips ) cos α  j

Then

tan α =

Rx 0.64 + 2.1sin α = Ry 1.57 + 2.1cos α

1.57 tan α + 2.1sin α = 0.64 + 2.1sin α tan α =

0.64 1.57

α = 22.178°

or (b)

From α = 22.178° Rx = − 0.64 kip − ( 2.1 kips ) sin 22.178° = −1.43272 kips

Ry = −1.57 kips − ( 2.1 kips ) cos 22.178° = − 3.5146 kips

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

α = 22.2°

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R=

( −1.43272 )2 + ( − 3.5146 )2

= 3.7954 kips

or Then

R = 3.80 kips

67.8°

M EF = ΣM EF

Where M EF = ( 0.52 kip )( 3.6 in.) + (1.05 kips )(1.6 in.) − ( 2.1 kips )( 0.8 in.) − ( 0.64 kip ) (1.6 in.) sin 22.178° + 1.6 in.

= 0.46146 kip ⋅ in.

To obtain distance d left of EF, Have

M EF = dRy = d ( − 3.5146 kips )

d =

0.46146 kip ⋅ in. − 3.5146 kips

= − 0.131298 in.

or

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

d = 0.1313 in. left of EF

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Chapter 3, Solution 111.

Equivalent force-couple at A due to belts on pulley A ΣF : −120 N − 160 N = RA

Have

∴ R A = 280 N ΣM A : − 40 N ( 0.02 m ) = M A

Have

∴ M A = 0.8 N ⋅ m Equivalent force-couple at B due to belts on pulley B ΣF :

Have

( 210 N + 150 N )

∴ R B = 360 N

25° = R B 25°

ΣM B : − 60 N ( 0.015 m ) = M B

Have

∴ M B = 0.9 N ⋅ m Equivalent force-couple at F Have

ΣF : R F = ( −280 N ) j + ( 360 N )( cos 25°i + sin 25° j) = ( 326.27 N ) i − (127.857 N ) j

R = RF =

2 2 RFx + RFy =

( 326.27 )2 + (127.857 )2

= 350.43 N

 RFy  −1  −127.857   = tan   = −21.399°  326.27   RFx 

θ = tan −1 

or R F = R = 350 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.

21.4°

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Have ΣM F : M F = − ( 280 N )( 0.06 m ) − 0.80 N ⋅ m − ( 360 N ) cos 25° ( 0.010 m ) + ( 360 N ) sin 25° ( 0.120 m ) − 0.90 N ⋅ m

M F = − ( 3.5056 N ⋅ m ) k To determine where a single resultant force will intersect line FE,

M F = dR y ∴ d =

MF −3.5056 N ⋅ m = = 0.027418 m = 27.418 mm Ry −127.857 N

or d = 27.4 mm

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 3, Solution 112. (a)

Have

R = ΣF

R = ( 25 N )( cos 40°i + sin 40° j) − (15 N ) i − (10 N ) j = ( 4.1511 N ) i + ( 6.0696 N ) j R =

( 4.1511)2 + ( 6.0696 )2

= 7.3533 N

 6.0696 

θ = tan −1    4.1511  = 55.631°

R = 7.35 N

or

55.6°

(a) and (b) (b)

From

M B = ΣM B = dRy

where

M B = − ( 25 N ) cos 40° ( 0.375 m ) sin 50° − ( 25 N ) sin 40° ( 0.375 m ) cos 50° + (15 N ) ( 0.150 m ) sin 50° − (10 N )( 0.150 m ) + 6.25 N ⋅ m ∴ M B = −2.9014 N ⋅ m

and

d =

=

MB Ry

−2.9014 N ⋅ m 6.0696 N

= 0.47802 m or

(c)

d = 478 mm to the left of B

From M B = rD/B × R

− ( 2.9014 N ⋅ m ) k = ( −d1 cos 50°i + d1 sin 50° j) × ( 4.1511 N ) i + ( 6.096 N ) j − ( 2.9014 N ⋅ m ) k = − ( 7.0814 d1 ) k ∴ d1 = 0.40972 m

or d1 = 410 mm from B along line AB or 34.7 mm above and to left of 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 3, Solution 113.

ΣFx = 0

Based on

P cos α − 15 N = 0 ∴ P cos α = 15 N

(1)

ΣFy = 0

and

P sin α − 10 N = 0 ∴ P sin α = 10 N

(2)

Dividing Equation (2) by Equation (1), tan α =

10 15

∴ α = 33.690° Substituting into Equation (1), P=

15 N = 18.0278 N cos33.690°

or

P = 18.03 N

(a) Based on

ΣM B = 0

33.7°

− (18.0278 N ) cos 33.690° ( d + 0.150 m ) sin 50° − (18.0278 N ) sin 33.690° ( d + 0.150 m ) cos 50° + (15 N ) ( 0.150 m ) sin 50° − (10 N )( 0.150 m ) + 6.25 N ⋅ m = 0

−17.9186d = −3.7858 ∴ d = 0.21128 m or d = 211 mm

Vector Mechanics for Engineers: Statics and Dynamics, 8/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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(b) Based on

ΣM D = 0

− (18.0278 N ) cos 33.690° ( d + 0.150 m ) sin 50° − (18.0278 N ) sin 33.690°  ( d + 0.150 m ) cos50° + 0.150 m  + (15 N ) ( 0.150 m ) sin 50° + 6.25 N ⋅ m = 0

−17.9186d = −3.7858 ∴ d = 0.21128 m or d = 211 mm This result is expected, since R = 0 and M RB = 0 for d = 211 mm implies that R = 0 and M = 0 at any other point for the value of d found in part 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 3, Solution 114.

(a)

Let ( R, M D ) be the equivalent force-couple system at D. First note...

At

x = b; y = h

For

y = k x2

We have h = k b 2 h or k = 2 b  h  ∴ y =  2  x2 b   h  y =  2  x2 For any contact point c alone the surface b  dy h =2 2x dx b

 b2  tan −1    2hx 

R = F and

ΣΜ D : M D = − ( x ) F sin θ + ( h − y ) F cosθ  b2 = −x F   b 4 + 4h 2 x 2 

or

MD

  2h x  +  h − h x 2  F 2 4   b  b + 4h 2 x 2 

  h 2  2  − xb +  h − 2 x  ( 2hx )  b    = F  4 2 2 b + 4h x  2h 2 x 3  2 2  − xb + 2h x −  b2   = F b 4 + 4h 2 x 2

or M D

 2h 2 x3  2 2  2h − b x −  b2  = F   b 4 + 4h 2 x 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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b = 1 ft, h = 2 ft

(b) With

 7 x − 8x3  MD =  F 2   1 + 16 x 

For M D to be a maximum

Then

dM D dx

  7 − 24 x 2 = 0 = F   

(

)

1 1 + 16 x 2 − 7 x − 8x3   ( 32 x ) 1 + 16 x 2 2 2 1 + 16 x

( (

) )

(

)

−

1 2

     

For the non-trivial solution:

(

)(

)

(

0 = 7 − 24 x 2 1 + 16 x 2 − 16 x 7 x − 8 x3

)

0 = 256 x 4 + 24 x 2 − 7

Solving by the quadratic formula gives for the positive root. x = 0.354 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 3, Solution 115.

For equivalence ΣF : FB + FC + FD = R A

R A = − ( 240 N ) j − (125 N ) k − ( 300 N ) i + (150 N ) k ∴ R A = − ( 300 N ) i − ( 240 N ) j + ( 25 N ) k

Also for equivalence ΣΜ Α : rB/ A × FB + rC/ A × FC + rD/ A × FD = M A

or M A

i j k i j k i j k = 0 0.12 m 0 + 0.06 m 0.03 m − 0.075 m + 0.06 m 0.08 m − 0.75 m 0 − 240 N −125 N −300 N 0 0 0 0 150 N =  − (15 N ⋅ m ) i  + ( 22.5 N ⋅ m ) j + ( 9 N ⋅ m ) k  + (12 N ⋅ m ) i − ( 9 N ⋅ m ) j

or M A = − ( 3 N ⋅ m ) i + (13.5 N ⋅ m ) j + ( 9 N ⋅ m ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 3, Solution 116.

Let ( R, M D ) be the equivalent force-couple system at O. Now ΣF : R = ΣF = (1.8 lb )( − sin 40°i − cos 40°k ) + (11 lb )( − sin12° j − cos12°k ) + (18 lb )( − sin15° j − cos15°k ) or R = − (1.157 lb ) i − ( 6.95 lb ) j − ( 29.5 lb ) k Note that each belt force may be replaced by a force-couple that is equivalent to the same force plus the moment of the force about the shaft (x axis) of the sander. Then ... ΣM O : M O = ΣM O i j k = (1.8 lb ) 0 0.75 in. 2.2 in. − sin 40° 0 − cos 40° − ( 2.5 in.)(11 lb )  i − ( 9 in.) i × (11 lb )( − sin12° j − cos12°k ) + ( 2.5 in.)(18 lb )  i − ( 9 in.) i × (18 lb )( − sin15° j − cos15°k ) = (1.8 )( −0.75cos 40°i − 2.2sin 40° j + 0.75sin 40°k ) − 27.5i + ( 99 )( sin12°k − cos12° j) + 45i + (162 )( sin15°k − cos15° j)  ( lb ⋅ in.) = ( −1.03416 − 27.5 + 45 ) i + ( −2.5454 − 96.837 − 156.480 ) j + ( 0.86776 + 20.583 + 41.929 ) k  ( lb ⋅ in.)

or M O = (16.47 lb ⋅ in.) i − ( 256 lb ⋅ in.) j + ( 63.4 lb ⋅ in.) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 3, Solution 117.

Have

and Now

ΣFx :

−10 N = Ax + Bx

ΣFy :

0 = Ay + By

ΣFz :

6 N = Az + Bz

ΣM O : d BA :

⇒ Bx = −10 − Ax ⇒ By = − Ay ⇒ Bz = 6 − Az

M O = rO/ A × A + rO/B × B 372 mm =

( 60 mm )2 + ( −72 mm )2 + ( d BA )2z

or ( d BA ) z = 360 mm

rO/ A = (135 mm ) i − ( 72 mm ) j + ( 310 mm ) k

Then

rO/B = ( 75 mm ) i − ( 50 mm ) k

( 60 N ⋅ m ) i + ( 0.05 N ⋅ m ) j − (10 N ⋅ m ) k i j k i j k = 0.135 − 0.72 0.310 ( N ⋅ m ) + 0.075 0 − 0.050 ( N ⋅ m ) Ax Ay Az Bx By Bz

i:

(

)

60 = −0.072 Az − 0.310 Ay + ( 0.050 ) By or 60 = −0.072 Az − 0.360 Ay

j:

0.05 = ( 0.310 Ax − 0.135 Az ) + ( −0.050 Bx − 0.075 Bz ) = 0.310 Ax − 0.050 ( −10 − Ax ) − 0.135 Az − 0.075 ( 6 − Az ) or 0 = 0.360 Ax − 0.060 Az Az = 6 Ax

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

(1)

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Now

Ax = 2 N

From equation (1)

∴ Az = 12.00 N 60 = −0.072 (12.00 ) − 0.360 Ay or Ay = −169.1 N

Then

Bx = −12.00 N By = 169.1 N Bz = −6.00 N ∴ A = ( 2.00 N ) i − (169.1 N ) j + (12.00 N ) k B = − (12.00 N ) i + (169.1 N ) j − ( 6.00 N ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 3, Solution 118.

ΣF : B + C = R

Have

ΣFx : Bx + C x = 3.9 lb

or

ΣFy : C y = Ry

(2)

ΣFz : C z = −1.1 lb

(3)

Bx = 3.9 lb − Cx

(1)

ΣM A : rB/ A × B + rC/ A × C + M B = M RA

Have

i j k i j k 1 1 4 0 2.0 + ( 2 lb ⋅ ft ) i = M xi + (1.5 lb ⋅ ft ) j − (1.1 lb ⋅ ft ) k ∴ x 0 4.5 + 12 12 Bx 0 0 C x C y −1.1

( 2 − 0.166667C y ) i + ( 0.375Bx + 0.166667Cx + 0.36667 ) j + ( 0.33333C y ) k = M xi + (1.5 ) j − (1.1) k From

(a)

i - coefficient

2 − 0.166667C y = M x

(4)

j - coefficient

0.375Bx + 0.166667Cx + 0.36667 = 1.5

(5)

k - coefficient

0.33333C y = −1.1

(6)

or

C y = −3.3 lb

From Equations (1) and (5): 0.375 ( 3.9 − Cx ) + 0.166667Cx = 1.13333 Cx =

0.32917 = 1.58000 lb 0.20833 From Equation (1):

Bx = 3.9 − 1.58000 = 2.32 lb

∴ B = ( 2.32 lb ) i C = (1.580 lb ) i − ( 3.30 lb ) j − (1.110 lb ) k (b)

From Equation (2): or R y = − ( 3.30 lb ) From Equation (4):

M x = −0.166667 ( −3.30 ) + 2.0 = 2.5500 lb ⋅ ft or M x = ( 2.55 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.

Ry = C y = −3.30 lb

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Chapter 3, Solution 119. (a) Duct AB will not have a tendency to rotate about the vertical or y-axis if:

(

)

R M By = j ⋅ ΣM BR = j ⋅ rF /B × FF + rE/B × FE = 0

where rF /B = (1.125 m ) i − ( 0.575 m ) j + ( 0.7 m ) k

rE/B = (1.35 m ) i − ( 0.85 m ) j + ( 0.7 m ) k

FF = 50 N ( sin α ) j + ( cos α ) k 

FE = − ( 25 N ) k

∴

ΣM RB

j k i j k i = ( 50 N ) 1.125 m −0.575 m 0.7 m + ( 25 N ) 1.35 m −0.85 m 0.70 0 sin α cos α 0 0 −1 = ( −28.75cos α − 35sin α + 21.25 ) i − ( 56.25cos α − 33.75 ) j + ( 56.25sin α ) k  N ⋅ m R M By = −56.25cos α + 33.75 = 0

Thus,

cos α = 0.60

α = 53.130° or α = 53.1°

(b) R = FE + FF where

FE = − ( 25 N ) k FF = ( 50 N )( sin 53.130° j + cos 53.130°k ) = ( 40 N ) j + ( 30 N ) k ∴ R = ( 40 N ) j + ( 5 N ) k and

M = ΣM RB = −  28.75 ( 0.6 ) + 35 ( 0.8 ) − 21.25 i − 56.25 ( 0.6 ) − 33.75 j + 56.25 ( 0.8 )  k = − ( 24 N ⋅ m ) i − ( 0 ) j + ( 45 N ⋅ m ) k

or M = − ( 24.0 N ⋅ m ) i + ( 45.0 N ⋅ m ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 3, Solution 120.

R = ΣF = FF + FE

(a) Have

FF = 50 N ( sin 60° ) j + ( cos 60° ) k  = ( 43.301 N ) j + ( 25 N ) k

where

FE = − ( 25 N ) k ∴ R = ( 43.301 N ) j or R = ( 43.3 N ) j M CR = Σ ( r × F ) = rF /C × FF + rE/C × FE

Have

rF /C = ( 0.225 m ) i − ( 0.050 m ) j

where

rE/C = ( 0.450 m ) i − ( 0.325 m ) j i ∴

M CR

j

k

i

j

k

= 0.225 −0.050 0 N ⋅ m + 0.450 −0.325 0 N ⋅ m 0 43.301 25 0 0 −25 = ( 6.875 N ⋅ m ) i + ( 5.625 N ⋅ m ) j + ( 9.7427 N ⋅ m ) k or M CR = ( 6.88 N ⋅ m ) i + ( 5.63 N ⋅ m ) j + ( 9.74 N ⋅ m ) k

(b) To determine which direction duct section CD has a tendency to turn, have R M CD = λ DC ⋅ M CR

where λ DC = Then

− ( 0.45 m ) i + ( 0.1 m ) j

( −0.45)2 + ( 0.1)2

= −0.97619i + 0.21693j

R M CD = ( −0.97619i + 0.21693j) ⋅ ( 6.875i + 5.625 j + 9.7427k ) N ⋅ m

= ( −6.7113 + 1.22023) N ⋅ m = −5.4911 N ⋅ m Since λ DC ⋅ M CR < 0, duct DC tends to rotate counterclockwise relative to elbow C as viewed from D to C.

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

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Chapter 3, Solution 121.

ΣF : R = R A = Rλ λ BC

Have

λ BC =

where

( 42 in.) i − ( 96 in.) j − (16 in.) k

∴ RA =

106 in.

21.2 lb ( 42i − 96j − 16k ) 106

or R A = ( 8.40 lb ) i − (19.20 lb ) j − ( 3.20 lb ) k ΣM A : rC/ A × R + M = M A

Have where

rC/ A = ( 42 in.) i + ( 48 in.) k =

1 ( 42i + 48k ) ft 12

= ( 3.5 ft ) i + ( 4.0 ft ) k R = ( 8.40 lb ) i − (19.20 lb ) j − ( 3.20 lb ) k M = −λ BC M

=

−42i + 96 j + 16k (13.25 lb ⋅ ft ) 106

= − ( 5.25 lb ⋅ ft ) i + (12 lb ⋅ ft ) j + ( 2 lb ⋅ ft ) k

Then

i j k 3.5 0 4.0 lb ⋅ ft + ( −5.25i + 12 j + 2k ) lb ⋅ ft = M A 8.40 −19.20 −3.20 ∴ M A = ( 71.55 lb ⋅ ft ) i + ( 56.80 lb ⋅ ft ) j − ( 65.20 lb ⋅ ft ) k

or M A = ( 71.6 lb ⋅ ft ) i + ( 56.8 lb ⋅ ft ) j − ( 65.2 lb ⋅ ft ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 3, Solution 122.

 − ( 0.6 ft ) i + ( 4.2 ft ) j − (1.5 ft ) k R C = R = ( 60 lb ) λ AB = 60 lb   2 2 2  ( − 0.6 ft ) + ( 4.2 ft ) + ( −1.5 ft )

From

   

R C = − ( 8.00 lb ) i + ( 56.0 lb ) j − ( 20.0 lb ) k M C = rA/C × R + M

From where

rA/C = ( 7.8 ft ) i + (1.5 ft ) k  M = ( 22.5 lb ⋅ ft ) λ BA = ( 22.5 lb ⋅ ft )   



( 0.6 ft ) i − ( 4.2 ft ) j + (1.5 ft ) k   ( 0.6 ft )2 + ( − 4.2 ft )2 + (1.5 ft )2 

= ( 3 lb ⋅ ft ) i − ( 21 lb ⋅ ft ) j + ( 7.5 lb ⋅ ft ) k ∴ MC

i j k = 7.8 0 1.5 lb ⋅ ft + ( 3 i − 21 j + 7.5 k ) lb ⋅ ft −8 56 −20 = ( − 84 + 3) lb ⋅ ft  i + (144 − 21) lb ⋅ ft  j + ( 436.8 + 7.5 ) lb ⋅ ft  k

or M C = − ( 81.0 lb ⋅ ft ) i + (123.0 lb ⋅ ft ) j + ( 444 lb ⋅ ft ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 3, Solution 123.

Have:

ΣF :

FA + FC + FD + FE = R R = − ( 80 kN ) j − ( 40 kN ) j − (100 kN ) j − ( 60 kN ) j = − ( 280 kN ) j or R = 280 kN

Have:

ΣM x :

FA ( z A ) + FC ( zC ) + FD ( z D ) + FE ( z E ) = R ( zG )

(80 kN )( 0 ) + ( 40 kN ) ( 3 m ) sin 60° + 60 kN ( 0 ) + ( 60 kN )  − ( 3 m ) sin 60° = ( 280 kN ) Z G

∴

Z G = − 0.185577 m

or Z G = − 0.1856 m ΣM z :

FA ( x A ) + FC ( xC ) + FD ( xD ) + FE ( xE ) = R ( xG )

(80 kN )  − ( 3 m ) cos 60° − 1.5 m  + ( 40 kN )(1.5 m ) + 60 kN (1.5 m ) + (100 kN ) ( 3 m ) cos 60° + 1.5 m  = ( 280 kN ) xG or xG = 0.750 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 3, Solution 124.

Have:

ΣM x :

FA ( z A ) + FB ( z B ) + FC ( zC ) + FD ( z D ) + FE ( z E ) + FF ( z F ) = R ( zG )

(80 kN )( 0 ) + FB ( 3 m ) sin 60° + ( 40 kN ) ( 3 m ) sin 60° + (100 kN )( 0 ) + ( 60 kN )  − ( 3 m ) sin 60° + FF  − ( 3 m ) sin 60° = R ( 0 )

FB − FF = 20 kN

Also

ΣM z :

(1)

FA ( x A ) + FB ( xB ) + FC ( xC ) + FD ( xD ) + FE ( xE ) + FF ( xF ) = R ( xG )

(80 kN ) − ( 3 m ) cos 60° − 1.5 m  + FB ( −1.5 m ) + ( 40 kN )(1.5 m ) + (100 kN ) ( 3 m ) cos 60° + 1.5 m  + ( 60 kN ) (1.5 m ) + FF ( − 1.5 m ) = R ( 0 ) FB + FF = 140 kN

Solving equations (1) and (2):

(2) FB = 80.0 kN FF = 60.0 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 3, Solution 125.

Have

ΣF : FA + FB + FC + FD = R − (116 kips ) j − ( 470 kips ) j − ( 66 kips ) j − ( 28 kips ) j = R ∴ R = − ( 680 kips ) j R = 680 kips

Have

ΣM x : FA ( z A ) + FB ( z B ) + FC ( zC ) + FD ( z D ) = R ( z E )

(116 kips )( 24 ft ) + ( 470 kips )( 48 ft ) + ( 66 kips )(18 ft ) + ( 28 kips )(100.5 ft ) = ( 680 kips )( zE ) ∴ z E = 43.156 ft or z E = 43.2 ft Have

ΣM z : FA ( x A ) + FB ( xB ) + FC ( xC ) + FD ( xD ) = R ( xE )

(116 kips )( 30 ft ) + ( 470 kips )( 96 ft ) + ( 66 kips )(162 ft ) + ( 28 kips )( 96 ft ) = ( 680 kips )( xE ) ∴ xE = 91.147 or xE = 91.1 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 3, Solution 126.

Have

ΣF : FB + FC + FD + FE = R − ( 470 kips ) j − ( 66 kips ) j − ( 28 kips ) j − (116 kips ) j = R ∴ R = − ( 680 kips ) j

Have

ΣM x : FB ( z B ) + FC ( zC ) + FD ( z D ) + FE ( z E ) = R ( z B )

( 470 kips )( 48 ft ) + ( 66 kips )(18 ft ) + ( 28 kips )(100.5 ft ) + (116 kips )( b ) = ( 680 kips )( 48 ft ) ∴ b = 52.397 ft or b = 52.4 ft Have

ΣM z : FB ( xB ) + FC ( xC ) + FD ( xD ) + FE ( xE ) = R ( xB )

( 470 kips )( 96 ft ) + ( 66 kips )(162 ft ) + ( 28 kips )( 96 ft ) + (116 kips )( a ) = ( 680 kips )( 96 ft ) ∴ a = 58.448 ft or a = 58.4 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 3, Solution 127.

For the smallest weight on the trailer so that the resultant force of the four weights acts over the axle at the intersection with the center line of the trailer, the added 0.6 × 0.6 × 1.2-m box should be placed adjacent to one of the edges of the trailer with the 0.6 × 0.6-m side on the bottom. The edges to be considered are based on the location of the resultant for the three given weights.

Have

ΣF : − ( 200 N ) j − ( 400 N ) j − (180 N ) j = R ∴ R = − ( 780 N ) j

Have

ΣM z :

( 200 N )( 0.3 m ) + ( 400 N )(1.7 m ) + (180 N )(1.7 m ) = ( 780 N )( x ) ∴ x = 1.34103 m

Have

ΣM x :

( 200 N )( 0.3 m ) + ( 400 N )( 0.6 m ) + (180 N )( 2.4 m ) = ( 780 N )( z ) ∴ z = 0.93846 m

From the statement of the problem, it is known that the resultant of R from the original loading and the lightest load W passes through G, the point of intersection of the two center lines. Thus, ΣM G = 0.

Further, since the lightest load W is to be as small as possible, the fourth box should be placed as far from G as possible without the box overhanging the trailer. These two requirements imply

( 0.3 m ≤

x ≤ 1 m ) (1.8 m ≤ z ≤ 3.7 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.

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Let x = 0.3 m,

ΣM Gz :

( 200 N )( 0.7 m ) − ( 400 N )( 0.7 m ) − (180 N )( 0.7 m ) + W ( 0.7 m ) = 0 ∴ W = 380 N

ΣM Gx : − ( 200 N )(1.5 m ) − ( 400 N )(1.2 m ) + (180 N )( 0.6 m ) + ( 380 N )( z − 1.8 m ) = 0 ∴ z = 3.5684 m < 3.7 m Let z = 3.7 m,

∴ acceptable

ΣM Gx : − ( 200 N )(1.5 m ) − ( 400 N )(1.2 m ) + (180 N )( 0.6 m ) + W (1.7 m ) = 0

∴ W = 395.29 N > 380 N Since the weight W found for x = 0.3 m is less than W found for z = 3.7 m, x = 0.3 m results in the smallest weight W. or W = 380 N at

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

( 0.3 m, 0, 3.57 m )

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Chapter 3, Solution 128.

For the largest additional weight on the trailer with the box having at least one side coinsiding with the side of the trailer, the box must be as close as possible to point G. For x = 0.6 m, with a small side of the box touching the z-axis, satisfies this condition.

Let x = 0.6 m,

ΣM Gz :

( 200 N )( 0.7 m ) − ( 400 N )( 0.7 m ) − (180 N )( 0.7 m ) + W ( 0.4 m ) = 0 ∴ W = 665 N

and

ΣM GX : − ( 200 N )(1.5 m ) − ( 400 N )(1.2 m ) + (180 N )( 0.6 m ) + ( 665 N )( z − 1.8 m ) = 0 ∴ z = 2.8105 m

(2 m <

z < 4 m)

∴ acceptable

or W = 665 N at

( 0.6 m, 0, 2.81 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 3, Solution 129.

First, reduce the given force system to a force-couple system at the origin.

Have

ΣF :

( 2P ) i − ( P ) j + ( P ) j = R ∴ R = ( 2P ) i

Have

ΣM O : Σ ( rO × F ) = M OR M OR

i j k i j k = Pa 2 2 2.5 + 0 0 4 = Pa ( −1.5i + 5j − 6k ) 2 −1 0 0 1 0 R = 2Pi

(a)

or Magnitude of R = 2P Direction of R : θ x = 0°, θ y = −90°, θ z = 90°

(b) Have

M1 = λ R ⋅ M OR

λR =

R R

= i ⋅ ( −1.5Pai + 5Paj − 6Pak )

= −1.5Pa and pitch

P=

M1 −1.5Pa = = −0.75a 2P R

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

or P = −0.75a

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M OR = M1 + M 2

(c) Have

∴ M 2 = M OR − M1 = ( 5Pa ) j − ( 6Pa ) k Require

M 2 = rQ/O × R

( 5Pa ) j − ( 6Pa ) k = ( yj + zk ) × ( 2Pi ) = − ( 2Py ) k + ( 2Pz ) j From

i : 5Pa = 2Pz

∴ z = 2.5a From

k : − 6Pa = −2Py ∴ y = 3a ∴ The axis of the wrench is parallel to the x-axis and intersects the yz-plane at y = 3a, z = 2.5a

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 3, Solution 130.

First, reduce the given force system to a force-couple at the origin. ΣF : Pi − Pi − Pk = R

Have

∴ R = − Pk ΣM O : − P ( 3a ) k − P ( 3a ) j + P ( −ai + 3aj) = MOR

Have

∴ MOR = Pa ( −i − 3k )

Then let vectors ( R, M1 ) represent the components of the wrench, where their directions are the same.

R = − Pk or Magnitude of R = P

(a)

Direction of R : θ x = 90°, θ y = 90°, θ z = −180° M1 = λ R ⋅ M OR

(b) Have

= −k ⋅  Pa ( −i − 3k ) 

= 3Pa and pitch (c) Have

P=

M1 3Pa = = 3a R P

or P = 3a

M OR = M1 + M 2 ∴ M 2 = M OR − M1 = Pa ( −i − 3k ) − ( −3Pak ) = − Pai

Require

M 2 = rQ/O × R − Pai = ( xi + yj) × ( − P ) k = Pxj − Pyi

From

i : − Pa = − Py

or

y =a

j: x = 0 ∴ The axis of the wrench is parallel to the z-axis and intersects the xy plane at x = 0, y = 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 3, Solution 131.

First, reduce the given force system to a force-couple at the origin.

Have ΣF : − (10 N ) j − (11 N ) j = R ∴ R = − ( 21 N ) j Have ΣM O : Σ ( rO × F ) + ΣM C = M OR

M OR

i j k i j k = 0 0 0.5 N ⋅ m + 0 0 −0.375 N ⋅ m − (12 N ⋅ m ) j 0 −10 0 0 −11 0 = ( 0.875 N ⋅ m ) i − (12 N ⋅ m ) j R = − ( 21 N ) j

(a) R = − ( 21.0 N ) j

Have M1 = λ R ⋅ M OR

(b)

or

λR =

R R

= ( − j) ⋅ ( 0.875 N ⋅ m ) i − (12 N ⋅ m ) j

= 12 N ⋅ m and pitch P = (c)

and

M1 = − (12 N ⋅ m ) j

M1 12 N ⋅ m = = 0.57143 m R 21 N Have

or P = 0.571 m M OR = M1 + M 2

∴ M 2 = M OR − M1 = ( 0.875 N ⋅ m ) i Require ∴

( 0.875 N ⋅ m ) i = ( xi + zk ) × − ( 21 N ) j 0.875i = − ( 21x ) k + ( 21z ) i

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

M 2 = rQ/O × R

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From i:

0.875 = 21z

∴ z = 0.041667 m From k:

0 = −21x ∴ z =0

∴ The axis of the wrench is parallel to the y-axis and intersects the xz-plane at x = 0, z = 41.7 mm

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 3, Solution 132.

(a) Have

First, reduce the given force system to a force-couple system. ΣF : − ( 50 N ) k − ( 50 N ) j + ( 50 N ) k = R R = − ( 50 N ) j;

R = 50 N

R = − ( 50.0 N ) j

Have ΣM O :

( 0.1 N ⋅ m ) k − ( 0.1 N ⋅ m ) j + ( 0.1 N ⋅ m ) k

M OR = − ( 0.1 N ⋅ m ) j + ( 0.2 N ⋅ m ) k (b)

Have

M1 = λ R × MOR =

R ⋅ M OR R

= − j ⋅  − ( 0.1 N ⋅ m ) j + ( 0.2 N ⋅ m ) k 

= 0.1 N ⋅ m

and pitch

P=

M1 0.1 N ⋅ m = = 0.002 m R 50 N

or P = 2.00 mm (c)

Have

M1 = PR = ( 0.002 m )  − ( 50 N ) j

= − ( 0.1 N ⋅ m ) j

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

= M OR

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Note that because M z = ( 0.2 N ⋅ m ) k , the line of action of the wrench must pass through the x-axis to compensate for M z as shown above: With

M1 + ( r × R ) = M OR Then

− ( 0.1 N ⋅ m ) j +  − ( d ) i × − ( 50N ) j

= − ( 0.1 N ⋅ m ) j + ( 0.2 N ⋅ m ) k

or ( 50 N )( d )  k = ( 0.2 N ⋅ m ) k and

d = 0.004 m

x = − d = − 0.004 m

or x = − 4.00 mm, z = 0

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

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Chapter 3, Solution 133.

(

)

First replace the given couples with an equivalent force-couple system R, M OB at the origin. ΣF : ΣM O :

R = − ( 35 lb ) i − (12 lb ) k M OR = − ( 200 lb ⋅ in.) i + ( 8 in.) j + ( 8 in.) k  ×  − ( 35 lb ) i  − (140 lb ⋅ in.) k + (10 in.) i + ( 4 in.) j ×  − (12 lb ) k 

= ( − 200 − 48 ) i + ( − 280 + 120 ) j + ( 280 − 140 ) k = − ( 248 lb ⋅ in.) i − (1600 lb ⋅ in.) j + (140 lb ⋅ in.) k Now

R=

( − 35 lb )2 + ( −12 lb )2

= 37 lb Then

λ axis =

1 ( − 35 i − 12 k ) 37 R = − ( 35.0 lb ) i − (12.00 ) k

(a) M1 = λ axis ⋅ M OR

(b)

Then

=

1 ( − 35 i − 12 k ) ⋅ ( − 248 i − 160 j + 140 k )( lb ⋅ in.)

37

=

1 ( 35 × 248 − 12 × 140 )( lb ⋅ in.)

37

=

7000 lb ⋅ in.

37

M P= 1 = R

7000 lb ⋅ in.

37 37 lb or P = 5.11 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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(c)

Have

M1 = λ axis M1

=

7000 lb ⋅ in. ( − 35 i − 12 k ) 37 2

Then

M OR = M1 + M 2

or

M 2 = ( − 248 i − 160 j + 140 k ) −

7000 ( − 35 i − 12 k ) 37 2

= − ( 69.037 lb ⋅ in.) i − (160 lb ⋅ in.) j + ( 201.36 lb ⋅ in.) k

M z = rO/P × R

Require

i j k or − 69.037i − 160 j + 201.36k = 0 y z −35 0 −12 j:

−160 = − 35 z z = 4.57 in.

or k:

201.36 = 35 y

or

y = 5.75 in.

∴

The point of intersection is defined by y = 5.75 in. z = 4.57 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 3, Solution 134.

First reduce the given force system to a force-couple at the origin at B.

15   8 Have ΣF : − ( 79.2 lb ) k − ( 51 lb )  i + j = R 17   17

(a)

∴ R = − ( 24.0 lb ) i − ( 45.0 lb ) j − ( 79.2 lb ) k and R = 94.2 lb ΣM B : rA/B

M RB

Have × FA + M A + M B = M RB

i j k 15   8 = 0 −20 0 − 660k − 714  i + j  = 1584i − 660k − 42 ( 8i + 15 j) 17 17   0 0 −79.2 ∴ M RB = (1248 lb ⋅ in.) i − ( 630 lb ⋅ in.) j − ( 660 lb ⋅ in.) k

(b)

Have

M1 = λ R ⋅ M OR

=

λR =

R R

−24.0i − 45.0 j − 79.2k ⋅ (1248 lb ⋅ in.) i − ( 630 lb ⋅ in.) j − ( 660 lb ⋅ in.) k  94.2 = 537.89 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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M1 = M1λ R

and

= − (137.044 lb ⋅ in.) i − ( 256.96 lb ⋅ in.) j − ( 452.24 lb ⋅ in.) k Then pitch p = or p = 5.71 in. (c)

M1 537.89 lb ⋅ in. = = 5.7101 in.

R 94.2 lb

Have

M RB = M1 + M 2

∴ M 2 = M RB − M1 = (1248i − 630 j − 660k ) − ( −137.044i − 256.96 j − 452.24k ) = (1385.04 lb ⋅ in.) i − ( 373.04 lb ⋅ in.) j − ( 207.76 lb ⋅ in.) k Require

M 2 = rQ/B × R

i j k 1385.04i − 373.04 j − 207.76k = x 0 z −24 −45 −79.2 = ( 45 z ) i − ( 24 z ) j + ( 79.2 x ) j − ( 45 x ) k From i: From k:

1385.04 = 45z −207.76 = −45x

∴ z = 30.779 in. ∴ x = 4.6169 in. ∴ The axis of the wrench intersects the xz-plane

at x = 4.62 in., z = 30.8 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 3, Solution 135.

(a)

First reduce the given force system to a force-couple at the origin. ΣF : Pλ BA + Pλ DC + Pλ DE = R

Have

 4 12  3  3 4  −9 4 R = P  j − k  +  i − j +  i − j+ k  5 25  5  5 5   25  5 ∴ R = R=

3P 25

( 2 )2 + ( 20 )2 + (1)2

3P ( 2i − 20 j − k ) 25 =

27 5 P 25

Have ΣM : Σ ( rO × P ) = M OR −4 P 3P  4P  4P 12 P   3P  −9 P j− k  + ( 20a ) j ×  i− j  + ( 20a ) j ×  i− j+ k  = M OR 5  5  5 25   5  5  25

( 24a ) j × 

∴ M OR = (b)

24 Pa ( −i − k ) 5 Have M1 = λ R ⋅M OR

where λ R =

3P 25 1 R = = ( 2i − 20 j − k ) ( 2i − 20 j − k ) R 25 27 5 P 9 5

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

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Then M 1 = and pitch

1 9 5

( 2i − 20 j − k ) ⋅

p=

24 Pa −8Pa ( −i − k ) = 5 15 5

M1 −8Pa  25  −8a =  = R 81 15 5  27 5 P 

or p = −0.0988a

(c) M1 = M 1λ R =

−8Pa  1  8Pa ( −2i + 20 j + k )   ( 2i − 20 j − k ) = 675 15 5  9 5 

Then M 2 = M OR − M1 =

24Pa 8Pa 8Pa ( −i − k ) − ( −2i + 20 j + k ) = ( −403i − 20 j − 406k ) 5 675 675 M 2 = rQ/O × R

Require  8Pa   3P    ( −403i − 20 j − 406k ) = ( xi + zk ) ×   ( 2i − 20 j − k )  675   25   3P  =   20 zi + ( x + 2 z ) j − 20 xk   25  From i:

8 ( −403)

Pa  3P  = 20 z   675  25 

∴ z = −1.99012a

From k:

8 ( −406 )

Pa  3P  = −20 x   675  25 

∴ x = 2.0049a

∴ The axis of the wrench intersects the xz-plane at x = 2.00a, z = −1.990a

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 3, Solution 136.

First reduce the given force-couple system to an equivalent force-couple system ( R, M B ) at point B. d BD =

( − 480 mm )2 + ( 560 mm )2 + ( − 480 mm )2

= 880 mm

FBD = FBDλBD =

132 N ( − 480i + 560 j − 480k ) 880

= (12 N )( − 6i + 7 j − 6k ) d EB =

( 240 mm )2 + ( − 220 mm )2 + ( 480 mm )2

= 580 mm

FEB = FEBλEB =

145 N ( 240i − 220 j + 480k ) 580

= ( 5 N )(12i − 11j + 24k )

ΣF :

R = FBD + FEB = (12 N )( − 6i + 7 j − 6k ) + 5 N (12i − 11j + 24k ) = − (12 N ) i + ( 29 N ) j + ( 48 N ) k d BF =

( 340 mm )2 + ( 240 mm )2 + ( − 60 mm )2

= 20 442 mm Then

MB = =

20 N ⋅ m ( 340i + 240 j − 60k ) 20 442 20 N ⋅ m (17i + 12 j − 3k ) 442

Vector Mechanics for Engineers: Statics and Dynamics, 8/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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Now determine whether R and M B are perpendicular R ⋅ M B = ( −12 j + 29 j + 48k ) ⋅

20 (17i + 12 j − 3k ) 442

20 ( −12 × 17 + 29 × 12 − 48 × 3) 442

=

=0 ∴ R and M B are perpendicular so that ( R, M B ) can be reduced to the single equivalent force

R = − (12.00 N ) i + ( 29.0 N ) j + ( 48.0 N ) k

Now require M B = rB/P × R or

j:

20 × 12 = −12 ( z − 0.480 ) + 0.480 ( 48 ) 442

or k:

i j k 20 N ⋅ m (17i + 12 j − 3k ) = − 0.480 y z − 0.480 ( N ⋅ m ) 442 −12 29 48

z = 1.449 m

− 20 × 3 = − 0.480 ( 29 ) + 12 y 442

or

y = 0.922 m

∴ The point of intersection is defined by y = 0.922 m z = 1.449 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 3, Solution 137.

First, reduce the given force system to a force-couple at the origin. ΣF : FA + FG = R

Have

 ( 4 in.) i + ( 6 in.) j − (12 in.) k  ∴ R = (10 lb ) k + 14 lb   = ( 4 lb ) i + ( 6 lb ) j − ( 2 lb ) k 14 in.   R=

and Have

56 lb

ΣM O : ∑ ( rO × F ) + ∑ M C = MOR

{

}

M OR = (12 in.) j × (10 lb ) k  + (16 in.) i × ( 4 lb ) i + ( 6 lb ) j − (12 lb ) k 

 (16 in.) i − (12 in.) j   ( 4 in.) i − (12 in.) j + ( 6 in.) k  + ( 84 lb ⋅ in.)   + ( 120 lb ⋅ in.)   20 in.

14 in.    

∴ M 0R = ( 221.49 lb ⋅ in.) i + ( 38.743 lb ⋅ in.) j + (147.429 lb ⋅ in.) k = (18.4572 lb ⋅ ft ) i + ( 3.2286 lb ⋅ ft ) j + (12.2858 lb ⋅ ft ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/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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The force-couple at O can be replaced by a single force if the direction of R is perpendicular to M OR .

To be perpendicular R ⋅ M OR = 0 Have

R ⋅ M OR = ( 4i + 6 j − 2k ) ⋅ (18.4572i + 3.2286 j + 12.2858k ) = 0? = 73.829 + 19.3716 − 24.572 ≠0

∴ System cannot be reduced to a single equivalent force.

To reduce to an equivalent wrench, the moment component along the line of action of P is found.

M1 = λ R ⋅ M OR

λR =

R R

 ( 4i + 6 j − 2k )  =  ⋅ (18.4572i + 3.2286 j + 12.2858k ) 56  

= 9.1709 lb ⋅ ft M1 = M1λ R = ( 9.1709 lb ⋅ ft )( 0.53452i + 0.80178 j − 0.26726k )

and And pitch

p=

M1 9.1709 lb ⋅ ft = = 1.22551 ft R 56 lb

or p = 1.226 ft Have M 2 = M OR − M1 = (18.4572i + 3.2286 j + 12.2858k ) − ( 9.1709 )( 0.53452i + 0.80178 j − 0.26726k ) = (13.5552 lb ⋅ ft ) i − ( 4.1244 lb ⋅ ft ) j + (14.7368 lb ⋅ ft ) k Require

M 2 = rQ/O × R

(13.5552i − 4.1244 j + 14.7368k ) = ( yj + zk ) × ( 4i + 6 j − 2k ) = − ( 2 y + 6z ) i + ( 4z ) j − ( 4 y ) k From j:

−4.1244 = 4 z

From k:

14.7368 = −4 y

or or

z = −1.0311 ft y = −3.6842 ft

∴ line of action of the wrench intersects the yz plane at y = −3.68 ft, z = 1.031 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 3, Solution 138.

FA = ( FA ) x i + ( FA ) y j

Define

FB = ( FB ) x i + ( FB ) y j ΣFx :

Then

( FA ) x + ( FB ) x = 0 ( FA ) x = − ( FB ) x

ΣFy :

( FA ) y + ( FB ) y = R ( FA ) y

bk × ( FB ) x i + ( FB ) y j = − ak + R j + M j  

ΣM A :

and i:

= R − ( FB ) y

− b ( FB ) y = aR

or

( FB ) y = −

Then

( FA ) y

a R b

 a  = R − − R  b  a  = R 1 +  b 

j:

b ( FB ) x = M M b

or

( FB ) x =

Then

( FA ) x = −

M b ∴ FA = −

M a  i + R 1 +  j b b  FB =

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 i − Rj b b

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Chapter 3, Solution 139.

First, choose a coordinate system so that the xy plane coincides with the given plane. Also, position the coordinate system so that the line of action of the wrench passes through the origin as shown in Figure a.

Since the orientation of the plane and the components (R, M) of the wrench are known, it follows that the scalar components of R and M are known relative to the shown coordinate system.

A force system to be shown as equivalent is illustrated in Figure b. Let A be the force passing through the given point P and B be the force that lies in the given plane. Let b be the x-axis intercept of B. The known components of the wrench can be expressed as R = Rxi + Ry j + Rzk

M = M xi + M y j + M zk

and

while the unknown forces A and B can be expressed as A = Axi + Ay j + Azk

and

B = Bxi + Bzk

Since the position vector of point P is given, it follows that the scalar components (x, y, z) of the position vector rP are also known.

Then, for equivalence of the two systems ΣFx : Rx = Ax + Bx ΣFy : Ry = Ay

(1)

(2)

ΣFz : Rz = Az + Bz

(3)

ΣM x : M x = yAz − zAy

(4)

ΣM y : M y = zAx − xAz − bBz

(5)

ΣM z : M z = xAy − yAx

(6) 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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(

)

Based on the above six independent equations for the six unknowns Ax , Ay , Az , Bx , Bz , b , there exists a unique solution for A and B.

Ay = Ry

From Equation (2) Equation (6)

1 Ax =   xRy − M z  y

Equation (1)

1 Bx = Rx −   xRy − M z  y

Equation (4)

1 Az =   M x + zRy  y

Equation (3)

1 Bz = Rz −   M x + zRy  y

Equation (5)

b=

(

)

(

(

)

)

(

)

( xM x + yM y + zM z ) ( M x − yRz + zRy )

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 3, Solution 140.

First, observe that it is always possible to construct a line perpendicular to a given line so that the constructed line also passes through a given point. Thus, it is possible to align one of the coordinate axes of a rectangular coordinate system with the axis of the wrench while one of the other axes passes through the given point.

See Figures a and b.

Have

R = Rj

M = Mj

and

and are known.

The unknown forces A and B can be expressed as A = Axi + Ay j + Azk

B = Bxi + By j + Bzk

and

The distance a is known. It is assumed that force B intersects the xz plane at (x, 0, z). Then for equivalence ∑ Fx : 0 = Ax + Bx

(1)

∑ Fy : R = Ay + By

(2)

∑ Fz : 0 = Az + Bz

(3)

∑ M x : 0 = − zBy (4) ∑ M y : M = −aAz − xBz + zBx

(5)

∑ M z : 0 = aAy + xBy

(6)

Since A and B are made perpendicular, A⋅B = 0 There are eight unknowns:

or

Ax Bx + Ay By + Az Bz = 0

(7)

Ax , Ay , Az , Bx , By , Bz , x, z

But only seven independent equations. Therefore, there exists an infinite number of solutions. 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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0 = − zBy

Next consider Equation (4): If By = 0, Equation (7) becomes

Ax Bx + Az Bz = 0 Ax2 + Az2 = 0

Using Equations (1) and (3) this equation becomes

Since the components of A must be real, a nontrivial solution is not possible. Thus, it is required that By ≠ 0, so that from Equation (4), z = 0.

To obtain one possible solution, arbitrarily let Ax = 0.

(Note: Setting Ay , Az , or Bz equal to zero results in unacceptable solutions.)

The defining equations then become.

0 = Bx

(1)′

R = Ay + By

(2)

0 = Az + Bz

(3)

M = −aAz − xBz

(5)′

0 = aAy + xBy

(6)

Ay By + Az Bz = 0 (7)′ Then

Equation (2) can be written

Ay = R − By

Equation (3) can be written

Bz = − Az x=−

Equation (6) can be written

aAy By

Substituting into Equation (5)′,  R − By  M = −aAz −  −a  ( − Az )   B y  

or

Az = −

M By (8) aR

Substituting into Equation (7)′,

M  M  By  By  = 0 ( R − By ) By +  − aR aR  

Vector Mechanics for Engineers: Statics and Dynamics, 8/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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By =

or

a 2 R3 a2R2 + M 2

Then from Equations (2), (8), and (3) Ay = R − Az = −

Bz =

a 2 R3 RM 2 = a2R2 + M 2 a2R2 + M 2

 M  a 2 R3 aR 2 M = −  2 2 2 2 aR  a R + M  a R2 + M 2

aR 2 M a R2 + M 2 2

In summary A=

RM ( Mj − aRk ) a R2 + M 2

B=

aR 2 ( aRj + Mk ) a R2 + M 2

2

2

Which shows that it is possible to replace a wrench with two perpendicular forces, one of which is applied at a given point.

Lastly, if R > 0 and M > 0, it follows from the equations found for A and B that Ay > 0 and By > 0.

From Equation (6), x < 0 (assuming a > 0). Then, as a consequence of letting Ax = 0, force A lies in a plane parallel to the yz plane and to the right of the origin, while force B lies in a plane parallel to the yz plane but to the left of the origin, as shown in the figure below.

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 3, Solution 141.

First, choose a rectangular coordinate system where one axis coincides with the axis of the wrench and another axis intersects the prescribed line of action ( AA′ ) . Note that it has been assumed that the line of action of force B intersects the xz plane at point P ( x, 0, z ) . Denoting the known direction of line AA′ by λ A = λxi + λ y j + λzk it follows that force A can be expressed as

(

A = Aλ A = A λxi + λ y j + λz k

)

Force B can be expressed as B = Bxi + By j + Bzk

Next, observe that since the axis of the wrench and the prescribed line of action AA′ are known, it follows that the distance a can be determined. In the following solution, it is assumed that a is known.

Then, for equivalence ΣFx : 0 = Aλx + Bx (1) ΣFy : R = Aλ y + By (2) ΣFz : 0 = Aλz + Bz (3) ΣM x : 0 = − zBy

Since there are six unknowns

(4)

ΣM y : M = −aAλz + zBx − xBz

(5)

ΣM z : 0 = aAλ y + xBy

(6)

( A, Bx , By , Bz , x, z )

and six independent equations, it will be possible to

obtain a solution.

Vector Mechanics for Engineers: Statics and Dynamics, 8/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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Case 1: Let z = 0 to satisfy Equation (4) Aλ y = R − By

Now Equation (2)

Bz = − Aλz

Equation (3) By

 a  = −  R − By  By   

  a M = −aAλz −  −    By

   R − By ( − Aλz )    

x=−

Equation (6)

aAλ y

(

)

Substitution into Equation (5)

∴ A=−

(

)

1 M  B λz  aR  y

Substitution into Equation (2) R=−

1 M  By λ y + By λz  aR 

∴ By = Then

λz aR 2 λz aR − λ y M

R aR λy − λz M λx MR Bx = − Aλx = λz aR − λ y M A=−

MR

λz aR − λ y M

Bz = − Aλz =

=

λz MR λz aR − λ y M

In summary A=

B=

and

P λA aR λy − λz M

R ( λ Mi + λz aRj + λz Mk ) λz aR − λ y M x

   λz aR − λ y M R  x = a 1 −  = a 1 − R    By  λz aR 2   

     or x =

Note that for this case, the lines of action of both A and B intersect the x axis. 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.

λy M λz R

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Case 2: Let By = 0 to satisfy Equation (4) A=

Now Equation (2)

R

λy

Equation (1)

λ  Bx = − R  x   λy   

Equation (3)

λ Bz = − R  z  λy  aAλ y = 0

Equation (6)

   

which requires a = 0

Substitution into Equation (5)  λ M = z −R  x   λ y

  λ    − x −R  z     λ y   

or

This last expression is the equation for the line of action of force B. In summary  R A =  λ A  λy     R B=  λy 

M 

λz x − λx z =   λ y  R

  ( −λ x i − λ z k )  

Assuming that λx , λ y , λz > 0, the equivalent force system is as shown below.

Note that the component of A in the xz plane is parallel to 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 3, Solution 142.

(a) Have

M B = rC/B FN = ( 0.1 m )( 800 N ) = 80.0 N ⋅ m or M B = 80.0 N ⋅ m

(b) By definition M B = rA/B P sin θ where

θ = 90° − ( 90° − 70° ) − α = 90° − 20° − 10° = 60° ∴ 80.0 N ⋅ m = ( 0.45 m ) P sin 60°

P = 205.28 N or

P = 205 N

(c) For P to be minimum, it must be perpendicular to the line joining points A and B. Thus, P must be directed as shown.

Thus or

M B = dPmin = rA/B Pmin 80.0 N ⋅ m = ( 0.45 m ) Pmin ∴ Pmin = 177.778 N or Pmin = 177.8 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.

20°

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Chapter 3, Solution 143.

M C = rB/C × FB

Have

Noting the direction of the moment of each force component about C is clockwise, M C = xFBy + yFBx where

x = 144 mm − 78 mm = 66 mm y = 86 mm + 108 mm = 194 mm

and

FBx =

FBy =

78

( 78)2 + (86 )2 86

( 78)

2

+ ( 86 )

2

( 580 N ) = 389.65 N

( 580 N ) = 429.62 N

∴ M C = ( 66 mm )( 429.62 N ) + (194 mm )( 389.65 N ) = 103947 N ⋅ mm = 103.947 N ⋅ m or M C = 103.9 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 3, Solution 144.

M A = rE/ A × TDE

(a) Have

rE/ A = ( 92 in.) j

where

TDE = λ DETDE =

( 24 in.) i + (132 in.) j − (120 in.) k 360 lb ( ) ( 24 )2 + (132 )2 + (120 )2 in.

= ( 48 lb ) i + ( 264 lb ) j − ( 240 lb ) k i j k ∴ M A = 0 92 0 lb ⋅ in. = − ( 22,080 lb ⋅ in.) i − ( 4416 lb ⋅ in ) k 48 264 −240 or M A = − (1840 lb ⋅ ft ) i − ( 368 lb ⋅ ft ) k

M A = rG/ A × TCG

(b) Have

rG/ A = (108 in.) i + ( 92 in.) j

where

TCG = λ CGTCG =

− ( 24 in.) i + (132 in.) j − (120 in.) k

( 24 )2 + (132 )2 + (120 )2 in.

( 360 lb )

= − ( 48 lb ) i + ( 264 lb ) j − ( 240 lb ) k i

j

k

∴ M A = 108 92 0 lb ⋅ in. −48 264 −240 = − ( 22, 080 lb ⋅ in.) i + ( 25,920 lb ⋅ in.) j + ( 32,928 lb ⋅ in.) k or M A = − (1840 lb ⋅ ft ) i + ( 2160 lb ⋅ ft ) j + ( 2740 lb ⋅ ft ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 3, Solution 145.

First note

and

AC = rC/ A =

( −2.4 )2 + (1.8)2

AD = rD/ A =

(1.2 )2 + ( −2.4 )2 + ( 0.3)2

m = 3m

m = 2.7 m

rC/ A = − ( 2.4 m ) j + (1.8 m ) k rD/ A = (1.2 m ) i − ( 2.4 m ) j + ( 0.3 m ) k

By definition rC/ A ⋅ rD/ A = rC/ A rD/ A cosθ or

( −2.4 j + 1.8k ) ⋅ (1.2i − 2.4 j + 0.3k ) = ( 3)( 2.7 ) cosθ ( 0 )(1.2 ) + ( −2.4 )( −2.4 ) + (1.8)( 0.3) = 8.1cosθ

and

cosθ =

6.3 = 0.77778 8.1

θ = 38.942° or θ = 38.9°

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 3, Solution 146.

M O = rA/O × TBA

Based on where

M O = M xi + M y j + M zk = M xi + (100 lb ⋅ ft ) j − ( 400 lb ⋅ ft ) k rA/O = ( 6 ft ) i + ( 4 ft ) j TBA = λ BATBA =

( 6 ft ) i − (12 ft ) j − ( a ) k T

BA

d BA

i

j

k

T ∴ M xi + 100 j − 400k = 6 4 0 BA d 6 −12 −a BA =

TBA  − ( 4a ) i + ( 6a ) j − ( 96 ) k  d BA  100 d BA 6a

From j-coefficient:

100d AB = 6aTBA or TBA =

From k -coefficient:

−400d AB = −96TBA or TBA =

Equating Equations (1) and (2) yields

a=

400 d BA 96

100 ( 96 ) 6 ( 400 ) or a = 4.00 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.

(1) (2)

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Chapter 3, Solution 147.

(

M DB = λ DB ⋅ rC/D × TCF

Have λ DB =

where

( 48 in.) i − (14 in.) j 50 in.

)

= 0.96i − 0.28 j

rC/D = ( 8 in.) j − (16 in.) k

( 24 in.) i − ( 36 in.) j − (8 in.) k

TCF = λ CF TCF =

44 in.

(132 lb )

= ( 72 lb ) i − (108 lb ) j − ( 24 lb ) k 0.96 −0.28 ∴ M DB =

0 72

0

8 −16 lb ⋅ in. −108 −24

= 0.96 ( 8 )( −24 ) − ( −16 )( −108 )  + ( −0.28 ) ( −16 )( 72 ) − 0 

= −1520.64 lb ⋅ in. or M DB = −1521 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 3, Solution 148.

(a) Based on

ΣF : FA = T = 1000 N or FA = 1000 N

20°

ΣM A : M A = (T sin 50° )( dA ) = (1000 N ) sin 50° ( 2.25 m ) = 1723.60 N ⋅ m or M A = 1724 N ⋅ m (b) Based on

ΣF : FB = T = 1000 N or FB = 1000 N

20°

ΣMB : M B = (T sin 50° )( d B ) = (1000 N ) sin 50° (1.25 m ) = 957.56 N ⋅ m or M B = 958 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 3, Solution 149.

Require the equivalent forces acting at A and C be parallel and at an angle of α with the vertical.

Then for equivalence, ΣFx :

( 250 lb ) sin 30° =

FA sin α + FB sin α

(1)

ΣFy : − ( 250 lb ) cos 30° = − FA cos α − FB cos α (2) Dividing Equation (1) by Equation (2),

( 250 lb ) sin 30° − ( 250 lb ) cos 30°

=

( FA + FB ) sin α − ( FA + FB ) cos α

Simplifying yields α = 30° Based on ΣM C : ( 250 lb ) cos 30° (12 ft ) = ( FA cos 30° )( 32 ft )

∴ FA = 93.75 lb or FA = 93.8 lb

60°

Based on ΣM A : − ( 250 lb ) cos 30° ( 20 ft ) = ( FC cos 30° ) ( 32 ft )

∴ FC = 156.25 lb or FC = 156.3 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.

60°

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Chapter 3, Solution 150.

Have ΣF : PAB = FC where PAB = λ AB PAB =

( 2.0 in.) i + ( 38 in.) j − ( 24 in.) k 44.989 in.

( 45 lb )

or FC = ( 2.00 lb ) i + ( 38.0 lb ) j − ( 24.0 lb ) k Have ΣM C : rB/C × PAB = M C i

j

k

M C = 2 29.5 −33 0 lb ⋅ in.

1 19 −12 = ( 2 lb ⋅ in.) {( −33)( −12 ) i − ( 29.5 )( −12 ) j

+ ( 29.5 )(19 ) − ( −33)(1)  k

}

or M C = ( 792 lb ⋅ in.) i + ( 708 lb ⋅ in.) j + (1187 lb ⋅ in.) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 3, Solution 151.

For equivalence ΣFx : − ( 90 N ) sin 30° + (125 N ) cos 40° = Rx or Rx = 50.756 N ΣFy : − ( 90 N ) cos30° − 200 N − (125 N ) sin 40° = Ry or Ry = −358.29 N

and

( 50.756 )2 + ( −358.29 )2

R=

Then tan θ =

Ry Rx

=

−358.29 = −7.0591 50.756

= 361.87 N ∴ θ = −81.937°

or R = 362 N

81.9°

Also ΣM A : M − ( 90 N ) sin 35° ( 0.6 m ) − ( 200 N ) cos 25°  ( 0.85 m ) − (125 N ) sin 65°  (1.25 m ) = 0

∴ M = 326.66 N ⋅ m or M = 327 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 3, Solution 152.

For equivalence Σ F: FA + FB + FC + FD = R C R C = − ( 5 lb ) j − ( 3 lb ) j − ( 4 lb ) k − ( 7 lb ) i ∴ R C = ( −7 lb ) i − ( 8 lb ) j − ( 4 lb ) k Also for equivalence ΣM C : rA′/C × FA + rB′/C × FB + rD′/C × FD = M C or MC

i j k i j k i j k = 0 0 −1.5 in. + 1 in. 0 −1.5 in. + 0 1.5 in. 1.5 in.

0 5 lb 0 0 −3 lb 0 −7 lb 0 0 = ( −7.50 lb ⋅ in. − 0 ) i  + ( 0 − 4.50 lb ⋅ in.) i + ( −3.0 lb ⋅ in. − 0 ) k  + (10.5 lb ⋅ in. − 0 ) j + ( 0 + 10.5 lb ⋅ in.) k 

or M C = − (12.0 lb ⋅ in.) i + (10.5 lb ⋅ in.) j + ( 7.5 lb ⋅ in.) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 3, Solution 153.

Have

ΣF : FA + FB + FC + FD = R − ( 85 lb ) j − ( 60 lb ) j − ( 90 lb ) j − ( 95 lb ) j = R ∴ R = − ( 330 lb ) j

Have

ΣM x : FA ( z A ) + FB ( z B ) + FC ( zC ) + FD ( z D ) = R ( z H )

(85 lb )( 9 ft ) + ( 60 lb )(1.5 ft ) + ( 90 lb )(14.25 ft ) + ( 95 lb )( zD ) = ( 330 lb )( 7.5 ft ) ∴ z D = 3.5523 ft Have

or z D = 3.55 ft

ΣM z : FA ( x A ) + FB ( xB ) + FC ( xC ) + FD ( xD ) = R ( xH )

(85 lb )( 3 ft ) + ( 60 lb )( 4.5 ft ) + ( 90 lb )(14.25 ft ) + ( 95 lb )( xD ) = ( 330 lb )( 7.5 ft ) ∴ xD = 7.0263 ft or xD = 7.03 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.