Solution Manual - Vector Mechanics Engineers Dynamics 8th Beer Chapter 03 PDF

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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 ) cos30 ° = 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 x 

α = θ = tan −1  

 4.8 in.  = tan−1    3.4 in. 

= 54.689° Then

M A = rAB Pmin

or

Pmin =

M A 19.5 lb⋅ in. = 5.8822 in. rAB

= 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 = xT BFy + yT BFx = ( 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 = xTBF + yTBF y x

= ( 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 = xF C

FC =

M A 386.41 N ⋅ m = 2m x

= 193.205 N ⊳

∴ FC = 193.2 N

(c)

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 (T ) 13 AB

=

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 = T ABx ( y ) + T ABy (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 T AB 25

and

TABy =

7 T 25 AB

M D = TABx ( y) + T ABy ( x )

Have

∴ 7840 lb ⋅in. =

24 7 TAB ( 0 ) + TAB (112 in. ) 25 25 TAB = 250 lb or T AB = 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 forceT

ABabout

D:

M D = ( TAB max ) y ( d ) M D = 1152 N ⋅ m

where

( TAB max ) y Now

sin θ =

= TAB max sinθ = ( 2880 N ) sinθ

1.05 m

( d + 0.24) 2 + (1.05) 2 m

 ∴ 1152 N ⋅m = 2880 N   

1.05

( d + 0.24) 2

  (d ) 2  + (1.05) 

( d + 0.24)2 + ( 1.05)2

or or

(d

or

= 2.625d

+ 0.24 ) + (1.05 ) = 6.8906d2 2

2

2

5.8906 d − 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 sinθ =

42 mm 150 mm

cos θ =

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 MC = 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.

MC = 140.0 N⋅ m

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Chapter 3, Solution 13.

( 42 mm )2 + (144 mm )2

with d AB =

= 150 mm sinθ =

42 mm 150 mm

cosθ =

144 mm 150 mm

FAB = − FAB sin θi − 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 MC = rB/C × FAB = ( −0.042 i − 0.056 j ) × ( −700 i − 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.

MC = 61.6 N ⋅ m

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Chapter 3, Solution 14.

ΣM D :

 88 80 N   105  M D = ( 0.090 m )  ×  − ( 0.280 m )  137 × 80 N  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 ( cosα i + sinα 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 sinα + sin β cosα )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 cosα )

(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 cosα )

(6)

Adding equations (5) and (6) gives: sin (α − β ) + sin ( α + β ) = 2cos β sinα 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 − (840 mm )i + (495 mm )j

Then λAB =

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 

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 −3 9

9 − 7.5

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

∴λ =

9 ( − 4i + 7j + 10k ) 9 165 or λ =

(b) where

λ =

A×B A ×B

A = −14i − 2j + 8k B = 3i + 1.5 j − k

Then

j k i A × B = −14 − 2 8 1.5 −1

3

= ( 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 =

5( − 2 i + 2 j − 3k) 5 17

or λλ =

1 17

( − 2i + 2j − 3k ) ⊳

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

1 165

( − 4i + 7j + 10k ) ⊳

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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  (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 

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 MO = − (18.00 N⋅ m )i − (15.75 Ν ⋅ m) j − ( 40.5 N⋅ m) k  (b)

Have

MO = r × F i = 2 7.5

j − 0.75 3

k −1 N ⋅ m − 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 MO = ( 6.38 N ⋅ m) i + (1.500 Ν ⋅ m) j + (11.63 Ν ⋅ m) k  (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  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 = −7.5 3

k −6 lb ⋅ft

−6

3

4

= [ (12 − 36)i + (−18 + 30) j + (45 − 9)k ] lb ⋅ ft or MO = − ( 24.0 lb⋅ ft) i + (12.00 lb⋅ ft) j + (36.0 lb⋅ ft) k  (b)

Have

MO = r × F i j k = −7.5 1.5 −1 lb ⋅ft −6 4

3

= [ (6 − 6)i + (−3 + 3) j + (4.5 − 4.5) k] lb ⋅ ft or M O = 0  (c)

Have

MO = r × F i j = −8 2 3

−6

k −14 lb ⋅ft 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 

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 = r A/C × R A

=

i 0

j 3.1

k 1.2 N ⋅ m

216 − 1017 − 108

= ( 885.6 N ⋅m ) i + ( 259.2 N ⋅m ) j − ( 669.6 N ⋅ m) k MC = (886 N ⋅m ) i + (259 N ⋅m ) j − (670 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 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

MA =

i 0.215

j − 0.050

k 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 

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 = r A/O × R

where

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