PROBLEM 3.1 A 13.2-N force P is applied to the lever which controls the auger of a snowblower. Determine the moment of P about A when α is equal to 30°.

SOLUTION

First note

( )sin 13.2 N sin 30 6.60 NxP P α= = ° =

( )cos 13.2 N cos30 11.4315 Nα= = ° =yP P

Noting that the direction of the moment of each force component about A is counterclockwise,

/ /A B A y B A xM x P y P= +

( )( ) ( )( )0.086 m 11.4315 N 0.122 m 6.60 N= +

1.78831 N m= ⋅

or 1.788 N mA = ⋅M

PROBLEM 3.2 The force P is applied to the lever which controls the auger of a snowblower. Determine the magnitude and the direction of the smallest force P which has a 2.20- N m⋅ counterclockwise moment about A.

SOLUTION

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

( ) ( )2 286 mm 122 mm 149.265 mm= + =ABr

1 1 122 mmtan tan 54.81986 mm

α θ − − = = = = °

yx

Then min=A ABM r P

or min = A

AB

MPr

2.20 N m 1000 mm149.265 mm 1 m

⋅ =

14.7389 N=

min 14.74 N∴ =P 54.8°

or min 14.74 N=P 35.2°

PROBLEM 3.3 A 13.1-N force P is applied to the lever which controls the auger of a snowblower. Determine the value of α knowing that the moment of P about A is counterclockwise and has a magnitude of 1.95 N m.⋅

SOLUTION

By definition / sinθ=A B AM r P

where ( )90θ φ α= + ° −

and 1 122 mmtan 54.81986 mm

φ − = = °

Also ( ) ( )2 2/ 86 mm 122 mm 149.265 mm= + =B Ar

Then ( )( ) ( )1.95 N m 0.149265 m 13.1 N sin 54.819 90 α⋅ = ° + ° −

or ( )sin 144.819 0.99725α° − =

or 144.819 85.752α° − = °

and 144.819 94.248α° − = °

50.6 , 59.1α∴ = ° °

PROBLEM 3.4 A foot valve for a pneumatic system is hinged at B. Knowing that α = 28°, determine the moment of the 4-lb force about point B by resolving the force into horizontal and vertical components.

SOLUTION

Note that 20 28 20 8θ α= − ° = ° − ° = °

and ( )4 lb cos8 3.9611 lb= ° =xF

( )4 lb sin8 0.55669 lb= ° =yF

Also ( )6.5 in. cos 20 6.1080 in.= ° =x

( )6.5 in. sin 20 2.2231 in.= ° =y

Noting that the direction of the moment of each force component about B is counterclockwise,

B y xM xF yF= +

( )( ) ( )( )6.1080 in. 0.55669 lb 2.2231 in. 3.9611 lb= +

12.2062 lb in.= ⋅

or 12.21 lb in.B = ⋅M

PROBLEM 3.5 A foot valve for a pneumatic system is hinged at B. Knowing that α = 28°, determine the moment of the 4-lb force about point B by resolving the force into components along ABC and in a direction perpendicular to ABC.

SOLUTION

First resolve the 4-lb force into components P and Q, where

( )4.0 lb sin 28 1.87787 lbQ = ° =

Then /=B A BM r Q

( )( )6.5 in. 1.87787 lb=

12.2063 lb in.= ⋅

or 12.21 lb in.B = ⋅M

PROBLEM 3.6 It is known that a vertical force of 800 N is required to remove the nail at C from the board. As the nail first starts moving, determine (a) the moment about B of the force exerted on the nail, (b) the magnitude of the force P which creates the same moment about B if α = 10°, (c) the smallest force P which creates the same moment about B.

SOLUTION

(a) Have /=B C B NM r F

( )( )0.1 m 800 N=

80.0 N m= ⋅

or 80.0 N mB = ⋅M

(b) By definition

/ sinθ=B A BM r P

where ( )90 90 70θ α= ° − ° − ° −

90 20 10= ° − ° − °

60= °

( ) 80.0 N m 0.45 m sin 60P∴ ⋅ = °

205.28 N=P

or 205 NP =

(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 min / min= =B A BM dP r P

or ( ) min80.0 N m 0.45 m⋅ = P

min 177.778 N∴ =P

or min 177.8 N=P 20°

PROBLEM 3.7 A sign is suspended from two chains AE and BF. Knowing that the tension in BF is 45 lb, determine (a) the moment about A of the force exert by the chain at B, (b) the smallest force applied at C which creates the same moment about A.

SOLUTION

(a) Have /A B A BF=M r T×

Noting that the direction of the moment of each force component about A is counterclockwise,

= +A BFy BFxM xT yT

( )( ) ( )( )6.5 ft 45 lb sin 60 4.4 ft 3.1 ft 45 lb cos60= ° + − °

282.56 lb ft= ⋅

or 283 lb ftA = ⋅M

(b) Have ( )/ minA C A C=M r F×

For CF to be minimum, it must be perpendicular to the line joining points A and C.

( )minA CM d F∴ =

where ( ) ( )2 2/ 6.5 ft 4.4 ft 7.8492 ftC Ad r= = + =

( )( )min 282.56 lb ft 7.8492 ft CF∴ ⋅ =

( )min 35.999 lbCF =

1 4.4 fttan 34.0956.5 ft

φ − = = °

90 90 34.095 55.905θ φ= ° − = ° − ° = °

or ( )min 36.0 lbC =F 55.9°

PROBLEM 3.8 A sign is suspended from two chains AE and BF. Knowing that the tension in BF is 45 lb, determine (a) the moment about A of the force exerted by the chain at B, (b) the magnitude and sense of the vertical force applied at C which creates the same moment about A, (c) the smallest force applied at B which creates the same moment about A.

SOLUTION

(a) Have /A B A BF=M r T×

Noting that the direction of the moment of each force component about A is counterclockwise,

= +A BFy BFxM xT yT

( )( ) ( )( )6.5 ft 45 lb sin 60 4.4 ft 3.1 ft 45 lb cos60= ° + − °

282.56 lb ft= ⋅

or 283 lb ftA = ⋅M

(b) Have /A C A C=M r F×

or =A CM xF

282.56 lb ft 43.471 lb6.5 ft

AC

MFx

⋅∴ = = =

or 43.5 lbC =F

(c) Have ( )/ minA B A B=M r F×

For BF to be minimum, it must be perpendicular to the line joining points A and B.

( )min ∴ =A BM d F

where ( ) ( )2 26.5 ft 4.4 ft 3.1 ft 6.6287 ftd = + − =

( )min282.56 lb ft 42.627 lb

6.6287 ft⋅

∴ = = =AB

MFd

and 1 6.5 fttan 78.6904.4 ft 3.1 ft

θ − = = ° −

or ( )min 42.6 lb=BF 78.7°

PROBLEM 3.9 The tailgate of a car is supported by the hydraulic lift BC. If the lift exerts a 125-N force directed along its center line on the ball and socket at B, determine the moment of the force about A.

SOLUTION

First note ( ) ( )2 2240 mm 46.6 mmCBd = +

244.48 mm=

Then 240 mmcos244.48 mm

θ =

46.6 mmsin244.48 mm

θ =

and cos sinCB CB CBF Fθ θ= −F i j

( ) ( )125 N 240 mm 46.6 mm244.48 mm

= − i j

Now /A B A CB=M r F×

where ( ) ( )/ 306 mm 240 mm 46.6 mmB A = − +r i j

( ) ( )306 mm 286.6 mm= −i j

Then ( ) ( ) ( )125 N306 mm 286.6 mm 240 46.6244.48A = − − M i j i j×

( ) ( )27878 N mm 27.878 N m= ⋅ = ⋅k k

or 27.9 N mA = ⋅M

PROBLEM 3.10 The tailgate of a car is supported by the hydraulic lift BC. If the lift exerts a 125-N force directed along its center line on the ball and socket at B, determine the moment of the force about A.

SOLUTION

First note ( ) ( )2 2344 mm 152.4 mm 376.25 mmCBd = + =

Then 344 mmcos376.25 mm

θ = 152.4 mmsin376.25 mm

θ =

and ( ) ( )cos sinCB CB CBF Fθ θ= −F i j

( ) ( )125 N 344 mm 152.4 mm376.25 mm

= + i j

Now /A B A CB=M r F×

where ( ) ( )/ 410 mm 87.6 mmB A = −r i j

Then ( ) ( ) ( )125 N410 mm 87.6 mm 344 152.4376.25A = − − M i j i j×

( )30770 N mm= ⋅ k

( )30.770 N m= ⋅ k

or 30.8 N mA = ⋅M

PROBLEM 3.11 A winch puller AB is used to straighten a fence post. Knowing that the tension in cable BC is 260 lb, length a is 8 in., length b is 35 in., and length d is 76 in., determine the moment about D of the force exerted by the cable at C by resolving that force into horizontal and vertical components applied (a) at point C, (b) at point E.

SOLUTION

(a) Slope of line 35 in. 576 in. 8 in. 12

EC = =+

Then ( )1213ABx ABT T=

( )12 260 lb 240 lb13

= =

and ( )5 260 lb 100 lb13AByT = =

Then ( ) ( )35 in. 8 in.D ABx AByM T T= −

( )( ) ( )( )240 lb 35 in. 100 lb 8 in.= −

7600 lb in.= ⋅

or 7600 lb in.D = ⋅M

(b) Have ( ) ( )D ABx AByM T y T x= +

( )( ) ( )( )240 lb 0 100 lb 76 in.= +

7600 lb in.= ⋅

or 7600 lb in.D = ⋅M

PROBLEM 3.12 It is known that a force with a moment of 7840 lb in.⋅ about D is required to straighten the fence post CD. If 8a = in., 35b = in., and 112d = in., determine the tension that must be developed in the cable of winch puller AB to create the required moment about point D.

SOLUTION

Slope of line 35 in. 7112 in. 8 in. 24

EC = =+

Then 2425ABx ABT T=

and 725ABy ABT T=

Have ( ) ( )D ABx AByM T y T x= +

( ) ( )24 77840 lb in. 0 112 in.25 25AB ABT T∴ ⋅ = +

250 lbABT =

or 250 lbABT =

PROBLEM 3.13 It is known that a force with a moment of 1152 N m⋅ about D is required to straighten the fence post CD. If the capacity of the winch puller AB is 2880 N, determine the minimum value of distance d to create the specified moment about point D knowing that 0.24a = m and

1.05b = m.

SOLUTION

The minimum value of d can be found based on the equation relating the moment of the force ABT about D:

( ) ( )maxD AB yM T d=

where 1152 N mDM = ⋅

( ) ( )max max sin 2880 N sinAB AByT T θ θ= =

Now ( ) ( )2 2

1.05 msin0.24 1.05 md

θ =+ +

( ) ( )

( )2 2

1.051152 N m 2880 N0.24 1.05

dd

∴ ⋅ =

+ +

or ( ) ( )2 20.24 1.05 2.625d d+ + =

or ( ) ( )2 2 20.24 1.05 6.8906d d+ + =

or 25.8906 0.48 1.1601 0d d− − =

Using the quadratic equation, the minimum values of d are 0.48639 m and −0.40490 m. Since only the positive value applies here, 0.48639 md =

or 486 mmd =

PROBLEM 3.14 A mechanic uses a piece of pipe AB as a lever when tightening an alternator belt. When he pushes down at A, a force of 580 N is exerted on the alternator B. Determine the moment of that force about bolt C if its line of action passes through O.

SOLUTION

Have /C B C B=M r F×

Noting the direction of the moment of each force component about C is clockwise,

C By BxM xF yF= +

where 144 mm 78 mm 66 mmx = − =

86 mm 108 mm 194 mmy = + =

and ( ) ( )

( )2 2

78 580 N 389.65 N78 86

BxF = =+

( ) ( )

( )2 2

86 580 N 429.62 N78 86

ByF = =+

( )( ) ( )( ) 66 mm 429.62 N 194 mm 389.65 NCM∴ = +

103947 N mm= ⋅

103.947 N m= ⋅

or 103.9 N mC = ⋅M

PROBLEM 3.15 Form the vector products B C× and ,′B C× where ,B B′= and use the results obtained to prove the identity

( ) ( )1 12 2sin cos sin sin .α β α β α β= + + −

SOLUTION

First note ( )cos sinB β β= +B i j

( )cos sinB β β′ = −B i j

( )cos sinC α α= +C i j

By definition ( )sinBC α β= −B C× (1)

( )sinBC α β′ = +B C× (2)

Now ( ) ( )cos sin cos sinB Cβ β α α= + +B C i j i j× ×

( )cos sin sin cosBC β α β α= − k (3)

( ) ( )cos sin cos sinB Cβ β α α= − +B C i j i j× ×

( )cos sin sin cosBC β α β α= + k (4)

Equating magnitudes of B C× from Equations (1) and (3), (5)

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

Similarly, equating magnitudes of ′B C× from Equations (2) and (4),

( )sin cos sin sin cosα β β α β α+ = + (6)

Adding Equations (5) and (6)

( ) ( )sin sin 2cos sinα β α β β α− + + =

( ) ( )1 1 sin cos sin sin2 2

α β α β α β∴ = + + −

PROBLEM 3.16 A line passes through the points (420 mm, −150 mm) and (−140 mm, 180 mm). Determine the perpendicular distance d from the line to the origin O of the system of coordinates.

SOLUTION

Have /AB O Ad = rλ ×

where /

/

B AAB

B A=

r

rλ

and ( ) ( )/ 140 mm 420 mm 180 mm 150 mmB A = − − + − − r i j

( ) ( )560 mm 330 mm= − +i j

( ) ( )2 2/ 560 330 mm 650 mmB A = − + =r

( ) ( ) ( )560 mm 330 mm 1 56 33650 mm 65AB

− +∴ = = − +

i ji jλ

( ) ( ) ( ) ( )/ 0 0 420 mm 150 mmO A A Ax y= − + − = − +r i j i j

( ) ( ) ( )1 56 33 420 mm 150 mm 84.0 mm65

d ∴ = − − − + = i j i j×

84.0 mmd =

PROBLEM 3.17 A plane contains the vectors A and B. Determine the unit vector normal to the plane when A and B are equal to, respectively, (a) 4i − 2j + 3k and −2i + 6j − 5k, (b) 7i + j − 4k and −6i − 3k + 2k.

SOLUTION

(a) Have =A BA B

×λ

×

where 4 2 3= − +A i j k

2 6 5= − + −B i j k

Then ( ) ( ) ( ) ( )4 2 3 10 18 6 20 24 4 2 4 7 102 6 5

= − = − + − + + − = − + +− −

i j kA B i j k i j k×

and ( ) ( ) ( )2 2 22 4 7 10 2 165= − + + =A B×

( )2 4 7 10

2 165− + +

∴ =i j k

λ or ( )1 4 7 10165

= − + +i j kλ

(b) Have =A BA B

×λ

×

where 7 4= + −A i j k

6 3 2= − − +B i j k

Then ( ) ( ) ( ) ( )7 1 4 2 12 24 14 21 6 5 2 2 36 3 2

= − = − + − + − + = − + −− −

i j kA B i j k i j k×

and ( ) ( ) ( )2 2 25 2 2 3 5 17= − + + − =A B×

( )5 2 2 3

5 17− + −

∴ =i j k

λ or ( )1 2 2 317

= − + −i j kλ

PROBLEM 3.18 The vectors P and Q are two adjacent sides of a parallelogram. Determine the area of the parallelogram when (a) P = (8 in.)i + (2 in.)j − (1 in.)k and Q = −(3 in.)i + (4 in.)j + (2 in.)k, (b) P = −(3 in.)i + (6 in.)j + (4 in.)k and Q = (2 in.)i + (5 in.)j − (3 in.)k.

SOLUTION

(a) Have A = P Q×

where ( ) ( ) ( )8 in. 2 in. 1 in.= + −P i j k

( ) ( ) ( )3 in. 4 in. 2 in.= − + +Q i j k

Then ( ) ( ) ( )2 28 2 1 in 4 4 3 16 32 6 in3 4 2

= − = + + − + + −

i j kP Q i j k×

( ) ( ) ( )2 2 28 in 13 in 38 in= − +i j k

( ) ( ) ( )2 2 2 2 2 8 13 38 in 40.951 inΑ∴ = + − + = or 241.0 inA =

(b) Have A = P Q×

where ( ) ( ) ( )3 in. 6 in. 4 in.= − + +P i j k

( ) ( ) ( )2 in. 5 in. 3 in.= + −Q i j k

Then ( ) ( ) ( )2 23 6 4 in 18 20 8 9 15 12 in2 5 3

= − = − − + − + − − −

i j kP Q i j k×

( ) ( ) ( )2 2 238 in 1 in 27 in= − − −i j k

( ) ( ) ( )2 2 2 2 2 38 1 27 in 46.626 inΑ∴ = − + − + − = or 246.6 inA =

PROBLEM 3.19 Determine the moment about the origin O of the force F = −(5 N)i − (2 N)j + (3 N)k which acts at a point A. Assume that the position vector of A is (a) r = (4 m)i − (2 m)j − (1 m)k, (b) r = −(8 m)i + (3 m)j + (4 m)k, (c) r = (7.5 m)i + (3 m)j − (4.5 m)k.

SOLUTION

(a) Have O =M r F×

where ( ) ( ) ( )5 N 2 N 3 N= − − +F i j k

( ) ( ) ( )4 m 2 m 1 m= − −r i j k

( ) ( ) ( ) 4 2 1 N m 6 2 5 12 8 10 N m5 2 3

O ∴ = − − ⋅ = − − + − + − − ⋅ − −

i j kM i j k

( )8 7 18 N m= − − − ⋅i j k

or ( ) ( ) ( )8 N m 7 N m 18 N mO = − ⋅ − ⋅ − ⋅M i j k

(b) Have O =M r F×

where ( ) ( ) ( )5 N 2 N 3 N= − − +F i j k

( ) ( ) ( )8 m 3 m 4 m= − + −r i j k

( ) ( ) ( ) 8 3 4 N m 9 8 20 24 16 15 N m5 2 3

O ∴ = − ⋅ = + + − + + + ⋅ − −

i j kM i j k

( )17 4 31 N m= + + ⋅i j k

or ( ) ( ) ( )17 N m 4 N m 31 N mO = ⋅ + ⋅ + ⋅M i j k

(c) Have O =M r F×

where ( ) ( ) ( )5 N 2 N 3 N= − − +F i j k

( ) ( ) ( )7.5 m 3 m 4.5 m= + −r i j k

PROBLEM 3.19 CONTINUED

( ) ( ) ( ) 7.5 3 4.5 N m 9 9 22.5 22.5 15 15 N m5 2 3

O ∴ = − ⋅ = − + − + − + ⋅ − −

i j kM i j k

or 0O =M

This answer is expected since r and F are proportional 2 .3

− =

F r Therefore, vector F has a line of action

passing through the origin at O.

PROBLEM 3.20 Determine the moment about the origin O of the force F = −(1.5 lb)i + (3 lb)j − (2 lb)k which acts at a point A. Assume that the position vector of A is (a) r = (2.5 ft)i − (1 ft)j + (2 ft)k, (b) r = (4.5 ft)i − (9 ft)j + (6 ft)k, (c) r = (4 ft)i − (1 ft)j + (7 ft)k.

SOLUTION

(a) Have O =M r F×

where ( ) ( ) ( )1.5 lb 3 lb 2 lb= − + +F i j k

( ) ( ) ( )2.5 ft 1 ft 2 ft= − +r i j k

Then ( ) ( ) ( )2.5 1 2 lb ft 2 6 3 5 7.5 1.5 lb ft1.5 3 2

O = − ⋅ = − + − + + − ⋅ − −

i j kM i j k

or ( ) ( ) ( )4 lb ft 2 lb ft 6 lb ftO = − ⋅ + ⋅ + ⋅M i j k

(b) Have O =M r F×

where ( ) ( ) ( )1.5 lb 3 lb 2 lb= − + −F i j k

( ) ( ) ( )4.5 ft 9 ft 6 ft= − +r i j k

Then ( ) ( ) ( )4.5 9 6 lb ft 18 18 9 9 13.5 13.5 lb ft1.5 3 2

O = − ⋅ = − + − + + − ⋅ − −

i j kM i j k

or 0O =M

This answer is expected since r and F are proportional 1 .3− =

F r

Therefore, vector F has a line of action passing through the origin at O.

(c) Have O =M r F×

where ( ) ( ) ( )1.5 lb 3 lb 2 lb= − − −F i j k

( ) ( ) ( )4 ft 1 ft 7 ft= − +r i j k

Then ( ) ( ) ( )4 1 7 lb ft 2 21 10.5 8 12 1.5 lb ft1.5 3 2

O = − ⋅ = − + − + + − ⋅ − −

i j kM i j k

or ( ) ( ) ( )19 lb ft 2.5 lb ft 10.5 lb ftO = − ⋅ − ⋅ + ⋅M i j k

PROBLEM 3.21 Before the trunk of a large tree is felled, cables AB and BC are attached as shown. Knowing that the tension in cables AB and BC are 777 N and 990 N, respectively, determine the moment about O of the resultant force exerted on the tree by the cables at B.

SOLUTION

Have /O B O B=M r F×

where ( )/ 8.4 mB O =r j

B AB BC= +F T T

( ) ( ) ( )( ) ( ) ( )

( )2 2 2

0.9 m 8.4 m 7.2 m777 N

0.9 8.4 7.2 mAB BA ABT

− − += =

+ +

i j kT λ

( ) ( ) ( )( ) ( ) ( )

( )2 2 2

5.1 m 8.4 m 1.2 m990 N

5.1 8.4 1.2 mBC BC BCT

− += =

+ +

i j kT λ

PROBLEM 3.21 CONTINUED

( ) ( ) ( ) ( ) ( ) ( ) 63.0 N 588 N 504 N 510 N 840 N 120 NB ∴ = − − + + − + F i j k i j k

( ) ( ) ( )447 N 1428 N 624 N= − +i j k

and ( ) ( )0 8.4 0 N m 5241.6 N m 3754.8 N m447 1428 624

O = ⋅ = ⋅ − ⋅−

i j kM i k

or ( ) ( )5.24 kN m 3.75 kN mO = ⋅ − ⋅M i k

PROBLEM 3.22 Before a telephone cable is strung, rope BAC is tied to a stake at B and is passed over a pulley at A. Knowing that portion AC of the rope lies in a plane parallel to the xy plane and that the tension T in the rope is 124 N, determine the moment about O of the resultant force exerted on the pulley by the rope.

SOLUTION

Have /O AO=M r R×

where ( ) ( ) ( )/ 0 m 9 m 1 mAO = + +r i j k

1 2= +R T T

( ) ( )1 124 N cos10 124 N sin10 = − ° − ° T i j

( ) ( )122.116 N 21.532 N= − −i j

( ) ( ) ( )( ) ( ) ( )

( )2 2 2 2 2

1.5 m 9 m 1.8 m124 N

1.5 m 9 m 1.8 mT

− + = = + +

i j kT λ

( ) ( ) ( )20 N 120 N 24 N= − +i j k

( ) ( ) ( ) 102.116 N 141.532 N 24 N∴ = − − +R i j k

0 9 1 N m102.116 141.532 24

O = ⋅− −

i j kM

( ) ( ) ( )357.523 N m 102.116 N m 919.044 N m= ⋅ − ⋅ + ⋅i j k

or ( ) ( ) ( )358 N m 102.1 N m 919 N mO = ⋅ − ⋅ + ⋅M i j k

PROBLEM 3.23 An 8-lb force is applied to a wrench to tighten a showerhead. Knowing that the centerline of the wrench is parallel to the x axis, determine the moment of the force about A.

SOLUTION

Have /A C A=M r F×

where ( ) ( ) ( )/ 8.5 in. 2.0 in. 5.5 in.C A = − +r i j k

( )8cos 45 sin12 lbxF = − ° °

( )8sin 45 lbyF = − °

( )8cos 45 cos12 lbzF = − ° °

( ) ( ) ( ) 1.17613 lb 5.6569 lb 5.5332 lb∴ = − − −F i j k

and 8.5 2.0 5.5 lb in.1.17613 5.6569 5.5332

A = − ⋅− − −

i j kM

( ) ( ) ( )42.179 lb in. 40.563 lb in. 50.436 lb in.= ⋅ + ⋅ − ⋅i j k

or ( ) ( ) ( )42.2 lb in. 40.6 lb in. 50.4 lb in.A = ⋅ + ⋅ − ⋅M i j k

PROBLEM 3.24 A wooden board AB, which is used as a temporary prop to support a small roof, exerts at point A of the roof a 228 N force directed along BA. Determine the moment about C of that force.

SOLUTION

Have /C AC BA=M r F×

where ( ) ( ) ( )/ 0.96 m 0.12 m 0.72 mAC = − +r i j k

and BA BA BAF=F λ

( ) ( ) ( )( ) ( ) ( )

( )2 2 2

0.1 m 1.8 m 0.6 m228 N

0.1 1.8 0.6 m

− + − = + +

i j k

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

0.96 0.12 0.72 N m12.0 216 72

C∴ = − ⋅− −

i j kM

( ) ( ) ( )146.88 N m 60.480 N m 205.92 N m= − ⋅ + ⋅ + ⋅i j k

or ( ) ( ) ( )146.9 N m 60.5 N m 206 N mC = − ⋅ + ⋅ + ⋅M i j k

PROBLEM 3.25 The ramp ABCD is supported by cables at corners C and D. The tension in each of the cables is 360 lb. Determine the moment about A of the force exerted by (a) the cable at D, (b) the cable at C.

SOLUTION

(a) Have /A E A DE=M r T×

where ( )/ 92 in.E A =r j

DE DE DET=T λ

( ) ( ) ( )( ) ( ) ( )

( )2 2 2

24 in. 132 in. 120 in.360 lb

24 132 120 in.

+ −=

+ +

i j k

( ) ( ) ( )48 lb 264 lb 240 lb= + −i j k

( ) ( ) 0 92 0 lb in. 22,080 lb in. 4416 lb in48 264 240

A∴ = ⋅ = − ⋅ − ⋅−

i j kM i k

or ( ) ( )1840 lb ft 368 lb ftA = − ⋅ − ⋅M i k

(b) Have /A G A CG=M r T×

where ( ) ( )/ 108 in. 92 in.G A = +r i j

( ) ( ) ( )( ) ( ) ( )

( )2 2 2

24 in. 132 in. 120 in.360 lb

24 132 120 in.CG CG CGT

− + −= =

+ +

i j kT λ

( ) ( ) ( )48 lb 264 lb 240 lb= − + −i j k

108 92 0 lb in.48 264 240

A∴ = ⋅− −

i j kM

( ) ( ) ( )22,080 lb in. 25,920 lb in. 32,928 lb in.= − ⋅ + ⋅ + ⋅i j k

or ( ) ( ) ( )1840 lb ft 2160 lb ft 2740 lb ftA = − ⋅ + ⋅ + ⋅M i j k

PROBLEM 3.26 The arms AB and BC of a desk lamp lie in a vertical plane that forms an angle of o30 with the xy plane. To reposition the light, a force of magnitude 8 N is applied at C as shown. Determine the moment of the force about O knowing that 450=AB mm, 325=BC mm, and line CD is parallel to the z axis.

SOLUTION

Have /O C O C=M r F×

where ( ) ( )/ cos30C O xz xzxr AB BC= + °

( )0.450 m sin 45 0.31820 mxzAB = ° =

( )0.325 m sin 50 0.24896 mxzBC = ° =

( ) ( ) ( )/ 0.150 m 0.450 m cos 45C O y y yyr OA AB BC= + − = + °

( )0.325 m cos50 0.25929 m− ° =

( ) ( )/ sin 30C O xz xzzr AB BC= + °

( )0.31820 m 0.24896 m sin 30 0.28358 m= + ° =

or ( ) ( ) ( )/ 0.49118 m 0.25929 m 0.28358 mC O = + +r i j k

( ) ( )8 N cos 45 sin 20 1.93476 NC xF = − ° ° = −

( ) ( )8 N sin 45 5.6569 NC yF = − ° = −

( ) ( )8 N cos 45 cos 20 5.3157 NC zF = ° ° =

or ( ) ( ) ( )1.93476 N 5.6569 N 5.3157 NC = − − +F i j k

0.49118 0.25929 0.28358 N m1.93476 5.6569 5.3157

O∴ = ⋅− −

i j kM

( ) ( ) ( )2.9825 N m 3.1596 N m 2.2769 N m= ⋅ − ⋅ − ⋅i j k

or ( ) ( ) ( )2.98 N m 3.16 N m 2.28 N mO = ⋅ − ⋅ − ⋅M i j k

PROBLEM 3.27 In Problem 3.21, determine the perpendicular distance from point O to cable AB.

Problem 3.21: Before the trunk of a large tree is felled, cables AB and BC are attached as shown. Knowing that the tension in cables AB and BC are 777 N and 990 N, respectively, determine the moment about O of the resultant force exerted on the tree by the cables at B.

SOLUTION

Have | O BAT d=M |

where perpendicular distance from O to line .d AB=

Now /O B O BA=Μ r T×

and ( )/ 8.4 mB O =r j

( ) ( ) ( )( ) ( ) ( )

( )2 2 2

0.9 m 8.4 m 7.2 m777 N

0.9 8.4 7.2 mBA BA ABT

− − += =

+ +

i j kT λ

( ) ( ) ( )63.0 N 588 N 504 N= − − +i j k

( ) ( )0 8.4 0 N m 4233.6 N m 529.2 N m63.0 588 504

O∴ = ⋅ = ⋅ + ⋅− −

i j kM i k

and ( ) ( )2 2| | 4233.6 529.2 4266.5 N mO = + = ⋅M

( )4266.5 N m 777 N d∴ ⋅ =

or 5.4911 md =

or 5.49 md =

PROBLEM 3.28 In Problem 3.21, determine the perpendicular distance from point O to cable BC.

Problem 3.21: Before the trunk of a large tree is felled, cables AB and BC are attached as shown. Knowing that the tension in cables AB and BC are 777 N and 990 N, respectively, determine the moment about O of the resultant force exerted on the tree by the cables at B.

SOLUTION

Have | |O BCT d=M

where perpendicular distance from to line .d O BC=

/O B O BC=M r T×

/ 8.4 mB O =r j

( ) ( ) ( )( ) ( ) ( )

( )2 2 2

5.1 m 8.4 m 1.2 m990 N

5.1 8.4 1.2 mBC BC BCT

− += =

+ +

i j kT λ

( ) ( ) ( )510 N 840 N 120 N= − +i j k

( ) ( ) 0 8.4 0 1008 N m 4284 N m510 840 120

O∴ = = ⋅ − ⋅−

i j kM i k

and ( ) ( )2 2| | 1008 4284 4401.0 N mO = + = ⋅M

( )4401.0 N m 990 N d∴ ⋅ =

4.4454 md =

or 4.45 md =

PROBLEM 3.29 In Problem 3.24, determine the perpendicular distance from point D to a line drawn through points A and B.

Problem 3.24: A wooden board AB, which is used as a temporary prop to support a small roof, exerts at point A of the roof a 228 N force directed along BA. Determine the moment about C of that force.

SOLUTION

Have | |D BAF d=M

where perpendicular distance from to line .d D AB=

/D A D BA=M r F×

( ) ( )/ 0.12 m 0.72 mA D = − +r j k

( ) ( ) ( )( )( ) ( ) ( )

( )2 2 2

0.1 m 1.8 m 0.6 m228 N

0.1 1.8 0.6 mBA BA BAF

− + −= =

+ +

i j kF λ

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

0 0.12 0.72 N m12.0 216 72

D∴ = − ⋅− −

i j kM

( ) ( ) ( )146.88 N m 8.64 N m 1.44 N m= − ⋅ − ⋅ − ⋅i j k

and ( ) ( ) ( )2 2 2| | 146.88 8.64 1.44 147.141 N mD = + + = ⋅M

( )147.141 N m 228 N d∴ ⋅ =

0.64536 md =

or 0.645 md =

PROBLEM 3.30 In Problem 3.24, determine the perpendicular distance from point C to a line drawn through points A and B.

Problem 3.24: A wooden board AB, which is used as a temporary prop to support a small roof, exerts at point A of the roof a 228 N force directed along BA. Determine the moment about C of that force.

SOLUTION

Have | |C BAF d=M

where perpendicular distance from to line .d C AB=

/C AC BA=M r F×

( ) ( ) ( )/ 0.96 m 0.12 m 0.72 mAC = − +r i j k

( ) ( ) ( )( )( ) ( ) ( )

( )2 2 2

0.1 m 1.8 m 0.6228 N

0.1 1.8 0.6 mBA BA BAF

− + −= =

+ +

i j kF λ

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

0.96 0.12 0.72 N m12.0 216 72

C∴ = − ⋅− −

i j kM

( ) ( ) ( )146.88 N m 60.48 N m 205.92 N m= − ⋅ − ⋅ + ⋅i j k

and ( ) ( ) ( )2 2 2| | 146.88 60.48 205.92 260.07 N mC = + + = ⋅M

( )260.07 N m 228 N d∴ ⋅ =

1.14064 md =

or 1.141 md =

PROBLEM 3.31 In Problem 3.25, determine the perpendicular distance from point A to portion DE of cable DEF.

Problem 3.25: The ramp ABCD is supported by cables at corners C and D. The tension in each of the cables is 360 lb. Determine the moment about A of the force exerted by (a) the cable at D, (b) the cable at C.

SOLUTION

Have A DET d=M

where perpendicular distance from to line .d A DE=

/A E A DE=M r T×

( )/ 92 in.E A =r j

( ) ( ) ( )( ) ( ) ( )

( )2 2 2

24 in. 132 in. 120 in.360 lb

24 132 120 in.DE DE DET

+ −= =

+ +

i j kT λ

( ) ( ) ( )48 lb 264 lb 240 lb= + −i j k

0 92 0 N m48 264 240

A∴ = ⋅−

i j kM

( ) ( )22,080 lb in. 4416 lb in.= − ⋅ − ⋅i k

PROBLEM 3.31 CONTINUED

and ( ) ( )2 222,080 4416 22,517 lb in.A = + = ⋅M

( )22,517 lb in. 360 lb d∴ ⋅ =

62.548 in.d =

or 5.21 ftd =

PROBLEM 3.32 In Problem 3.25, determine the perpendicular distance from point A to a line drawn through points C and G.

Problem 3.25: The ramp ABCD is supported by cables at corners C and D. The tension in each of the cables is 360 lb. Determine the moment about A of the force exerted by (a) the cable at D, (b) the cable at C.

SOLUTION

Have A CGT d=M

where perpendicular distance from to line .d A CG=

/A G A CG=M r T×

( ) ( )/ 108 in. 92 in.G A = +r i j

CG CG CGT=T λ

( ) ( ) ( )( ) ( ) ( )

( )2 2 2

24 in. 132 in. 120 in.360 lb

24 132 120 in.

− + −=

+ +

i j k

( ) ( ) ( )48 lb 264 lb 240 lb= − + −i j k

108 92 0 lb in.48 264 240

A∴ = ⋅− −

i j kM

( ) ( ) ( )22,080 lb in. 25,920 lb in. 32,928 lb in.= − ⋅ + ⋅ + ⋅i j k

and ( ) ( ) ( )2 2 222,080 25,920 32,928 47,367 lb in.A = + + = ⋅M

( )47,367 lb in. 360 lb d∴ ⋅ =

131.575 in.d = or 10.96 ftd =

PROBLEM 3.33 In Problem 3.25, determine the perpendicular distance from point B to a line drawn through points D and E.

Problem 3.25: The ramp ABCD is supported by cables at corners C and D. The tension in each of the cables is 360 lb. Determine the moment about A of the force exerted by (a) the cable at D, (b) the cable at C.

SOLUTION

Have B DET d=M

where perpendicular distance from to line .d B DE=

/B E B DE=M r T×

( ) ( )/ 108 in. 92 in.E B = − +r i j

( ) ( ) ( )( ) ( ) ( )

( )2 2 2

24 in. 132 in. 120 in.360 lb

24 132 120 in.DE DE DET

+ −= =

+ +

i j kT λ

( ) ( ) ( )48 lb 264 lb 240 lb= + −i j k

108 92 0 lb in.48 264 240

B∴ = − ⋅−

i j kM

( ) ( ) ( )22,080 lb in. 25,920 lb in. 32,928 lb in.= − ⋅ − ⋅ − ⋅i j k

and ( ) ( ) ( )2 2 222,080 25,920 32,928 47,367 lb in.B = + + = ⋅M

( )47,367 lb in. 360 lb d∴ ⋅ =

131.575 in.d =

or 10.96 ftd =

PROBLEM 3.34 Determine the value of a which minimizes the perpendicular distance from point C to a section of pipeline that passes through points A and B.

SOLUTION

Assuming a force F acts along AB,

( )/C AC F d= =M r F×

where

perpendicular distance from to line d C AB=

( ) ( ) ( )( ) ( ) ( )2 2 2

8 m 7 m 9 m

8 7 9 mABF F

+ −= =

+ +

i j kF λ

( ) ( ) ( )0.57437 0.50257 0.64616F= + −i j k

( ) ( ) ( )/ 1 m 2.8 m 3 mAC a= − − −r i j k

1 2.8 30.57437 0.50257 0.64616

C a F∴ = − −−

i j kM

( ) ( )0.30154 0.50257 2.3693 0.57437a a= + + − i j

]2.1108 F+ k

Since ( )22 2/ /or C AC AC dF= × =M r F r F×

( ) ( ) ( )2 2 2 20.30154 0.50257 2.3693 0.57437 2.1108a a d∴ + + − + =

Setting ( )2 0d dda

= to find a to minimize d

( )( )2 0.50257 0.30154 0.50257a+

( )( )2 0.57437 2.3693 0.57437 0a+ − − =

Solving 2.0761 ma =

or 2.08 ma =

PROBLEM 3.35

Given the vectors P = 7i − 2j + 5k, Q = −3i − 4j + 6k, and S = 8i + j − 9k, compute the scalar products ,P Q⋅ ,P S⋅ and .Q S⋅

SOLUTION

( ) ( )7 2 5 3 4 6= − + − − +P Q i j k i j k⋅ ⋅

( )( ) ( )( ) ( )( )7 3 2 4 5 6= − + − − +

17=

or 17=P Q⋅

( ) ( )7 2 5 8 9= − + + −P S i j k i j k⋅ ⋅

( )( ) ( )( ) ( )( )7 8 2 1 5 9= + − + −

9=

or 9=P S⋅

( ) ( )3 4 6 8 9= − − + + −Q S i j k i j k⋅ ⋅

( )( ) ( )( ) ( )( )3 8 4 1 6 9= − + − + −

82= −

or 82= −Q S⋅

PROBLEM 3.36 Form the scalar products B C⋅ and ,′B C⋅ where ,B B′= and use the results obtained to prove the identity

( ) ( )1 12 2cos cos cos cos .α β α β α β= + + −

SOLUTION

By definition

( )cosBC α β= −B C⋅

where ( ) ( )cos sinB β β = + B i j

( ) ( )cos sinC α α = + C i j

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

or ( )cos cos sin sin cosβ α β α α β+ = − (1)

By definition

( )cosBC α β′ = +B C⋅

where ( ) ( )cos sinβ β′ = − B i j

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

or ( )cos cos sin sin cosβ α β α α β− = + (2)

Adding Equations (1) and (2),

( ) ( )2 cos cos cos cosβ α α β α β= − + +

or ( ) ( )1 1cos cos cos cos2 2

α β α β α β= + + −

PROBLEM 3.37 Consider the volleyball net shown. Determine the angle formed by guy wires AB and AC.

SOLUTION

First note

( ) ( ) ( )2 2 2/ 1.95 m 2.4 m 0.6 m B AAB = = − + − +r

3.15 m=

( ) ( ) ( )2 2 2/ 0 m 2.4 m 1.8 m C AAC = = + − +r

3.0 m=

and

( ) ( ) ( )/ 1.95 m 2.40 m 0.6 m B A = − − +r i j k

( ) ( )/ 2.40 m 1.80 m C A = − +r j k

By definition

/ / / / cosB A C A B A C A θ=r r r r⋅

or ( ) ( ) ( )( )1.95 2.40 0.6 2.40 1.80 3.15 3.0 cosθ− − + − + =i j k j k⋅

( )( ) ( )( ) ( )( )1.95 0 2.40 2.40 0.6 1.8 9.45cosθ− + − − + =

cos 0.72381θ∴ =

and 43.630θ = °

or 43.6θ = °

PROBLEM 3.38 Consider the volleyball net shown. Determine the angle formed by guy wires AC and AD.

SOLUTION

First note

( ) ( )2 2/ 2.4 1.8 m 3 mC AAC = = − + =r

( ) ( ) ( )2 2 2/ 1.2 2.4 0.3 m 2.7 mD AAD = = + − + =r

and ( ) ( )/ 2.4 m 1.8 m C A = − +r j k

( ) ( ) ( )/ 1.2 m 2.4 m 0.3 m D A = − +r i j k

By definition

/ / / / cosC A D A C A D A θ=r r r r⋅

or ( ) ( ) ( )( )2.4 1.8 1.2 2.4 0.3 3 2.7 cosθ− + − + =j k i j k⋅

( )( ) ( )( ) ( )( )0 1.2 2.4 2.4 1.8 0.3 8.1cosθ+ − − + =

and 6.3cos 0.777788.1

θ = =

38.942θ = °

or 38.9θ = °

PROBLEM 3.39 Steel framing members AB, BC, and CD are joined at B and C and are braced using cables EF and EG. Knowing that E is at the midpoint of BC and that the tension in cable EF is 330 N, determine (a) the angle between EF and member BC, (b) the projection on BC of the force exerted by cable EF at point E.

SOLUTION

(a) By definition ( )( )1 1 cosBC EF θ=λ ⋅ λ

where ( ) ( ) ( )( ) ( ) ( )

( )2 2 2

16 m 4.5 m 12 m 1 16 4.5 1220.516 4.5 12 m

BC− −

= = − −+ +

i j ki j kλ

( ) ( ) ( )( ) ( ) ( )

( )2 2 2

7 m 6 m 6 m 1 7 6 611.07 6 6 m

EF− − +

= = − − ++ +

i j ki j kλ

( ) ( )16 4.5 12 7 6 6cos

20.5 11.0θ

− − − − +∴ =

i j k i j k⋅

( )( ) ( )( ) ( )( ) ( )( )16 7 4.5 6 12 6 20.5 11.0 cosθ− + − − + − =

and 1 157cos 134.125225.5

θ − − = = °

or 134.1θ = °

(b) By definition ( ) cosEF EFBCT T θ=

( )330 N cos134.125= °

229.26 N= −

or ( ) 230 NEF BCT = −

PROBLEM 3.40 Steel framing members AB, BC, and CD are joined at B and C and are braced using cables EF and EG. Knowing that E is at the midpoint of BC and that the tension in cable EG is 445 N, determine (a) the angle between EG and member BC, (b) the projection on BC of the force exerted by cable EG at point E.

SOLUTION

(a) By definition ( )( )1 1 cosBC EG θ=λ ⋅ λ

where ( ) ( ) ( )( ) ( ) ( )2 2 2

16 m 4.5 m 12 m 16 4.5 1220.516 m 4.5 12 m

BC− − − −

= =+ +

i j k i j kλ

0.78049 0.21951 0.58537= − −i j k

( ) ( ) ( )( ) ( ) ( )2 2 2

8 m 6 m 4.875 m 8 6 4.87511.1258 6 4.875 m

EG− + − +

= =+ +

i j k i j kλ

0.71910 0.53933 0.43820= − +i j k

( ) ( )( ) ( )( )( )( )

16 8 4.5 6 12 4.875cos

20.5 11.25BC EG θ+ − − + −

∴ = =λ ⋅ λ

and 1 96.5cos 64.967228.06

θ − = = °

or 65.0θ = °

(b) By definition ( ) cosEG EGBCT T θ=

( )445 N cos64.967= °

188.295 N=

or ( ) 188.3 NEG BCT =

PROBLEM 3.41 Slider P can move along rod OA. An elastic cord PC is attached to the slider and to the vertical member BC. Knowing that the distance from O to P is 0.12 m and the tension in the cord is 30 N, determine (a) the angle between the elastic cord and the rod OA, (b) the projection on OA of the force exerted by cord PC at point P.

SOLUTION

(a) By definition ( )( )1 1 cosOA PC θ=λ ⋅ λ

where ( ) ( ) ( )( ) ( ) ( )2 2 2

0.24 m 0.24 m 0.12 m

0.24 0.24 0.12 mOA

+ −=

+ +

i j kλ

2 2 13 3 3

= + −i j k

Knowing that /| | 0.36 mA O OAL= =r and that P is located 0.12 m from O, it follows that the coordinates

of P are 13

the coordinates of A.

( )0.08 m, 0.08 m, 0.040 mP∴ −

Then ( ) ( ) ( )( ) ( ) ( )2 2 2

0.10 m 0.22 m 0.28 m

0.10 0.22 0.28 mPC

+ +=

+ +

i j kλ

0.27037 0.59481 0.75703= + +i j k

( )2 2 1 0.27037 0.59481 0.75703 cos3 3 3

θ ∴ + − + + =

i j k i j k⋅

and ( )1cos 0.32445 71.068θ −= = °

or 71.1θ = °

(b) ( ) ( )cos 30 N cos71.068PC PCOAT T θ= = °

( ) 9.7334 NPC OAT =

or ( ) 9.73 NPC OAT =

PROBLEM 3.42 Slider P can move along rod OA. An elastic cord PC is attached to the slider and to the vertical member BC. Determine the distance from O to P for which cord PC and rod OA are perpendicular.

SOLUTION

The requirement that member OA and the elastic cord PC be perpendicular implies that

0OA PC =λ ⋅ λ or / 0OA C P =rλ ⋅

where ( ) ( ) ( )( ) ( ) ( )2 2 2

0.24 m 0.24 m 0.12 m

0.24 0.24 0.12 mOA

+ −=

+ +

i j kλ

2 2 13 3 3

= + −i j k

Letting the coordinates of P be ( ), , ,P x y z we have

( ) ( ) ( )/ 0.18 0.30 0.24 mC P x y z = − + − + − r i j k

( ) ( ) ( )2 2 1 0.18 0.30 0.24 03 3 3

x y z ∴ + − − + − + − = i j k i j k⋅ (1)

Since ( )/ 2 2 ,3OP

P O OA OPdd= = + −r i j kλ

Then ,2 2 1,3 3 3OP OP OPx d y d z d−

= = = (2)

Substituting the expressions for x, y, and z from Equation (2) into Equation (1),

( )1 2 2 12 2 0.18 0.30 0.24 03 3 3 3OP OP OPd d d

+ − − + − + + =

i j k i j k⋅

or 3 0.36 0.60 0.24 0.72OPd = + − =

0.24 mOPd∴ =

or 240 mmOPd =

PROBLEM 3.43 Determine the volume of the parallelepiped of Figure 3.25 when (a) P = −(7 in.)i − (1 in.)j + (2 in.)k, Q = (3 in.)i − (2 in.)j + (4 in.)k, and S = −(5 in.)i + (6 in.)j − (1 in.)k, (b) P = (1 in.)i + (2 in.)j − (1 in.)k, Q = −(8 in.)i − (1 in.)j + (9 in.)k, and S = (2 in.)i + (3 in.)j + (1 in.)k.

SOLUTION

Volume of a parallelepiped is found using the mixed triple product.

(a) ( )Vol = P Q S⋅ ×

( )3 37 1 23 2 4 in 14 168 20 3 36 20 in5 6 1

− −= − = − + + − + −

− −

3187 in=

3or Volume 187 in=

(b) ( )Vol = P Q S⋅ ×

( )3 31 2 18 1 9 in 1 27 36 16 24 2 in

2 3 1

−= − − = − − + + + −

346 in=

or 3Volume 46 in=

PROBLEM 3.44 Given the vectors P = 4i − 2j + Pzk, Q = i + 3j − 5k, and S = −6i + 2j − k, determine the value of Pz for which the three vectors are coplanar.

SOLUTION

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

( ) 0=P Q S⋅ ×

( )4 21 3 5 12 40 60 2 2 186 2 1

z

z

PO P

−∴ = − = − + − − + +

− −

so that 34 1.7020zP = =

or 1.700zP =

PROBLEM 3.45 The 0.732 1.2-m× lid ABCD of a storage bin is hinged along side AB and is held open by looping cord DEC over a frictionless hook at E. If the tension in the cord is 54 N, determine the moment about each of the coordinate axes of the force exerted by the cord at D.

SOLUTION

First note ( ) ( )2 20.732 0.132 mz = −

0.720 m=

Then ( ) ( ) ( )2 2 20.360 0.720 0.720 mDEd = + +

1.08 m=

and ( ) ( ) ( )/ 0.360 m 0.720 m 0.720 mE D = + −r i j k

Have ( )/OE

DE E DDE

Td

=T r

( )54 N 0.360 0.720 0.7201.08

= + −i j k

( ) ( ) ( )18.0 N 36.0 N 36.0 N= + −i j k

Now /A D A DE=M r T×

where ( ) ( )/ 0.132 m 0.720 mD A = +r j k

Then 0 0.132 0.720 N m18.0 36.0 36.0

A = ⋅−

i j kM

PROBLEM 3.45 CONTINUED

( )( ) ( )( ) ( )( ){ 0.132 36.0 0.720 36.0 0.720 18.0 0A ∴ = − − + − M i j

( )( ) }0 0.132 18.0 N m + − ⋅ k

or ( ) ( ) ( )30.7 N m 12.96 N m 2.38 N mA = − ⋅ + ⋅ − ⋅M i j k

30.7 N m, 12.96 N m,x yM M∴ = − ⋅ = ⋅ 2.38 N mzM = − ⋅

PROBLEM 3.46 The 0.732 1.2-m× -m lid ABCD of a storage bin is hinged along side AB and is held open by looping cord DEC over a frictionless hook at E. If the tension in the cord is 54 N, determine the moment about each of the coordinate axes of the force exerted by the cord at C.

SOLUTION

First note ( ) ( )2 20.732 0.132 mz = −

0.720 m=

Then ( ) ( ) ( )2 2 20.840 0.720 0.720 mCEd = + +

1.32 m=

and ( )/E CCE CE

CET

d=

rT

( ) ( ) ( ) ( )0.840 m 0.720 m 0.720 m54 N

1.32 m− + −

=i j k

( ) ( ) ( )36.363 N 29.454 N 29.454 N= − + −i j k

Now /A E A CE=M r T×

where ( ) ( )/ 0.360 m 0.852 mE A = +r i j

Then 0.360 0.852 0 N m34.363 29.454 29.454

A = ⋅− −

i j kM

( ) ( ) ( )25.095 N m 10.6034 N m 39.881 N m= − ⋅ + ⋅ + ⋅i j k

25.1 N m, 10.60 N m, 39.9 N mx y zM M M∴ = − ⋅ = ⋅ = ⋅

PROBLEM 3.47 A fence consists of wooden posts and a steel cable fastened to each post and anchored in the ground at A and D. Knowing that the sum of the moments about the z axis of the forces exerted by the cable on the posts at B and C is −66 N · m, determine the magnitude CDT when 56 N.BAT =

SOLUTION

Based on ( ) ( )| |z B BA C CDy y = + M k r T k r T⋅ × ⋅ ×

where

( )66 N mz = − ⋅M k

( ) ( ) ( )1 mB Cy y= =r r j

BA BA BAT=T λ

( ) ( ) ( ) ( )1.5 m 1 m 3 m56 N

3.5 m− +

=i j k

( ) ( ) ( )24 N 16 N 48 N= − +i j k

CD CD CDT=T λ

( ) ( ) ( )2 m 1 m 2 m3.0 m CDT

− −=

i j k

( )1 2 23 CDT= − −i j k

( ) ( ) ( ) ( ) ( ){ }66 N m 1 m 24 N 16 N 48 N ∴ − ⋅ = − + k j i j k⋅ ×

( ) ( )11 m 2 23 CDT

+ − − k j i j k⋅ ×

or 266 243 CDT− = − −

( )3 66 24 N2CDT∴ = −

or 63.0 NCDT =

PROBLEM 3.48 A fence consists of wooden posts and a steel cable fastened to each post and anchored in the ground at A and D. Knowing that the sum of the moments about the y axis of the forces exerted by the cable on the posts at B and C is 212 N · m, determine the magnitude of BAT when

33 N.CDT =

SOLUTION

Based on ( ) ( )| |y B BA C CDz z = + M j r T r T⋅ × ×

where

( )212 N my = ⋅M j

( ) ( )8 mB z =r k

( ) ( )2 mC z =r k

BA BA BAT=T λ

( ) ( ) ( )1.5 m 1 m 3 m3.5 m BAT

− −=

i j k

( )1.5 33.5

BAT= − +i j k

CD CD CDT=T λ

( ) ( ) ( ) ( )2 m 1 m 2 m33 N

3.0 m− −

=i j k

( )22 11 22 N= − −i j k

( ) ( ) ( )212 N m 8 m 1.5 33.5

BAT ∴ ⋅ = − + j k i j k⋅ ×

( ) ( )2 m 22 11 22 N + − − j k i j k⋅ ×

or ( ) ( )8 1.5212 2 22

3.5 BAT= +

16818.6667BAT∴ =

or 49.0 NBAT =

PROBLEM 3.49 To lift a heavy crate, a man uses a block and tackle attached to the bottom of an I-beam at hook B. Knowing that the moments about the y and z axes of the force exerted at B by portion AB of the rope are, respectively, 100 lb ft⋅ and 400 lb ft− ⋅ , determine the distance a.

SOLUTION

Based on /O A O BA=M r T×

where

O x y zM M M= + +M i j k

( ) ( )100 lb ft 400 lb ftxM= + ⋅ − ⋅i j k

( ) ( )/ 6 ft 4 ftA O = +r i j

BA BA BAT=T λ

( ) ( ) ( )6 ft 12 ftBA

BA

aT

d− −

=i j k

100 400 6 4 06 12

BAx

BA

TMd

a∴ + − =

− −

i j ki j k

( ) ( ) ( )4 6 96BA

BA

T a ad

= − + − i j k

From -coefficient:j 100 6AB BAd aT= 100or 6BA BAT d

a= (1)

From -coefficient:k 400 96AB BAd T− = − 400or 96BA BAT d= (2)

Equating Equations (1) and (2) yields ( )( )

100 966 400

a =

or 4.00 fta =

PROBLEM 3.50 To lift a heavy crate, a man uses a block and tackle attached to the bottom of an I-beam at hook B. Knowing that the man applies a 200-lb force to end A of the rope and that the moment of that force about the y axis is 175 lb ft⋅ , determine the distance a.

SOLUTION

Based on ( )/| |y A O BA=M j r T⋅ ×

where ( ) ( )/ 6 ft 4 ftA O = +r i j

/A BBA BA BA BA

BAT T

d= =

rT λ

( ) ( ) ( ) ( )6 ft 12 ft200 lb

BA

ad

− −=

i j k

( )200 6 12BA

ad

= − −i j k

0 1 0

200175 lb ft 6 4 06 12 BAd

a∴ ⋅ =

− −

( ) 200175 0 6BA

ad

= − −

where ( ) ( ) ( )2 2 26 12 ftBAd a= + +

2180 fta= +

2175 180 1200a a∴ + =

or 2180 6.8571a a+ =

Squaring each side

2 2180 47.020a a+ =

Solving 1.97771 fta =

or 1.978 fta =

PROBLEM 3.51 A force P is applied to the lever of an arbor press. Knowing that P lies in a plane parallel to the yz plane and that 230 lb in.,xM = ⋅

200 lb in.,yM = − ⋅ and 35 lb in.,zM = − ⋅ determine the magnitude of P

and the values of φ and θ.

SOLUTION

Based on ( ) ( ) ( ) ( )cos 9 in. sin sin 9 in. cosxM P Pφ θ φ θ = − (1)

( )( )cos 5 in.yM P φ= − (2)

( )( )sin 5 in.zM P φ= − (3)

Then ( )( )

sin (5)Equation (3) :Equation (2) cos (5)

z

y

PMM P

φφ

−=

−

or 35tan 0.175 9.9262200

φ φ−= = = °

−

or 9.93φ = °

Substituting into Equation (2)φ

( )200 lb in. cos9.9262 (5 in.)P− ⋅ = − °

40.608 lbP =

or 40.6 lbP =

Then, from Equation (1)

( ) ( )230 lb in. 40.608 lb cos9.9262 9 in. sinθ ⋅ = °

( ) ( )40.608 lb sin 9.9262 9 in. cosθ − °

or 0.98503sin 0.172380cos 0.62932θ θ− =

Solving numerically, 48.9θ = °

PROBLEM 3.52 A force P is applied to the lever of an arbor press. Knowing that P lies in a plane parallel to the yz plane and that 180 lb in.yM = − ⋅ and

30 lb in.,zM = − ⋅ determine the moment xM of P about the x axis when 60 .θ = °

SOLUTION

Based on ( ) ( ) ( ) ( )cos 9 in. sin sin 9 in. cosxM P Pφ θ φ θ = − (1)

( )( )cos 5 in.yM P φ= − (2)

( )( )sin 5 in.zM P φ= − (3)

Then ( )( )( )( )

sin 5Equation (3) :Equation (2) cos 5

z

y

PMM P

φφ

−=

−

or 30 tan180

φ−=

−

9.4623φ∴ = °

From Equation (3),

( )( )30 lb in. sin 9.4623 5 in.P− ⋅ = − °

36.497 lbP∴ =

From Equation (1),

( )( )( )36.497 lb 9 in. cos9.4623 sin 60 sin 9.4623 cos60xM = ° ° − ° °

253.60 lb in.= ⋅

or 254 lb in.xM = ⋅

PROBLEM 3.53 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable AE at A is 220 lb, determine the moment of that force about the line joining points D and B.

SOLUTION

Have ( )/DB DB A D AEM = r Tλ ⋅ ×

where ( ) ( )48 in. 14 in.0.96 0.28

50 in.DB−

= = −i j

i jλ

( ) ( )/ 4 in. 8 in.A D = − +r j k

( ) ( ) ( ) ( )36 in. 24 in. 8 in.

220 lb44 in.AE AE AET

− + = =i j k

T λ

( ) ( ) ( )180 lb 120 lb 40 lb= − +i j k

0.960 0.280 0

0 4 8 lb in.180 120 40

DBM−

∴ = − ⋅−

( ) ( )( ) ( )( ) ( ) ( )0.960 4 40 8 120 0.280 8 180 0 = − − − + − −

364.8 lb in.= ⋅

or 365 lb in.DBM = ⋅

PROBLEM 3.54 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable CF at C is 132 lb, determine the moment of that force about the line joining points D and B.

SOLUTION

Have ( )/DB DB C D CFM = r Tλ ⋅ ×

where ( ) ( )48 in. 14 in.0.96 0.28

50 in.DB−

= = −i j

i jλ

( ) ( )/ 8 in. 16 in.C D = −r j k

( ) ( ) ( ) ( )24 in. 36 in. 8 in.132 lb

44 in.CF CF CFT− −

= =i j k

T λ

( ) ( ) ( )72 lb 108 lb 24 lb= − −i j k

0.96 0.28 0

0 8 16 lb in.72 108 24

DBM−

∴ = − ⋅− −

( )( ) ( )( ) ( ) ( )( )0.96 8 24 16 108 0.28 16 72 0 = − − − − + − − −

1520.64 lb in.= − ⋅

or 1521 lb in.DBM = − ⋅

PROBLEM 3.55 A mast is mounted on the roof of a house using bracket ABCD and is guyed by cables EF, EG, and EH. Knowing that the force exerted by cable EF at E is 66 N, determine the moment of that force about the line joining points D and I.

SOLUTION

Have /DI DI F I EFM = r Tλ ⋅ ×

where ( ) ( )( ) ( )

( )2 2

1.6 m 0.4 m 1 4171.6 0.4 m

DI−

= = −+

i ji jλ

( ) ( )/ 4.6 m 0.8 m 5.4 mF I = + =r k k

EF EF EFT=T λ

( ) ( ) ( ) ( )1.2 m 3.6 m 5.4 m66 N

6.6 m− +

=i j k

( ) ( ) ( )12 N 36 N 54 N= − +i j k

( ) ( ) ( )6 2 N 6 N 9 N = − + i j k

( )( )4 1 0

6 N 5.4 m0 0 1

172 6 9

DIM−

∴ =−

( ) ( )7.8582 0 24 2 0 = + + − −

172.879 N m= ⋅

or 172.9 N mDIM = ⋅

PROBLEM 3.56 A mast is mounted on the roof of a house using bracket ABCD and is guyed by cables EF, EG, and EH. Knowing that the force exerted by cable EG at E is 61.5 N, determine the moment of that force about the line joining points D and I.

SOLUTION

Have /DI DI G I EGM = r Tλ ⋅ ×

where ( ) ( )1.6 m 0.4 m0.4 17 mDI

−=

i jλ

( )1 417

= −i j

( ) ( )/ 10.9 m 0.8 m 11.7 mG I = − + = −r k k

EG EG EGT=T λ

( ) ( ) ( ) ( )1.2 m 3.6 m 11.7 m61.5 N

12.3 m− −

=i j k

( ) ( ) ( )5 1.2 N 3.6 N 11.7 N = − − i j k

( )4 1 0

5 N 11.7 m 0 0 1

171.2 3.6 11.7

DIM−

∴ = −− −

( ) ( )( )( ) ( )( )( ){ }14.1883 N m 0 4 1 3.6 1 1 1.2 0 = ⋅ − − − + − − −

187.286 N m= − ⋅

or 187.3 N mDIM = − ⋅

PROBLEM 3.57 A rectangular tetrahedron has six edges of length a. A force P is directed as shown along edge BC. Determine the moment of P about edge OA.

SOLUTION

Have ( )/OA OA C OM = r Pλ ⋅ ×

where

From triangle OBC

( )2xaOA =

( ) ( ) 1tan302 3 2 3z xa aOA OA

= ° = =

Since ( ) ( ) ( ) ( )2 2 2 2zx yOA OA OA OA= + +

or ( )22

22

2 2 3ya aa OA = + +

( )2 2

2 2 4 12 3ya aOA a a∴ = − − =

Then /2

2 3 2 3A Oa aa= + +r i j k

and 1 2 12 3 2 3OA = + +i j kλ

BCP=P λ

( ) ( ) ( )sin30 cos30a aP

a° − °

=i k

( )32P

= −i k

/C O a=r i

PROBLEM 3.57 CONTINUED

( )

1 2 12 3 2 3

1 0 0 21 0 3

OAPM a ∴ =

−

( )( )2 1 32 3

aP = − −

2

aP=

2OAaPM =

PROBLEM 3.58 A rectangular tetrahedron has six edges of length a. (a) Show that two opposite edges, such as OA and BC, are perpendicular to each other. (b) Use this property and the result obtained in Problem 3.57 to determine the perpendicular distance between edges OA and BC.

SOLUTION

(a) For edge OA to be perpendicular to edge BC,

0OA BC =⋅

where

From triangle OBC

( )2xaOA =

( ) ( ) 1tan302 3 2 3z xa aOA OA

= ° = =

( ) 2 2 3ya aOA OA ∴ = + +

i j k

and ( ) ( )sin 30 cos30BC a a= ° − °i k

32 2a a

= −i k

( )32a

= −i k

Then ( ) ( )3 02 22 3ya a aOA

+ + − =

i j k i k⋅

or ( ) ( )2 2

0 04 4ya aOA+ − =

0OA BC∴ =⋅

so that OA is perpendicular to .BC

PROBLEM 3.58 CONTINUED

(b) Have ,OAM Pd= with P acting along BC and d the perpendicular distance from to .OA BC

From the results of Problem 3.57,

2OAPaM =

2

Pa Pd∴ =

or 2

ad =

PROBLEM 3.59 The 8-ft-wide portion ABCD of an inclined, cantilevered walkway is partially supported by members EF and GH. Knowing that the compressive force exerted by member EF on the walkway at F is 5400 lb, determine the moment of that force about edge AD.

SOLUTION

Having ( )/AD AD E A EFM = r Tλ ⋅ ×

where ( ) ( )( ) ( )

( )2 2

24 ft 3 ft 1 86524 3 ft

AD+

= = ++

i ji jλ

( ) ( )/ 7 ft 3 ftE A = −r i j

( ) ( )( ) ( )

( ) ( ) ( )( )

824

2 2 2

8 ft 7 ft 3 ft 3 ft 8 ft5400 lb

1 4 8 ftEF EF EFT

− + + + = =+ +

i j kT λ

( ) ( ) ( )600 1 lb 4 lb 8 lb = + + i j k

( )8 1 0

600 600 7 3 0 lb ft 192 56 lb ft65 65

1 4 8ADM∴ = − ⋅ = − − ⋅

18,456.4 lb ft= − ⋅ or 18.46 kip ftADM = − ⋅

PROBLEM 3.60 The 8-ft-wide portion ABCD of an inclined, cantilevered walkway is partially supported by members EF and GH. Knowing that the compressive force exerted by member GH on the walkway at H is 4800 lb, determine the moment of that force about edge AD.

SOLUTION

Having ( )/AD AD G A GHM = r Tλ ⋅ ×

where ( ) ( )( ) ( )

( )2 2

24 ft 3 ft 1 86524 3 ft

AD+

= = ++

i ji jλ

( ) ( ) ( ) ( )/ 20 ft 6 ft 2 10 ft 3 ftG A = − = − r i j i j

( ) ( )( ) ( )

( ) ( ) ( )( )

1624

2 2 2

16 ft 20 ft 6 ft 3 ft 8 ft4800 lb

4 8 8 ftGH GH GHT

− + + + = =+ +

i j kT λ

( ) ( ) ( )1600 1 lb 2 lb 2 lb = − + + i j k

( )( ) ( )8 1 0

1600 lb 2 ft 3200 lb ft 10 3 0 48 2065 65

1 2 2ADM ⋅

∴ = − = − −−

26,989 lb ft= − ⋅ or 27.0 kip ftADM = − ⋅

PROBLEM 3.61

Two forces F1 and F2 in space have the same magnitude F. Prove that the moment of F1 about the line of action of F2 is equal to the moment of F2 about the line of action of F1.

SOLUTION

First note that 1 1 1F=F λ and 2 2 2F=F λ

Let 1 2 moment of M = F about the line of action of 1M

and 2 1 moment of M = F about the line of action of 2M

Now, by definition

( ) ( )1 1 / 2 1 / 2 2B A B AM F= =r F rλ ⋅ × λ ⋅ × λ

( ) ( )2 2 / 1 2 / 1 1A B A BM F= =r F rλ ⋅ × λ ⋅ × λ

Since 1 2F F F= = and / /A B B A= −r r

( )1 1 / 2B AM F= rλ ⋅ × λ

( )2 2 / 1B AM F= −rλ ⋅ × λ

Using Equation (3.39)

( ) ( )1 / 2 2 / 1B A B A= −r rλ ⋅ × λ λ ⋅ × λ

so that ( )2 1 / 2B AM F= rλ ⋅ × λ

12 21 M M∴ =

PROBLEM 3.62 In Problem 3.53, determine the perpendicular distance between cable AE and the line joining points D and B.

Problem 3.53: The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable AE at A is 220 lb, determine the moment of that force about the line joining points D and B.

SOLUTION

Have ( )/DB DB A D AEM = r Tλ ⋅ ×

where ( ) ( )48 in. 14 in.0.96 0.28

50 in.DB−

= = −i j

i jλ

( ) ( )/ 4 in. 8 in.A D = − +r j k

AE AE AET=T λ

( ) ( ) ( ) ( )36 in. 24 in. 8 in.220 lb

44 in.− +

=i j k

( ) ( ) ( )180 lb 120 lb 40 lb= − +i j k

0.96 0.28 0 0 4 8 lb in.

180 120 40DBM

−∴ = − ⋅

−

364.8 lb in.= ⋅

Only the perpendicular component of AET contributes to the moment of

AET about line DB. The parallel component of AET will be used to find the perpendicular component.

PROBLEM 3.62 CONTINUED

Have

( )parallelAE DB AET = Tλ ⋅

( ) ( ) ( ) ( )0.96 0.28 180 lb 120 lb 40 lb = − − + i j i j k⋅

( )( ) ( )( ) ( )( )0.96 180 0.28 120 0 40 lb = + − − +

( )172.8 33.6 lb= +

206.4 lb=

Since ( ) ( )perpendicular parallelAE AE AE= +T T T

( ) ( ) ( )2 2perpendicular parallel AE AE AET T T∴ = −

( ) ( )2 2220 206.41= −

76.151 lb=

Then ( ) ( )perpendicularDB AEM T d=

( )364.8 lb in. 76.151 lb d⋅ =

4.7905 in.d =

or 4.79 in.d =

PROBLEM 3.63 In Problem 3.54, determine the perpendicular distance between cable CF and the line joining points D and B.

Problem 3.54: The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable CF at C is 132 lb, determine the moment of that force about the line joining points D and B.

SOLUTION

Have ( ) ( )/DB DB C D CFM = r Tλ ⋅ ×

where ( ) ( )48 in. 14 in.50 in.DB−

=i j

λ

0.96 0.28= −i j

( ) ( )/ 8 in. 16 in.C D = −r j k

CF CF CFT=T λ

( ) ( ) ( ) ( )24 in. 36 in. 8 in.132 lb

44 in.− −

=i j k

( ) ( ) ( )72 lb 108 lb 24 lb= − −i j k

0.96 0.28 0 0 8 16 lb in

72 108 24DBM

−∴ = − ⋅

− −

1520.64 lb in.= − ⋅

Only the perpendicular component of CFT contributes to the moment of

CFT about line DB. The parallel component of CFT will be used to obtain the perpendicular component.

PROBLEM 3.63 CONTINUED

Have

( )parallelCF DB CFT = Tλ ⋅

( ) ( ) ( ) ( )0.96 0.28 72 lb 108 lb 24 lb = − − − i j i j k⋅

( )( ) ( )( ) ( )( )0.96 72 0.28 108 0 24 lb = + − − + −

99.36 lb=

Since ( ) ( )perp. parallelCF CF CF= +T T T

( ) ( ) ( )2 2perp. parallel CF CF CFT T T∴ = −

( ) ( )2 2132 99.36= −

86.900 lb=

Then ( ) ( )perp.DB CFM T d=

( )1520.64 lb in. 86.900 lb d− ⋅ =

17.4988 in.d =

or 17.50 in.d =

PROBLEM 3.64 In Problem 3.55, determine the perpendicular distance between cable EF and the line joining points D and I.

Problem 3.55: A mast is mounted on the roof of a house using bracket ABCD and is guyed by cables EF, EG, and EH. Knowing that the force exerted by cable EF at E is 66 N, determine the moment of that force about the line joining points D and I.

SOLUTION

Have ( )/DI DI F I EFM = r Tλ ⋅ ×

where ( ) ( ) ( )1.6 m 0.4 m 1 40.4 17 m 17DI

−= = −

i ji jλ

( )/ 5.4 mF I =r k

( ) ( ) ( ) ( )1.2 m 3.6 m 5.4 m66 N

6.6 mEF EF EFT− +

= =i j k

T λ

( ) ( ) ( )6 2 N 6 N 9 N = − + i j k

( )( )4 1 0

6 N 5.4 m 0 0 1 172.879 N m

172 6 9

DIM−

∴ = = ⋅−

Only the perpendicular component of EFT contributes to the moment of

EFT about line DI. The parallel component of EFT will be used to find the perpendicular component.

Have

( )parallelEF DI EFT = Tλ ⋅

( ) ( ) ( ) ( )1 4 12 N 36 N 54 N17

= − − + i j i j k⋅

( )1 48 36 N17

= +

84 N17

=

PROBLEM 3.64 CONTINUED

Since ( ) ( )perp. parallelEF EF EF= +T T T

( ) ( ) ( )2 2perp. parallel EF EF EFT T T∴ = −

( )2

2 846617

= −

62.777 N=

Then ( ) ( )perp.DI EFM T d=

( )( )172.879 N m 62.777 N d⋅ =

2.7539 md =

or 2.75 md =

PROBLEM 3.65 In Problem 3.56, determine the perpendicular distance between cable EG and the line joining points D and I.

Problem 3.56: A mast is mounted on the roof of a house using bracket ABCD and is guyed by cables EF, EG, and EH. Knowing that the force exerted by cable EG at E is 61.5 N, determine the moment of that force about the line joining points D and I.

SOLUTION

Have /DI DI G I EGM = r Tλ ⋅ ×

where ( ) ( ) ( )1.6 m 0.4 m 1 40.4 17 m 17DI

−= = −

i ji jλ

( ) ( )/ 10.9 m 0.8 m 11.7 mG I = − + = −r k k

( ) ( ) ( ) ( )1.2 m 3.6 m 11.7 m61.5 N

12.3 mEG EG EGT− −

= =i j k

T λ

( ) ( ) ( )5 1.2 N 3.6 N 11.7 N = − − i j k

( )( )4 1 0

5 N 11.7 m 0 0 1

171.2 3.6 11.7

DIM−

∴ = −− −

187.286 N m= − ⋅

Only the perpendicular component of EGT contributes to the moment of

EGT about line DI. The parallel component of EGT will be used to find the perpendicular component.

Have

( )parallelEG DI EGT = Tλ ⋅

( ) ( ) ( ) ( )1 4 5 1.2 N 3.6 N 11.7 N17

= − − − i j i j k⋅

( )5 4.8 3.6 N17

= +

42 N17

=

PROBLEM 3.65 CONTINUED

Since ( ) ( )perp. parallelEF EG EG= +T T T

( ) ( ) ( )2 2perp. parallel EG EG EGT T T∴ = −

( )2

2 4261.517

= −

60.651 N=

Then ( ) ( )perp.DI EGM T d=

( )( )187.286 N m 60.651 N d⋅ =

3.0880 md =

or 3.09 md =

PROBLEM 3.66 In Problem 3.41, determine the perpendicular distance between post BC and the line connecting points O and A.

Problem 3.41: Slider P can move along rod OA. An elastic cord PC is attached to the slider and to the vertical member BC. Knowing that the distance from O to P is 0.12 m and the tension in the cord is 30 N, determine (a) the angle between the elastic cord and the rod OA, (b) the projection on OA of the force exerted by cord PC at point P.

SOLUTION

Assume post BC is represented by a force of magnitude BCF

where BC BCF=F j

Have ( )/OA OA B O BCM = r Fλ ⋅ ×

where ( ) ( ) ( )0.24 m 0.24 m 0.12 m 2 2 10.36 m 3 3 3OA

+ −= = + −

i j ki j kλ

( ) ( )/ 0.18 m 0.24 mB O = +r i k

( )2 2 1

1 0.18 0 0.24 0.48 0.18 0.223 3

0 1 0

BCOA BC BC

FM F F−

∴ = = − − = −

Only the perpendicular component of BCF contributes to the moment of BCF about line OA. The parallel component will be found first so that the perpendicular component of BCF can be determined.

( )parallel2 2 13 3 3OA BC BCBCF F = = + −

F i j k jλ ⋅ ⋅

23 BCF=

Since ( ) ( )parallel perp.BC BC BC= +F F F

( ) ( ) ( ) ( )2

2 2 2perp. parallel

23BC

BC BC BC BCFF F F F = − = −

0.74536 BCF=

Then ( ) ( )perp.OA BCM F d=

( )0.22 0.74536BC BCF F d=

0.29516 md =

or 295 mmd =

PROBLEM 3.67 In Problem 3.45, determine the perpendicular distance between cord DE and the y axis.

Problem 3.45: The 0.732 1.2× -m lid ABCD of a storage bin is hinged along side AB and is held open by looping cord DEC over a frictionless hook at E. If the tension in the cord is 54 N, determine the moment about each of the coordinate axes of the force exerted by the cord at D.

SOLUTION

First note

( ) ( )2 20.732 0.132 mz = −

0.720 m=

Have ( )/y D A DEM = j r T⋅ ×

where ( )/ 0.132 0.720 mD A = +r j k

DE DE DET=T λ

( ) ( ) ( ) ( )0.360 m 0.732 m 0.720 m54 N

1.08 m+ −

=i j k

( ) ( ) ( )18 N 36 N 36 N= + −i j k

0 1 0 0 0.132 0.720 12.96 N m

18 36 36yM∴ = = ⋅

−

Only the perpendicular component of DET contributes to the moment of

DET about the y-axis. The parallel component will be found first so that the perpendicular component of DET can be determined.

( )parallel 36 NDEDET = =j T⋅

PROBLEM 3.67 CONTINUED

Since ( ) ( ) ( )parallel perp.DE DE DE= +T T T

( ) ( ) ( )2 2perp. parallelDE DE DET T T= −

( ) ( )2 254 36 40.249 N= − =

Then ( ) ( )perp.y DEM T d=

( )( )12.96 N m 40.249 N d⋅ =

0.32199 md =

or 322 mmd =

PROBLEM 3.68 A plate in the shape of a parallelogram is acted upon by two couples. Determine (a) the moment of the couple formed by the two 21-N forces, (b) the perpendicular distance between the 12-N forces if the resultant of the two couples is zero, (c) the value of α if the resultant couple is 1.8 N m⋅ clockwise and d is 1.05 m.

SOLUTION

(a) Have 1 1 1M d F=

where 1 0.4 md =

1 21 NF =

( )( )1 0.4 m 21 N 8.4 N mM∴ = = ⋅

1or 8.40 N m= ⋅M

(b) Have 1 2 0+ =M M

or ( )28.40 N m 12 N 0d⋅ − =

2 0.700 md∴ =

(c) Have total 1 2= +M M M

or ( )( )( )1.8 N m 8.40 N m 1.05 m sin 12 Nα⋅ = ⋅ −

sin 0.52381α∴ =

and 31.588α = °

or 31.6α = °

PROBLEM 3.69 A couple M of magnitude 10 lb ft⋅ is applied to the handle of a screwdriver to tighten a screw into a block of wood. Determine the magnitudes of the two smallest horizontal forces that are equivalent to M if they are applied (a) at corners A and D, (b) at corners B and C, (c) anywhere on the block.

SOLUTION

(a) Have M Pd=

or ( ) 1 ft10 lb ft 10 in.12 in.

P ⋅ =

12 lbP∴ = minor 12.00 lbP =

(b) ( ) ( )2 2BCd BE EC= +

( ) ( )2 210 in. 6 in. 11.6619 in.= + =

Have M Pd=

( ) 1 ft10 lb ft 11.6619 in.12 in.

P ⋅ =

10.2899 lbP = or 10.29 lbP =

(c) ( ) ( )2 2ACd AD DC= +

( ) ( )2 210 in. 16 in. 2 89 in.= + =

Have ACM Pd=

( ) 1 ft10 lb ft 2 89 in.12 in.

P ⋅ =

6.3600 lbP = or 6.36 lbP =

PROBLEM 3.70 Two 60-mm-diameter pegs are mounted on a steel plate at A and C, and two rods are attached to the plate at B and D. A cord is passed around the pegs and pulled as shown, while the rods exert on the plate 10-N forces as indicated. (a) Determine the resulting couple acting on the plate when T = 36 N. (b) If only the cord is used, in what direction should it be pulled to create the same couple with the minimum tension in the cord? (c) Determine the value of that minimum tension.

SOLUTION

(a) Have ( )M Fd= Σ

( )( ) ( )( )36 N 0.345 m 10 N 0.380 m= −

8.62 N m= ⋅

8.62 N m= ⋅M

(b)

Have 8.62 N mM Td= = ⋅

For T to be minimum, d must be maximum.

min T∴ must be perpendicular to line AC

0.380 mtan 1.333330.285 m

θ = =

and 53.130θ = °

or 53.1θ = °

(c) Have min maxM T d=

where 8.62 N mM = ⋅

( ) ( ) ( )2 2max 0.380 0.285 2 0.030 m 0.535 md = + + =

( )min 8.62 N m 0.535 mT∴ ⋅ =

min 16.1121 NT =

minor 16.11 NT =

PROBLEM 3.71 The steel plate shown will support six 50-mm-diameter idler rollers mounted on the plate as shown. Two flat belts pass around the rollers, and rollers A and D will be adjusted so that the tension in each belt is 45 N. Determine (a) the resultant couple acting on the plate if a = 0.2 m, (b) the value of a so that the resultant couple acting on the plate is 54 N m⋅ clockwise.

SOLUTION

(a) Note when /0.2 m, C Fa = r is perpendicular to the inclined 45 N forces.

Have

( )M Fd= Σ

( ) ( )45 N 0.2 m 2 0.025 ma = − + +

( ) ( )45 N 2 2 2 0.025 ma − +

For 0.2 m,a =

( )( )45 N 0.450 m 0.61569 mM = − +

47.956 N m= − ⋅

or 48.0 N m= ⋅M

(b) 54.0 N m= ⋅M

Moment of couple due to horizontal forces at and M A D=

Moment of force-couple systems at and about .C F C+

( )54.0 N m 45 N 0.2 m 2 0.025 ma − ⋅ = − + +

( ) ( )0.2 m 2C F x yM M F a F a + + + + +

where ( )( )45 N 0.025 m 1.125 N mCM = − = − ⋅

1.125 N mF CM M= = − ⋅

PROBLEM 3.71 CONTINUED

45 N2xF −

=

45 N2yF −

=

( ) 54.0 N m 45 N 0.25 m 1.125 N m 1.125 N ma∴ − ⋅ = − + − ⋅ − ⋅

( ) ( )45 N 45 N0.2 m 22 2

a a−+ −

0.20 21.20 0.25 0.025 0.0252 2 2

a aa= + + + + + +

3.1213 0.75858a =

0.24303 ma =

or 243 mma =

PROBLEM 3.72 The shafts of an angle drive are acted upon by the two couples shown. Replace the two couples with a single equivalent couple, specifying its magnitude and the direction of its axis.

SOLUTION

Based on 1 2= +M M M

where

( )1 8 N m= − ⋅M j

( )2 6 N m= − ⋅M k

( ) ( ) 8 N m 6 N m∴ = − ⋅ − ⋅M j k

and ( ) ( )2 28 6 10 N m= + = ⋅M

or 10.00 N mM = ⋅

( ) ( )8 N m 6 N m0.8 0.6

10 N m− ⋅ − ⋅

= = = − −⋅

j kM j kM

λ

or ( )( )10 N m 0.8 0.6= = ⋅ − −M M j kλ

cos 0xθ = 90xθ∴ = °

cos 0.8 143.130y yθ θ= − ∴ = °

cos 0.6 126.870z zθ θ= − ∴ = °

or 90.0 , 143.1 , 126.9x y zθ θ θ= ° = ° = °

PROBLEM 3.73 Knowing that P = 0, replace the two remaining couples with a single equivalent couple, specifying its magnitude and the direction of its axis.

SOLUTION

Have 1 2= +M M M

where 1 / 1C B C=M r P×

( ) ( )/ 0.96 m 0.40 mC B = −r i j

( )1 100 NC = −P k

( ) ( )1 0.96 0.40 0 40 N m 96 N m0 0 100

∴ = − = ⋅ + ⋅−

i j kM i j

Also, 2 / 2D A E=M r P×

( ) ( )/ 0.20 m 0.55 mD A = −r j k

2 2E ED EP=P λ

( ) ( )( ) ( )

( )2 2

0.48 m 0.55 m146 N

0.48 0.55 m

− +=

+

i k

( ) ( )96 N 110 N= − +i k

2 0 0.20 0.55 N m96 0 110

∴ = − ⋅−

i j kM

( ) ( ) ( )22.0 N m 52.8 N m 19.2 N m= ⋅ + ⋅ + ⋅i j k

PROBLEM 3.73 CONTINUED

and ( ) ( ) ( )40 N m 96 N m 22.0 N m = ⋅ + ⋅ + ⋅ M i j i

( ) ( )52.8 N m 19.2 N m + ⋅ + ⋅ j k

( ) ( ) ( )62.0 N m 148.8 N m 19.2 N m= ⋅ + ⋅ + ⋅i j k

( ) ( ) ( )2 2 22 2 2 62.0 148.8 19.2x y zM M M= + + = + +M

162.339 N m= ⋅

or 162.3 N mM = ⋅

62.0 148.8 19.2162.339+ +

= =M i j kM

λ

0.38192 0.91660 0.118271= + +i j k

cos 0.38192 67.547x xθ θ= ∴ = °

or 67.5xθ = °

cos 0.91660 23.566y yθ θ= ∴ = °

or 23.6yθ = °

cos 0.118271 83.208z zθ θ= ∴ = °

or 83.2zθ = °

PROBLEM 3.74 Knowing that P = 0, replace the two remaining couples with a single equivalent couple, specifying its magnitude and the direction of its axis.

SOLUTION

Have 4 7= +M M M

where 4 / 4G C G=M r F×

( )/ 10 in.G C = −r i

( )4 4 lbG =F k

( ) ( ) ( )4 10 in. 4 lb 40 lb in.∴ = − = ⋅M i k j×

Also, 7 / 7D F D=M r F×

( ) ( )/ 5 in. 3 in.D F = − +r i j

7 7D ED DF=F λ

( ) ( ) ( )( ) ( ) ( )

( )2 2 2

5 in. 3 in. 7 in.7 lb

5 3 7 in.

− + +=

+ +

i j k

( )7 lb 5 3 783

= − + +i j k

( )77 lb in. 7 lb in. 5 3 0 21 35 0

83 835 3 7

⋅ ⋅∴ = − = + +

−

i j kM i j k

( )0.76835 21 35 lb in.= + ⋅i j

PROBLEM 3.74 CONTINUED

and ( ) ( )40 lb in. 0.76835 21 35 lb in. = ⋅ + + ⋅ M j i j

( ) ( )16.1353 lb in. 66.892 lb in.= ⋅ + ⋅i j

( ) ( ) ( ) ( )22 2 216.1353 66.892x yM M= + = +M

68.811 lb in.= ⋅

or 68.8 lb in.M = ⋅

( ) ( )16.1353 lb in. 66.892 lb in.68.811 lb in.⋅ + ⋅

= =⋅

i jMM

λ

0.23449 0.97212= +i j

cos 0.23449 76.438x xθ θ= ∴ = °

or 76.4xθ = °

cos 0.97212 13.5615y yθ θ= ∴ = °

or 13.56yθ = °

cos 0.0zθ = 90zθ∴ = °

or 90.0zθ = °

PROBLEM 3.75 Knowing that P = 5 lb, replace the three couples with a single equivalent couple, specifying its magnitude and the direction of its axis.

SOLUTION

Have 4 7 5= + +M M M M

where

( )4 / 4 10 0 0 lb in. 40 lb in.0 0 4

G C G= = − ⋅ = ⋅i j k

M r F j×

( )7 / 775 3 0 lb in. 0.76835 21 35 lb in.83

5 3 7D F D

= = − ⋅ = + ⋅

−

i j kM r F i j×

(See Solution to Problem 3.74.)

( ) ( )5 / 5 10 6 7 lb in. 35 lb in. 50 lb in.0 5 0

C A C= = − ⋅ = − ⋅ + ⋅i j k

M r F i k×

( ) ( ) ( ) 16.1353 35 40 26.892 50 lb in. ∴ = − + + + ⋅ M i j k

( ) ( ) ( )18.8647 lb in. 66.892 lb in. 50 lb in.= − ⋅ + ⋅ + ⋅i j k

( ) ( ) ( )2 2 22 2 2 18.8647 66.892 50 85.618 lb in.x y zM M M= + + = + + = ⋅M

or 85.6 lb in.M = ⋅

18.8647 66.892 50 0.22034 0.78129 0.5839985.618

− + += = = − + +

M i j k i j kM

λ

cos 0.22034xθ = − 102.729xθ∴ = ° or 102.7xθ = °

cos 0.78129 38.621y yθ θ= ∴ = ° or 38.6yθ = °

cos 0.58399 54.268z zθ θ= ∴ = ° or 54.3zθ = °

PROBLEM 3.76 Knowing that P = 210 N, replace the three couples with a single equivalent couple, specifying its magnitude and the direction of its axis.

SOLUTION

Have 1 2 P= + +M M M M

where ( ) ( )1 / 1 0.96 0.40 0 40 N m 96 N m0 0 100

C B C= = − = ⋅ + ⋅−

i j kM r P i j×

( ) ( ) ( )2 / 2 0 0.20 0.55 22.0 N m 52.8 N m 19.2 N m96 0 110

D A E= = − = ⋅ + ⋅ + ⋅−

i j kM r P i j k×

(See Solution to Problem 3.73.)

( ) ( )/ 0.48 0.20 1.10 231 N m 100.8 N m0 210 0

P E A E= = − = ⋅ + ⋅i j k

M r P i k×

( ) ( ) ( ) 40 22 231 96 52.8 19.2 100.8 N m ∴ = + + + + + + ⋅ M i j k

( ) ( ) ( )293 N m 148.8 N m 120 N m= ⋅ + ⋅ + ⋅i j k

( ) ( ) ( )2 2 22 2 2 293 148.8 120 349.84 N mx y zM M M= + + = + + = ⋅M

or 350 N mM = ⋅

293 148.8 120 0.83752 0.42533 0.34301349.84

+ += = = + +

M i j k i j kM

λ

cos 0.83752 33.121x xθ θ= ∴ = ° or 33.1xθ = °

cos 0.42533 64.828y yθ θ= ∴ = ° or 64.8yθ = °

cos 0.34301 69.940z zθ θ= ∴ = ° or 69.9zθ = °

PROBLEM 3.77 In a manufacturing operation, three holes are drilled simultaneously in a workpiece. Knowing that the holes are perpendicular to the surfaces of the workpiece, replace the couples applied to the drills with a single equivalent couple, specifying its magnitude and the direction of its axis.

SOLUTION

Have 1 2 3= + +M M M M

where ( )( )1 1.1 lb ft cos 25 sin 25= − ⋅ ° + °M j k

( )2 1.1 lb ft= − ⋅M j

( )( )3 1.3 lb ft cos 20 sin 20= − ⋅ ° − °M j k

( ) ( ) 0.99694 1.1 1.22160 0.46488 0.44463∴ = − − − + − +M j k

( ) ( )3.3185 lb ft 0.020254 lb ft= − ⋅ − ⋅j k

and ( ) ( ) ( )2 2 22 2 2 0 3.3185 0.020254x y zM M M= + + = + +M

3.3186 lb ft= ⋅

or 3.32 lb ftM = ⋅

( )0 3.3185 0.0202543.3186

− −= =

i j kMM

λ

0.99997 0.0061032= − −j k

cos 0xθ = 90xθ∴ = ° or 90.0xθ = °

cos 0.99997yθ = − 179.555yθ∴ = ° or 179.6yθ = °

cos 0.0061032 90.349z zθ θ= − ∴ = ° or 90.3zθ = °

PROBLEM 3.78 The tension in the cable attached to the end C of an adjustable boom ABC is 1000 N. Replace the force exerted by the cable at C with an equivalent force-couple system (a) at A, (b) at B.

SOLUTION

(a) Based on : 1000 NAF F TΣ = =

or 1000 NA =F 20°

( )( ): sin 50A A AM M T dΣ = °

( ) ( )1000 N sin 50 2.25 m= °

1723.60 N m= ⋅

or 1724 N mA = ⋅M

(b) Based on : 1000 NBF F TΣ = =

or 1000 NB =F 20°

( )( ): sin 50B B BM M T dΣ = °

( ) ( )1000 N sin 50 1.25 m= °

957.56 N m= ⋅

or 958 N mB = ⋅M

PROBLEM 3.79 The 20-lb horizontal force P acts on a bell crank as shown. (a) Replace P with an equivalent force-couple system at B. (b) Find the two vertical forces at C and D which are equivalent to the couple found in part a.

SOLUTION

(a) Based on : 20 lbBF P PΣ = =

or 20 lbB =P

: B BM M PdΣ =

( )20 lb 5 in.=

100 lb in.= ⋅

or 100 lb in.B = ⋅M

(b) If the two vertical forces are to be equivalent to ,BM they must be a couple. Further, the sense of the moment of this couple must be counterclockwise.

Then, with CP and DP acting as shown,

: D CM M P dΣ =

( )100 lb in. 4 in.CP⋅ =

25 lbCP∴ =

or 25 lbC =P

: 0y D CF P PΣ = −

25 lbDP∴ =

or 25 lbD =P

PROBLEM 3.80 A 700-N force P is applied at point A of a structural member. Replace P with (a) an equivalent force-couple system at C, (b) an equivalent system consisting of a vertical force at B and a second force at D.

SOLUTION

(a) Based on : 700 NCF P PΣ = =

or 700 NC =P 60°

:C C x Cy y CxM M P d P dΣ = − +

where ( )700 N cos60 350 NxP = ° =

( )700 N sin 60 606.22 NyP = ° =

1.6 mCxd =

1.1 mCyd =

( )( ) ( )( ) 350 N 1.1 m 606.22 N 1.6 mCM∴ = − +

385 N m 969.95 N m= − ⋅ + ⋅

584.95 N m= ⋅

or 585 N mC = ⋅M

(b) Based on : cos60x DxF P PΣ = °

( )700 N cos60= °

350 N=

( )( ) ( ): cos60D DA B DBM P d P dΣ ° =

( ) ( ) ( )700 N cos60 0.6 m 2.4 mBP ° =

87.5 NBP =

or 87.5 NB =P

PROBLEM 3.80 CONTINUED

: sin 60y B DyF P P PΣ ° = +

( )700 N sin 60 87.5 N DyP° = +

518.72 NDyP =

( ) ( )22D Dx DyP P P= +

( ) ( )2 2350 518.72 625.76 N= + =

1 1 518.72tan tan 55.991350

Dy

Dx

PP

θ − − = = = °

or 626 NDP = 56.0°

PROBLEM 3.81 A landscaper tries to plumb a tree by applying a 240-N force as shown. Two helpers then attempt to plumb the same tree, with one pulling at B and the other pushing with a parallel force at C. Determine these two forces so that they are equivalent to the single 240-N force shown in the figure.

SOLUTION

Based on

( ): 240 N cos30 cos cosx B CF F Fα αΣ − ° = − −

or ( ) ( )cos 240 N cos30B CF F α− + = − ° (1)

( ): 240 N sin30 sin siny B CF F Fα αΣ ° = +

or ( ) ( )sin 240 N sin 30B CF F α+ = ° (2)

From

( )Equation (2) : tan tan 30Equation 1

α = °

30α∴ = °

Based on

( ) ( ) ( ) ( )( ): 240 N cos 30 20 0.25 m cos10 0.60 mC BM F Σ ° − ° = °

100 NBF∴ =

or 100.0 NB =F 30°

From Equation (1), ( )100 N cos30 240cos30CF− + ° = − °

140 NCF =

or 140.0 NC =F 30°

PROBLEM 3.82 A landscaper tries to plumb a tree by applying a 240-N force as shown. (a) Replace that force with an equivalent force-couple system at C. (b) Two helpers attempt to plumb the same tree, with one applying a horizontal force at C and the other pulling at B. Determine these two forces if they are to be equivalent to the single force of part a.

SOLUTION

(a) Based on ( ): 240 N cos30 cos30x CF FΣ − ° = − °

240 NCF∴ =

or 240 NC =F 30°

( ) ( ): 240 N cos10 0.25 mC A C AM d M d Σ ° = =

59.088 N mCM∴ = ⋅

or 59.1 N mC = ⋅M

(b) Based on ( ): 240 N sin30 siny BF F αΣ ° =

or sin 120BF α = (1)

( ) ( ) ( ): 59.088 N m 240 N cos10 cos 20B C C CM d F d Σ ⋅ − ° = − °

( ) ( ) ( )59.088 N m 240 N cos10 0.60 m 0.60 m cos 20CF ⋅ − ° = − °

0.56382 82.724CF =

146.722 NCF =

or 146.7 NC =F

and ( ): 240 N cos30 146.722 N cosx BF F αΣ − ° = − −

cos 61.124BF α = (2)

From

Equation (1) :Equation (2)

120tan 1.9632361.124

α = =

63.007α = ° or 63.0α = °

From Equation (1), 120 134.670 Nsin 63.007BF = =

°

or 134.7 NB =F 63.0°

PROBLEM 3.83 A dirigible is tethered by a cable attached to its cabin at B. If the tension in the cable is 250 lb, replace the force exerted by the cable at B with an equivalent system formed by two parallel forces applied at A and C.

SOLUTION

Require the equivalent forces acting at A and C be parallel and at an angle ofα with the vertical.

Then for equivalence,

( ): 250 lb sin 30 sin sinx A BF F Fα αΣ ° = + (1)

( ): 250 lb cos30 cos cosy A BF F Fα αΣ − ° = − − (2)

Dividing Equation (1) by Equation (2),

( )( )

( )( )

250 lb sin 30 sin250 lb cos30 cos

A B

A B

F FF F

αα

° +=

− ° − +

Simplifying yields 30α = °

Based on

( ) ( ) ( )( ): 250 lb cos30 12 ft cos30 32 ftC AM F Σ ° = °

93.75 lbAF∴ =

or 93.8 lbA =F 60°

Based on

( ) ( ) ( )( ): 250 lb cos30 20 ft cos30 32 ftA CM F Σ − ° = °

156.25 lbCF∴ =

or 156.3 lbC =F 60°

PROBLEM 3.84 Three workers trying to move a 3 3 4-ft× × crate apply to the crate the three horizontal forces shown. (a) If 60 lb,P = replace the three forces with an equivalent force-couple system at A. (b) Replace the force-couple system of part a with a single force, and determine where it should be applied to side AB. (c) Determine the magnitude of P so that the three forces can be replaced with a single equivalent force applied at B.

SOLUTION

(a)

(b)

(c)

(a) Based on

: 50 lb 50 lb 60 lbz AF FΣ − + + =

60 lbAF =

( )or 60.0 lbA =F k

Based on

( )( ) ( )( ): 50 lb 2 ft 50 lb 0.6 ftA AM MΣ − =

70 lb ftAM = ⋅

( )or 70.0 lb ftA = ⋅M j

(b) Based on

: 50 lb 50 lb 60 lbzF FΣ − + + =

60 lbF =

( )or 60.0 lb=F k

Based on

( ): 70 lb ft 60 lbAM xΣ ⋅ =

1.16667 ftx =

or 1.167 ft from along x A AB=

(c) Based on

( ) ( ) ( ) ( ) ( ) ( ): 50 lb 1 ft 50 lb 2.4 ft 3 ft 0BM P RΣ − + − =

70 23.333 lb3

P = =

or 23.3 lbP =

PROBLEM 3.85 A force and a couple are applied to a beam. (a) Replace this system with a single force F applied at point G, and determine the distance d. (b) Solve part a assuming that the directions of the two 600-N forces are reversed.

SOLUTION (a)

Have :y C D EF F F F FΣ + + =

800 N 600 N 600 NF = − + −

800 NF = − or 800 N=F

Have ( ) ( ): 1.5 m 2 m 0G C DM F d FΣ − − =

( )( ) ( )( )800 N 1.5 m 600 N 2 m 0d − − =

1200 1200800

d +=

3 md = or 3.00 md =

(b)

Changing directions of the two 600 N forces only changes sign of the couple.

800 NF∴ = − or 800 N=F

and ( ) ( ): 1.5 m 2 m 0G C DM F d FΣ − + =

( )( ) ( )( )800 N 1.5 m 600 N 2 md − +

1200 1200 0800

d −= =

or 0d =

PROBLEM 3.86 Three cables attached to a disk exert on it the forces shown. (a) Replace the three forces with an equivalent force-couple system at A. (b) Determine the single force which is equivalent to the force-couple system obtained in part a, and specify its point of application on a line drawn through points A and D.

SOLUTION

(a) Have : B C D AΣ + + =F F F F F

Since B D= −F F

110 NA C∴ = =F F 20°

or 110.0 NA =F 20.0°

Have ( ) ( ) ( ):A BT CT DT AM F r F r F r MΣ − − + =

( ) ( ) ( ) ( ) ( ) ( )140 N sin15 0.2 m 110 N sin 25 0.2 m 140 N sin 45 0.2 m AM − ° − ° + ° =

3.2545 N mAM = ⋅

or 3.25 N mA = ⋅M

(b) Have : A EΣ =F F F or 110.0 NE =F 20.0°

[ ]( ): cos 20A EM M F aΣ = °

( ) ( ) 3.2545 N m 110 N cos 20 a ∴ ⋅ = °

0.031485 ma =

or 31.5 mm below a A=

PROBLEM 3.87 While tapping a hole, a machinist applies the horizontal forces shown to the handle of the tap wrench. Show that these forces are equivalent to a single force, and specify, if possible, the point of application of the single force on the handle.

SOLUTION

Since the forces at A and B are parallel, the force at B can be replaced with the sum of two forces with one of the forces equal in magnitude to the force at A except with an opposite sense, resulting in a force-couple.

Have 26.5 N 2.5 N,BF = + where the 26.5 N force be part of the couple. Combining the two parallel forces,

( ) ( )couple 26.5 N 0.080 m 0.070 m cos 25M = + °

3.60 N m= ⋅

coupleand, 3.60 N m= ⋅M

A single equivalent force will be located in the negative z-direction.

Based on ( ) ( ): 3.60 N m 2.5 N cos 25BM a Σ − ⋅ = °

1.590 ma = −

( )( )2.5 N cos 25 sin 25′ = ° + °F i j

and is applied on an extension of handle BD at a distance of 1.590 m to the right of B

PROBLEM 3.88 A rectangular plate is acted upon by the force and couple shown. This system is to be replaced with a single equivalent force. (a) For 40 ,α = ° specify the magnitude and the line of action of the equivalent force. (b) Specify the value of α if the line of action of the equivalent force is to intersect line CD 12 in. to the right of D.

SOLUTION

(a) Have ( ) ( ): 3 lb sin 40 3 lb sin 40x xF FΣ − ° + ° =

0xF∴ =

Have ( ) ( ): 3 lb cos40 10 lb 3 lb cos40y yF FΣ − ° − + ° =

10 lbyF∴ = −

or 10.00 lbF =

Note: The two 3-lb forces form a couple

and / / /:A C A C B A B X AΣ + =M r P r P r F× × ×

3 16 10 0 160 1 0 0 10 0 0sin 40 cos 40 0 0 1 0 0 1 0

d− + =° ° − −

i j k i j k i j k

( ) ( ): 3 16 cos 40 10 3sin 40 160 10d° − − ° − = −k

36.770 19.2836 160 10d+ − = −

10.3946 in.d∴ =

or 10.00 lb=F at 10.39 in. right of A or at 5.61 in. left of B

(b) From part (a), 10.00 lb=F

Have ( )/ /: 12 in.A C A C B A BΣ + =M r P r P i F× × ×

3 16 10 0 160 1 0 0 120 1 0 0sin cos 0 0 1 0 0 1 0α α

− + =− −

i j k i j k i j k

: 48cos 30sin 160 120α α+ − = −k

24cos 20 15sinα α= −

PROBLEM 3.88 CONTINUED

Squaring both sides of the equation, and

using the identity 2 2cos 1 sin ,α α= − results in

2sin 0.74906sin 0.21973 0α α− − =

Using quadratic formula

sin 0.97453 sin = 0.22547α α= −

so that

77.0 and 13.03α α= ° = − °

PROBLEM 3.89 A hexagonal plate is acted upon by the force P and the couple shown. Determine the magnitude and the direction of the smallest force P for which this system can be replaced with a single force at E.

SOLUTION

Since the minimum value of P acting at B is realized when minP is perpendicular to a line connecting B and E, 30α = °

Then,

/ min /: 0E B E D A DΣ + =M r P r P× ×

where

( ) ( )/ 0.30 m 2 0.30 m cos30B E = − + ° r i j

( ) ( )0.30 m 0.51962 m= − +i j

( )/ 0.30 m 2 0.3 m sin 30D A = + ° r i

( )0.60 m= i

( )450 ND =P j

( ) ( )min min cos30 sin 30P = ° + ° P i j

min 0.30 0.51962 0 0.60 0 0 N m 00.86603 0.50 0 0 450 0

P∴ − + ⋅ =i j k i j k

( ) ( )min 0.15 m 0.45 m 270 N m 0P − − + ⋅ =k k

min 450 NP∴ =

minor 450 N=P 30°

PROBLEM 3.90 An eccentric, compressive 270-lb force P is applied to the end of a cantilever beam. Replace P with an equivalent force-couple system at G.

SOLUTION

Have

( ): 270 lbΣ − =F i F

( ) 270 lb∴ = −F i

Also, have

/:G A GΣ =M r P M×

270 0 4 2.4 lb in.1 0 0

− − ⋅ =−

i j kM

( ) ( )( ) ( )( ) 270 lb in. 2.4 1 4 1 ∴ = ⋅ − − − − − M j k

( ) ( )or 648 lb in. 1080 lb in.= ⋅ − ⋅M j k

PROBLEM 3.91 Two workers use blocks and tackles attached to the bottom of an I-beam to lift a large cylindrical tank. Knowing that the tension in rope AB is 324 N, replace the force exerted at A by rope AB with an equivalent force-couple system at E.

SOLUTION

Have : ABΣ =F T F

where

AB AB ABT=T λ

( ) ( ) ( ) ( )0.75 m 6.0 m 3.0 m324 N

6.75 m− +

=i j k

( ) 36 N 8 4AB∴ = − +T i j k

so that ( ) ( ) ( )36.0 N 288 N 144.0 N= − +F i j k

Have

/:E A E ABΣ =M r T M×

or ( )( )7.5 m 36 N 0 1 01 8 4

=−

i j kM

( )( ) 270 N m 4∴ = ⋅ −M i k

( ) ( )or 1080 N m 270 N m= ⋅ − ⋅M i k

PROBLEM 3.92 Two workers use blocks and tackles attached to the bottom of an I-beam to lift a large cylindrical tank. Knowing that the tension in rope CD is 366 N, replace the force exerted at C by rope CD with an equivalent force-couple system at O.

SOLUTION

Have : CDΣ =F T F

where

CD CD CDT=T λ

( ) ( ) ( ) ( )0.3 m 5.6 m 2.4 m366 N

6.1 m− − +

=i j k

( )( ) 6.0 N 3 56 24CD∴ = − − +T i j k

so that ( ) ( ) ( )18.00 N 336 N 144.0 N= − − +F i j k

Have

/:O C O CDΣ =M r T M×

or ( )( )7.5 m 6 N 0 1 03 56 24

=− −

i j kM

( )( ) 45 N m 24 3∴ = ⋅ +M i k

( ) ( )or 1080 N m 135.0 N m= ⋅ + ⋅M i k

PROBLEM 3.93 To keep a door closed, a wooden stick is wedged between the floor and the doorknob. The stick exerts at B a 45-lb force directed along line AB. Replace that force with an equivalent force-couple system at C.

SOLUTION

Have

: AB CΣ =F P F

where

AB AB ABP=P λ

( ) ( ) ( ) ( )2.0 in. 38 in. 24 in.45 lb

44.989 in.+ −

=i j k

( ) ( ) ( )or 2.00 lb 38.0 lb 24.0 lbC = + −F i j k

Have

/:C B C AB CΣ =M r P M×

2 29.5 33 0 lb in.1 19 12

C = − ⋅−

i j kM

( ) ( )( ) ( )( ){2 lb in. 33 12 29.5 12= ⋅ − − − −i j

( )( ) ( )( ) }29.5 19 33 1 + − − k

( ) ( ) ( )or 792 lb in. 708 lb in. 1187 lb in.C = ⋅ + ⋅ + ⋅M i j k

PROBLEM 3.94 A 25-lb force acting in a vertical plane parallel to the yz plane is applied to the 8-in.-long horizontal handle AB of a socket wrench. Replace the force with an equivalent force-couple system at the origin O of the coordinate system.

SOLUTION

Have

: BΣ =F P F

where

( ) ( )25 lb sin 20 cos20B = − ° + ° P j k

( ) ( )8.5505 lb 23.492 lb= − +j k

( ) ( )or 8.55 lb 23.5 lb= − +F j k

Have

/:O B O B OΣ =M r P M×

where

( ) ( ) ( )/ 8cos30 15 8sin 30 in.B O = ° + − ° r i j k

( ) ( ) ( )6.9282 in. 15 in. 4 in.= + −i j k

6.9282 15 4 lb in.0 8.5505 23.492

O∴ − ⋅ =−

i j kM

( ) ( ) ( )318.18 162.757 59.240 lb in.O = − − ⋅ M i j k

( ) ( ) ( )or 318 lb in. 162.8 lb in. 59.2 lb in.O = ⋅ − ⋅ − ⋅M i j k

PROBLEM 3.95 A 315-N force F and 70-N · m couple M are applied to corner A of the block shown. Replace the given force-couple system with an equivalent force-couple system at corner D.

SOLUTION

Have

: DΣ =F F F

AI F= λ

( ) ( ) ( ) ( )0.360 m 0.120 m 0.180 m315 N

0.420 m− +

=i j k

( )( )750 N 0.360 0.120 0.180= − +i j k

( ) ( ) ( )or 270 N 90.0 N 135.0 ND = − +F i j k

Have

/:D I D DΣ + =M M r F M×

where

ACM=M λ

( ) ( ) ( )0.240 m 0.180 m70.0 N m

0.300 m−

= ⋅i k

( )( )70.0 N m 0.800 0.600= ⋅ −i k

( )/ 0.360 mI D =r k

( )( ) ( ) 70.0 N m 0.8 0.6 0 0 0.36 750 N m0.36 0.12 0.18

D∴ = ⋅ − + ⋅−

i j kM i k

( ) ( )56.0 N m 42.0 N m= ⋅ − ⋅i k ( ) ( )32.4 N m 97.2 N m + ⋅ + ⋅ i j

( ) ( ) ( )or 88.4 N m 97.2 N m 42.0 N mD = ⋅ + ⋅ − ⋅M i j k

PROBLEM 3.96 The handpiece of a miniature industrial grinder weighs 2.4 N, and its center of gravity is located on the y axis. The head of the handpiece is offset in the xz plane in such a way that line BC forms an angle of 25° with the x direction. Show that the weight of the handpiece and the two couples 1M and 2M can be replaced with a single equivalent force. Further assuming that 1 0.068 N mM = ⋅ and 2 0.065 N m,M = ⋅ determine (a) the magnitude and the direction of the equivalent force, (b) the point where its line of action intersects the xz plane.

SOLUTION

First assume that the given force W and couples 1M and 2M act at the origin.

Now W= − jW

and ( ) ( )1 2 2 1 2cos 25 sin 25M M M= + = − ° + − °M M M i k

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

(a) Have ( ) or 2.4 NF W= = − = −jW F j

( )or 2.40 N= −F j

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

Then for equivalence

/P O= ×M r F

where /P O x z= +r i k

( ) ( )2 1 2 cos25 sin 25M M M∴ − ° + − °i k

( ) ( )00 0x z Wz Wx

W= = −

−

i j ki k

PROBLEM 3.96 CONTINUED

Equating the i and k coefficients,

1 2cos25 sin 25 and zM M Mz xW W

− ° − ° = = −

(b) For 1 22.4 N, 0.068 N m, 0.065 N mW M M= = ⋅ = ⋅

0.068 0.065sin 25 0.0168874 m2.4

x − °= = −

−

or 16.89 mmx = −

0.065cos25 0.024546 m2.4

z − °= = −

or 24.5 mmz = −

PROBLEM 3.97 A 20-lb force 1F and a 40- lb ft⋅ couple 1M are applied to corner E of the bent plate shown. If 1F and 1M are to be replaced with an equivalent force-couple system ( )2 2, F M at corner B and if ( )2 0,zM = determine

(a) the distance d, (b) 2F and 2.M

SOLUTION

(a) Have 2: 0Bz zM MΣ =

( )/ 1 1 0H B zM+ =k r F⋅ × (1)

where ( ) ( )/ 31 in. 2 in.H B = −r i j

1 1EH F=F λ

( ) ( ) ( ) ( )6 in. 6 in. 7 in.20 lb

11.0 in.+ −

=i j k

( )20 lb 6 6 711.0

= + −i j k

1 1zM = k M⋅

1 1EJ M=M λ

( ) ( ) ( )2

3 in. 7 in.480 lb in.

58 in.

d

d

− + −= ⋅

+

i j k

Then from Equation (1),

( )( )2

0 0 17 480 lb in.20 lb in.31 2 0 0

11.0 586 6 7 d

− ⋅⋅− + =

+−

PROBLEM 3.97 CONTINUED

Solving for d, Equation (1) reduces to

( )2

20 lb in. 3360 lb in.186 12 011.0 58d

⋅ ⋅+ − =

+

From which 5.3955 in.d =

or 5.40 in.d =

(b) ( )2 120 lb 6 6 711.0

= = + −F F i j k

( )10.9091 10.9091 12.7273 lb= + −i j k

( ) ( ) ( )2or 10.91 lb 10.91 lb 12.73 lb= + −F i j k

2 / 1 1H B= +M r F M×

( ) ( )5.3955 3 720 lb in.31 2 0 480 lb in.11.0 9.3333

6 6 7

− + −⋅= − + ⋅

−

i j ki j k

( )25.455 394.55 360 lb in.= + + ⋅i j k

( )277.48 154.285 360 lb in.+ − + − ⋅i j k

( ) ( )2 252.03 lb in. 548.84 lb in.= − ⋅ + ⋅M i j

( ) ( )2or 21.0 lb ft 45.7 lb ft= − ⋅ + ⋅M i j

PROBLEM 3.98 A 4-ft-long beam is subjected to a variety of loadings. (a) Replace each loading with an equivalent force-couple system at end A of the beam. (b) Which of the loadings are equivalent?

SOLUTION

(a)

(a) Have : 200 lb 100 lby aF RΣ − − =

or 300 lba =R

and ( )( ): 900 lb ft 100 lb 4 ftA aM MΣ ⋅ − =

or 500 lb fta = ⋅M

(b) Have : 300 lby bF RΣ − =

or 300 lbb =R

and : 450 lb ftA bM MΣ − ⋅ =

or 450 lb ftb = ⋅M

(c) Have : 150 lb 450 lby cF RΣ − =

or 300 lbc =R

and ( )( ): 2250 lb ft 450 lb 4 ftA cM MΣ ⋅ − =

or 450 lb ftc = ⋅M

(d) Have : 200 lb 400 lby dF RΣ − + =

or 200 lbd =R

and ( )( ): 400 lb 4 ft 1150 lb ftA dM MΣ − ⋅ =

or 450 lb ftd = ⋅M

(e) Have : 200 lb 100 lby eF RΣ − − =

or 300 lbe =R

and ( )( ): 100 lb ft 200 lb ft 100 lb 4 ftA eM MΣ ⋅ + ⋅ − =

or 100 lb fte = ⋅M

(b)

PROBLEM 3.98 CONTINUED

(f) Have : 400 lb 100 lby fF RΣ − + =

or 300 lbf =R

and ( )( ): 150 lb ft 150 lb ft 100 lb 4 ftA fM MΣ − ⋅ + ⋅ + =

or 400 lb ftf = ⋅M

(g) Have : 100 lb 400 lby gF RΣ − − =

or 500 lbg =R

and ( )( ): 100 lb ft 2000 lb ft 400 lb 4 ftA gM MΣ ⋅ + ⋅ − =

or 500 lb ftg = ⋅M

(h) Have : 150 lb 150 lby hF RΣ − − =

or 300 lbh =R

and ( )( ): 1200 lb ft 150 lb ft 150 lb 4 ftA hM MΣ ⋅ − ⋅ − =

or 450 lb fth = ⋅M

Therefore, loadings (c) and (h) are equivalent

PROBLEM 3.99 A 4-ft-long beam is loaded as shown. Determine the loading of Problem 3.98 which is equivalent to this loading.

SOLUTION

Have : 100 lb 200 lby RΣ − − =F

or 300 lb=R

and ( )( ): 200 lb ft 1400 lb ft 200 lb 4 ftAM MΣ − ⋅ + ⋅ − =

or 400 lb ft= ⋅M

Equivalent to case (f) of Problem 3.98

Problem 3.98 Equivalent force-couples at A

case R M

( )a 300 lb 500 lb ft⋅

( )b 300 lb 450 lb ft⋅

( )c 300 lb 450 lb ft⋅

( )d 200 lb 450 lb ft⋅

( )e 300 lb 100 lb ft⋅

( )f 300 lb 400 lb ft⋅

( )g 500 lb 500 lb ft⋅

( )h 300 lb 450 lb ft⋅

PROBLEM 3.100 Determine the single equivalent force and the distance from point A to its line of action for the beam and loading of (a) Problem 3.98b, (b) Problem 3.98d, (c) Problem 3.98e.

Problem 3.98: A 4-ft-long beam is subjected to a variety of loadings. (a) Replace each loading with an equivalent force-couple system at end A of the beam. (b) Which of the loadings are equivalent?

SOLUTION

(a)

(b)

(c)

For equivalent single force at distance d from A

Have : 300 lbyF RΣ − =

or 300 lb=R

and ( )( ): 300 lb 450 lb ft 0CM dΣ − ⋅ =

or 1.500 ftd =

Have : 200 lb 400 lbyF RΣ − + =

or 200 lb=R

and ( )( ) ( )( ): 200 lb 400 lb 4 1150 lb ft 0CM d dΣ + − − ⋅ =

or 2.25 ftd =

Have : 200 lb 100 lbyF RΣ − − =

or 300 lb=R

and ( )( ) ( )( ): 100 lb ft 200 lb 100 lb 4 200 lb ft 0CM d dΣ ⋅ + − − + ⋅ =

or 0.333 ftd =

PROBLEM 3.101 Five separate force-couple systems act at the corners of a metal block, which has been machined into the shape shown. Determine which of these systems is equivalent to a force ( )10 N=F j and a couple of

moment ( ) ( )6 N m 4 N m= ⋅ + ⋅M i k located at point A.

SOLUTION

The equivalent force-couple system at A for each of the five force-couple systems will be determined. Each will then be compared to the given force-couple system to determine if they are equivalent.

Force-couple system at B

Have ( ): 10 NΣ =F j F

or ( )10 N=F j

and ( )/:A B B AΣ Σ + =M M r F M×

( ) ( ) ( ) ( )4 N m 2 N m 0.2 m 10 N⋅ + ⋅ + =i k i j M×

( ) ( )4 N m 4 N m= ⋅ + ⋅M i k

Comparing to given force-couple system at A, Is Not Equivalent

Force-couple system at C

Have ( ): 10 NΣ =F j F

or ( )10 N=F j

and ( )/:A C C AΣ + =M M r F M×

( ) ( ) ( ) ( )8.5 N m 0.2 m 0.25 m 10 N ⋅ + + = i i k j M×

( ) ( )6 N m 2.0 N m= ⋅ + ⋅M i k

Comparing to given force-couple system at A, Is Not Equivalent

PROBLEM 3.101 CONTINUED

Force-couple system at E

Have ( ): 10 NΣ =F j F

or ( )10 N=F j

and ( )/:A E E AMΣ + =M r F M×

( ) ( ) ( ) ( )6 N m 0.4 m 0.08 m 10 N ⋅ + − = i i j j M×

( ) ( )6 N m 4 N m= ⋅ + ⋅M i k

Comparing to given force-couple system at A, Is Equivalent

Force-couple system at G

Have ( ) ( ): 10 N 10 NΣ + =F i j F

or ( ) ( )10 N 10 N= +F i j

F has two force components

force-couple system at G∴ Is Not Equivalent

Force-couple system at I

Have ( ): 10 NΣ =F j F

or ( )10 N=F j

and ( )/:A I I AΣ Σ + =M M r F M×

( ) ( )10 N m 2 N m⋅ − ⋅i k

( ) ( ) ( ) ( )0.4 m 0.2 m 0.4 m 10 N + − + = i j k j M×

or ( ) ( )6 N m 2 N m= ⋅ + ⋅M i k

Comparing to given force-couple system at A, Is Not Equivalent

PROBLEM 3.102 The masses of two children sitting at ends A and B of a seesaw are 38 kg and 29 kg, respectively. Where should a third child sit so that the resultant of the weights of the three children will pass through C if she has a mass of (a) 27 kg, (b) 24 kg.

SOLUTION

First ( )38 kgA AW m g g= =

( )29 kgB BW m g g= =

(a) ( )27 kgC CW m g g= =

For resultant weight to act at C, 0CMΣ =

Then ( ) ( ) ( ) ( ) ( ) ( )38 kg 2 m 27 kg 29 kg 2 m 0g g d g − − =

76 58 0.66667 m27

d −∴ = =

or 0.667 md =

(b) ( )24 kgC CW m g g= =

For resultant weight to act at C, 0CMΣ =

Then ( ) ( ) ( ) ( ) ( ) ( )38 kg 2 m 24 kg 29 kg 2 m 0g g d g − − =

76 58 0.75 m24

d −∴ = =

or 0.750 md =

PROBLEM 3.103 Three stage lights are mounted on a pipe as shown. The mass of each light is 1.8 kgA Bm m= = and 1.6 kgCm = . (a) If 0.75d = m, determine the distance from D to the line of action of the resultant of the weights of the three lights. (b) Determine the value of d so that the resultant of the weights passes through the midpoint of the pipe.

SOLUTION

First ( )1.8 kgA B AW W m g g= = =

( )1.6 kgC CW m g g= =

(a) 0.75 md =

Have A B CR W W W= + +

( )1.8 1.8 1.6 kgR g = + +

or ( )5.2 Ng=R

Have

( ) ( ) ( ) ( ): 1.8 0.3 m 1.8 1.3 m 1.6 2.05 m 5.2DM g g g g DΣ − − − = −

1.18462 mD∴ =

or 1.185 mD =

(b) 1.25 m2LD = =

Have

( )( ) ( )( ) ( )( ): 1.8 0.3 m 1.8 1.3 m 1.6 1.3 mDM g g g dΣ − − − +

( )( )5.2 1.25 mg= −

0.9625 md∴ =

or 0.963 md =

PROBLEM 3.104 Three hikers are shown crossing a footbridge. Knowing that the weights of the hikers at points C, D, and E are 800 N, 700 N, and 540 N, respectively, determine (a) the horizontal distance from A to the line of action of the resultant of the three weights when 1.1 m,a = (b) the value of a so that the loads on the bridge supports at A and B are equal.

SOLUTION

(a)

(b)

(a) 1.1 ma =

Have : C D EF W W W RΣ − − − =

800 N 700 N 540 NR∴ = − − −

2040 NR =

or 2040 N=R

Have

( )( ) ( )( ) ( )( ): 800 N 1.5 m 700 N 2.6 m 540 N 4.25 mAMΣ − − −

( )R d= −

( ) 5315 N m 2040 N d∴ − ⋅ = −

and 2.6054 md =

or 2.61 m to the right of d A=

(b) For equal reaction forces at A and B, the resultant, ,R must act at the center of the span.

From 2ALM R Σ = −

( )( ) ( )( ) ( )( ) 800 N 1.5 m 700 N 1.5 m 540 N 1.5 m 2.5a a∴ − − + − +

( )( )2040 N 3 m= −

3060 2050 6120a+ =

and 1.49268 ma =

or 1.493 ma =

PROBLEM 3.105 Gear C is rigidly attached to arm AB. If the forces and couple shown can be reduced to a single equivalent force at A, determine the equivalent force and the magnitude of the couple M.

SOLUTION

For equivalence

( ) ( ): 90 N sin30 125 N cos40x xF RΣ − ° + ° =

or 50.756 NxR =

( ) ( ): 90 N cos30 200 N 125 N sin 40y yF RΣ − ° − − ° =

or 358.29 NyR = −

Then ( ) ( )2 250.756 358.29 361.87 NR = + − =

and 358.29tan 7.0591 81.93750.756

y

x

RR

θ θ−= = = − ∴ = − °

or 362 N=R 81.9°

Also

( ) ( ) ( ) ( ) ( ) ( ): 90 N sin 35 0.6 m 200 N cos 25 0.85 m 125 N sin 65 1.25 m 0AM M Σ − ° − ° − ° =

326.66 N mM∴ = ⋅

or 327 N mM = ⋅

PROBLEM 3.106 To test the strength of a 25 20-in.× suitcase, forces are applied as shown. If 18 lb,P = (a) determine the resultant of the applied forces, (b) locate the two points where the line of action of the resultant intersects the edge of the suitcase.

SOLUTION

(a) 18 lbP =

Have ( ) ( ) ( ) ( )42 lb: 20 lb + 3 2 18 lb 36 lb13 x yR RΣ − − + + + = +F i i j j i i j

( ) ( )18.9461 lb 41.297 lb x yR R∴ − + = +i j i j

or ( ) ( )18.95 lb 41.3 lb= − +R i j

( ) ( )2 22 2 18.9461 41.297 45.436 lbx yR R R= + = + =

1 1 41.297tan tan 65.35518.9461

yx

x

RR

θ − − = = = − ° −

or 45.4 lb=R 65.4°

(b) Have B BΣ =M M

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )42 lb4 in. 20 lb 21 in. 3 2 12 in. 36 lb 3 in. 18 lb13B

= − + − + + +

M j i i i j j i i j× × × ×

( ) 191.246 lb in.B∴ = ⋅M k

PROBLEM 3.106 CONTINUED

Since B B=M r R×

( ) ( )191.246 lb in. 0 41.297 18.946118.9461 41.297 0

x y x y∴ ⋅ = = +−

i j kk k

For 191.2460, 4.6310 in.41.297

y x= = = or 4.63 in.x =

For 191.2460, 10.0942 in.18.9461

x y= = = or 10.09 in.y =

PROBLEM 3.107 Solve Problem 3.106 assuming that 28 lb.P =

Problem 3.106: To test the strength of a 25 20-in.× suitcase, forces are applied as shown. If 18 lb,P = (a) determine the resultant of the applied forces, (b) locate the two points where the line of action of the resultant intersects the edge of the suitcase.

SOLUTION

(a) P = 28 lb

Have ( ) ( ) ( ) ( )42: 20 lb 3 2 28 lb 36 lb13 x yR RΣ − + − + + + = +F i i j j i i j

( ) ( )18.9461 lb 51.297 lb x yR R∴ − + = +i j i j

or ( ) ( )18.95 lb 51.3 lb= − +R i j

( ) ( )2 22 2 18.9461 51.297 54.684 lbx yR R R= + = + =

1 1 51.297tan tan 69.72918.9461

yx

x

RR

θ − − = = = − ° −

or 54.7 lb=R 69.7°

(b) Have B BΣ =M M

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )42 lb4 in. 20 lb 21 in. 3 2 12 in. 36 lb 3 in. 28 lb13B

= − + − + + +

M j i i i j j i i j× × × ×

( ) 221.246 lb in.B∴ = ⋅M k

PROBLEM 3.107 CONTINUED

Since B B=M r R×

( ) ( )221.246 lb in. 0 51.297 18.946118.9461 51.297 0

x y x y∴ ⋅ = = +−

i j kk k

For 221.2460, 4.3130 in.51.297

y x= = = or 4.31 in.x =

For 221.2460, 11.6776 in.18.9461

x y= = = or 11.68 in.y =

PROBLEM 3.108 As four holes are punched simultaneously in a piece of aluminum sheet metal, the punches exert on the piece the forces shown. Knowing that the forces are perpendicular to the surfaces of the piece, determine (a) the resultant of the applied forces when 45α = ° and the point of intersection of the line of action of that resultant with a line drawn through points A and B, (b) the value of α so that the line of action of the resultant passes through fold EF.

SOLUTION

Position the origin for the coordinate system along the centerline of the sheet metal at the intersection with line EF.

(a) Have Σ =F R

( )2.6 5.25 10.5 cos 45 sin 45 3.2 kN = − − − ° + ° − R j j i j i

( ) ( )10.6246 kN 15.2746 kN∴ = − −R i j

( ) ( )2 22 2 10.6246 15.2746x yR R R= + = +

18.6064 kN=

1 1 15.2746tan tan 55.17910.6246

y

x

RR

θ − − − = = = ° −

or 18.61 kN=R 55.2°

Have EF EFM M= Σ

where

( )( ) ( )( )2.6 kN 90 mm 5.25 kN 40 mmEFM = +

( )( ) ( ) ( )10.5 kN 20 mm 3.2 kN 40 mm sin 45 40 mm − − ° +

15.4903 N mEFM∴ = ⋅

To obtain distance d left of EF,

Have ( )15.2746 kNEF yM dR d= = −

33

15.4903 N m 1.01412 10 m15.2746 10 N

d −−⋅

∴ = = − ×− ×

or 1.014 mm left of d EF=

PROBLEM 3.108 CONTINUED

(b) Have 0EFM =

( )( ) ( )( )0 2.6 kN 90 mm 5.25 kN 40 mmEFM = = +

( )( )10.5 kN 20 mm−

( ) ( )3.2 kN 40 mm sin 40 mmα − +

( )128 N m sin 106 N mα∴ ⋅ = ⋅

sin 0.828125α =

55.907α = °

or 55.9α = °

PROBLEM 3.109 As four holes are punched simultaneously in a piece of aluminum sheet metal, the punches exert on the piece the forces shown. Knowing that the forces are perpendicular to the surfaces of the piece, determine (a) the value of α so that the resultant of the applied forces is parallel to the 10.5 N force, (b) the corresponding resultant of the applied forces and the point of intersection of its line of action with a line drawn through points A and B.

SOLUTION

(a) For the resultant force, R, to be parallel to the 10.5 kN force,

α φ=

tan tan y

x

RR

α φ∴ = =

where

( )3.2 kN 10.5 kN sinxR α= − −

( )2.6 kN 5.25 kN 10.5 kN cosyR α= − − −

3.2 10.5 sintan7.85 10.5cos

ααα

+∴ =

+

and 3.2tan 0.407647.85

α = =

22.178α = ° or 22.2α = °

(b) From 22.178α = °

( )3.2 kN 10.5 kN sin 22.178 7.1636 kNxR = − − ° = −

( )7.85 kN 10.5 kN cos 22.178 17.5732 kNyR = − − ° = −

( ) ( )2 22 2 7.1636 17.5732 18.9770 kNx yR R R= + = + =

or 18.98 kN=R 67.8°

Then

EF EFM M= Σ

where

( )( ) ( )( ) ( )( )2.6 kN 90 mm 5.25 kN 40 mm 10.5 kN 20 mmEFM = + −

( ) ( )3.2 kN 40 mm sin 22.178 40 mm − ° +

57.682 N m= ⋅

PROBLEM 3.109 CONTINUED To obtain distance d left of EF,

Have ( )17.5732EF yM dR d= = −

33

57.682 N m 3.2824 10 m17.5732 10 N

d −⋅∴ = = − ×

− ×

or 3.28 mm left of d EF=

PROBLEM 3.110 A truss supports the loading shown. Determine the equivalent force acting on the truss and the point of intersection of its line of action with a line through points A and G.

SOLUTION

Have = ΣR F

( )( ) ( )240 N cos70 sin 70 160 N= ° − ° −R i j j

( )( ) ( )300 N cos 40 sin 40 180 N+ − ° − ° −i j j

( ) ( )147.728 N 758.36 N∴ = − −R i j

( ) ( )2 22 2 147.728 758.36x yR R R= + = +

772.62 N=

1 1 758.36tan tan 78.977147.728

y

x

RR

θ − − − = = = ° −

or 773 N=R 79.0°

Have A yM dRΣ =

where

[ ]( ) [ ]( )240 N cos70 6 m 240 Nsin 70 4 mAMΣ = − ° − °

( )( ) [ ]( )160 N 12 m 300 N cos 40 6 m− + °

[ ]( ) ( )( )300 Nsin 40 20 m 180 N 8 m− ° −

7232.5 N m= − ⋅

7232.5 N m 9.5370 m758.36 N

d − ⋅∴ = =

−

or 9.54 m to the right of d A=

PROBLEM 3.111 Three forces and a couple act on crank ABC. For 5 lbP = and 40 ,α = ° (a) determine the resultant of the given system of forces, (b) locate the point where the line of action of the resultant intersects a line drawn through points B and C, (c) locate the point where the line of action of the resultant intersects a line drawn through points A and B.

SOLUTION

(a) 5 lb, 40P α= = °

Have = ΣR F

( )( ) ( ) ( )5 lb cos 40 sin 40 3 lb 2 lb= ° + ° − −i j i j

( ) ( )0.83022 lb 1.21394 lb∴ = +R i j

( ) ( )2 22 2 0.83022 1.21394x yR R R= + = +

1.47069 lb=

1 1 1.21394tan tan 55.6320.83022

y

x

RR

θ − − = = = °

or 1.471 lb=R 55.6°

(b) From B B yM M dR= Σ =

where

( ) ( ) ( )5 lb cos 40 15 in. sin 50 5 lb sin 40BM = − ° ° − °

( ) ( ) ( )15 in. sin 50 3 lb 6 in. sin 50 × ° + °

( )( )2 lb 6 in. 50 lb in.− + ⋅

23.211 lb in.BM∴ = − ⋅

and 23.211 lb in. 19.1205 in.1.21394 lb

B

y

MdR

− ⋅= = = −

or 19.12 in. to the left of d B=

PROBLEM 3.111 CONTINUED (c) From /B D B=M r R×

( ) ( )1 123.211 lb in. cos50 sin 50d d− ⋅ = − ° + °k i j

( ) ( )0.83022 lb 1.21394 lb × − + i j

( ) ( )1 123.211 lb in. 0.78028 0.63599d d− ⋅ = − −k k

123.211 16.3889 in.

1.41627d∴ = =

or 1 16.39 in. from along line d B AB=

or 1.389 in. above and to the left of A

PROBLEM 3.112 Three forces and a couple act on crank ABC. Determine the value of d so that the given system of forces is equivalent to zero at (a) point B, (b) point D.

SOLUTION

Based on 0xFΣ =

cos 3 lb 0P α − =

cos 3 lbP α∴ = (1)

and 0yFΣ =

sin 2 lb 0P α − =

sin 2 lbP α∴ = (2)

Dividing Equation (2) by Equation (1),

2tan3

α =

33.690α∴ = °

Substituting into Equation (1),

3 lb 3.6056 lbcos33.690

P = =°

or 3.61 lb=P 33.7°

(a) Based on 0BMΣ =

( ) ( )3.6056 lb cos33.690 6 in. sin 50d − ° + °

( ) ( )3.6056 lb sin 33.690 6 in. cos50d − ° + °

( ) ( ) ( )( )3 lb 6 in. sin 50 2 lb 6 in. 50 lb in. 0 + ° − + ⋅ =

3.5838 30.286d− = −

8.4509 in.d∴ =

or 8.45 in.d =

PROBLEM 3.112 CONTINUED

(b) Based on 0DMΣ =

( ) ( )3.6056 lb cos33.690 6 in. sin 50d − ° + °

( ) ( )3.6056 lb sin 33.690 6 in. cos50 6 in.d − ° + ° +

( ) ( )3 lb 6 in. sin 50 50 lb in. 0 + ° + ⋅ =

3.5838 30.286d− = −

8.4509 in.d∴ =

or 8.45 in.d =

This result is expected, since 0=R and 0RB =M for

8.45 in.d = implies that 0 and 0= =R M at any other point for the value of d found in part a.

PROBLEM 3.113 Pulleys A and B are mounted on bracket CDEF. The tension on each side of the two belts is as shown. Replace the four forces with a single equivalent force, and determine where its line of action intersects the bottom edge of the bracket.

SOLUTION

Equivalent force-couple at A due to belts on pulley A

Have : 120 N 160 N ARΣ − − =F

280 NA∴ =R

Have ( ): 40 N 0.02 mA AMΣ − =M

0.8 N mA∴ = ⋅M

Equivalent force-couple at B due to belts on pulley B

Have ( ): 210 N 150 NΣ +F 25 B° = R

360 NB∴ =R 25°

Have ( ): 60 N 0.015 mB BMΣ − =M

0.9 N mB∴ = ⋅M

Equivalent force-couple at F

Have ( ) ( )( ): 280 N 360 N cos 25 sin 25FΣ = − + ° + °F R j i j

( ) ( )326.27 N 127.857 N= −i j

( ) ( )2 22 2 326.27 127.857 350.43 NF Fx FyR R R R= = + = + =

1 1 127.857tan tan 21.399326.27

Fy

Fx

RR

θ − − − = = = − °

or 350 NF = =R R 21.4°

PROBLEM 3.113 CONTINUED

Have

( )( ): 280 N 0.06 m 0.80 N mF FMΣ = − − ⋅M

( ) ( )360 N cos 25 0.010 m − °

( ) ( )360 N sin 25 0.120 m 0.90 N m + ° − ⋅

( )3.5056 N mF = − ⋅M k

To determine where a single resultant force will intersect line FE,

F yM dR=

3.5056 N m 0.027418 m 27.418 mm127.857 N

F

y

MdR

− ⋅∴ = = = =

−

or 27.4 mmd =

PROBLEM 3.114 As follower AB rolls along the surface of member C, it exerts a constant force F perpendicular to the surface. (a) Replace F with an equivalent force-couple system at the point D obtained by drawing the perpendicular from the point of contact to the x axis (b) For 1 ma = and 2 m,b = determine the value of x for which the moment of the equivalent force-couple system at D is maximum.

SOLUTION

(a) The slope of any tangent to the surface of member C is

2

2 221dy d x bb x

dx dx a a

−= − =

Since the force F is perpendicular to the surface,

1 2 1tan2

dy adx b x

α−

= − =

For equivalence

:FΣ =F R

( )( ): cosD A DM F y MαΣ =

where

( ) ( )

2

22 22

2cos , 12

Abx xy b

aa bxα

= = −

+

32

2

4 2 2

2

4D

xxFba

Ma b x

−

∴ =+

Therefore, the equivalent force-couple system at D is

F=R 2

1tan2abx

−

32

2

4 2 2

2

a 4

xFb xa

b x

−

=+

M

PROBLEM 3.114 CONTINUED

(b) To maximize M, the value of x must satisfy 0dMdx

=

where, for 1 m, 2 ma b= =

( )3

2

8

1 16

F x xM

x

−=

+

( ) ( ) ( )( )( )

12 2 3 2 2

2

11 16 1 3 32 1 162

8 01 16

x x x x x xdM Fdx x

− + − − − +

∴ = =+

( )( ) ( )2 2 31 16 1 3 16 0x x x x x+ − − − =

or 4 232 3 1 0x x+ − =

( )( )( )

2 2 23 9 4 32 10.136011 m and 0.22976 m

2 32x

− ± − −= = −

Using the positive value of 2,x

0.36880 mx =

or 369 mmx =

PROBLEM 3.115 As plastic bushings are inserted into a 3-in.-diameter cylindrical sheet metal container, the insertion tool exerts the forces shown on the enclosure. Each of the forces is parallel to one of the coordinate axes. Replace these forces with an equivalent force-couple system at C.

SOLUTION

For equivalence

: A B C D CΣ + + + =F F F F F R

( ) ( ) ( ) ( )5 lb 3 lb 4 lb 7 lbC = − − − −R j j k i

( ) ( ) ( ) 7 lb 8 lb 4 lbC∴ = − − −R i j k

Also for equivalence

/ / /:C A C A B C B D C D C′ ′ ′Σ + + =M r F r F r F M× × ×

or

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

C = − + − +− −

i j k i j k i j kM

( ) ( ) ( )7.50 lb in. 0 0 4.50 lb in. + 3.0 lb in. 0 = − ⋅ − + − ⋅ − ⋅ − i i k

( ) ( )10.5 lb in. 0 0 10.5 lb in. + ⋅ − + + ⋅ j k

( ) ( ) ( )or 12.0 lb in. 10.5 lb in. 7.5 lb in.C = − ⋅ + ⋅ + ⋅M i j k

PROBLEM 3.116 Two 300-mm-diameter pulleys are mounted on line shaft AD. The belts B and C lie in vertical planes parallel to the yz plane. Replace the belt forces shown with an equivalent force-couple system at A.

SOLUTION

Equivalent force-couple at each pulley

Pulley B

( )( )290 N cos 20 sin 20 430 NB = − ° + ° −R j k j

( ) ( )702.51 N 99.186 N= − +j k

( )( )430 N 290 N 0.15 mB = − −M i

( )21 N m= − ⋅ i

Pulley C

( )( )310 N 480 N sin10 cos10C = + − ° − °R j k

( ) ( )137.182 N 778.00 N= − −j k

( )( )480 N 310 N 0.15 mC = −M i

( )25.5 N m= ⋅ i

Then ( ) ( )839.69 N 678.81 NB C= + = − −R R R j k

or ( ) ( )840 N 679 N= − −R j k

/ /A B C B A B C A C= + + +M M M r R r R× ×

( ) ( )21 N m 25.5 N m 0.45 0 0 N m0 702.51 99.186

= − ⋅ + ⋅ + ⋅−

i j ki i

0.90 0 0 N m0 137.182 778.00

+ ⋅− −

i j k

( ) ( ) ( )4.5 N m 655.57 N m 439.59 N m= ⋅ + ⋅ − ⋅i j k

( ) ( ) ( )or 4.50 N m + 656 N m 440 N mA = ⋅ ⋅ − ⋅M i j k

PROBLEM 3.117 A mechanic uses a crowfoot wrench to loosen a bolt at C. The mechanic holds the socket wrench handle at points A and B and applies forces at these points. Knowing that these forces are equivalent to a force-couple system at C consisting of the force ( ) ( )40 N 20 N= − +C i k and the

couple ( )40 N mC = ⋅M i , determine the forces applied at A and B when 10zA = N.

SOLUTION

Have :Σ + =F A B C

or : 40 Nx x xF A B+ = −

( ) 40 Nx xB A∴ = − + (1)

: 0y y yF A BΣ + =

or y yA B= − (2)

: 10 N 20 Nz zF BΣ + =

or 10 NzB = (3)

Have / /:C B C A C CΣ + =M r B r A M× ×

( ) 0.2 0 0.05 0.2 0 0.2 N m 40 N m10 10x y x yB B A A

∴ − + ⋅ = ⋅i j k i j k

i

or ( ) ( )0.05 0.2 0.05 2 0.2 2y x x xB A B A− + − − + −i j

( ) ( )0.2 0.2 40 N my yB A+ + = ⋅k i

From - coefficient 0.05 0.2 40 N my yB A− = ⋅i (4)

- coefficient 0.05 0.2 4 N mx xB A− + = ⋅j (5)

- coefficient 0.2 0.2 0y yB A+ =k (6)

PROBLEM 3.117 CONTINUED

From Equations (2) and (4): ( )0.05 0.2 40y yB B− − =

160 N, 160 Ny yB A= = −

From Equations (1) and (5): ( )0.05 40 0.2 4x xA A− − − + =

8 NxA =

From Equation (1): ( )8 40 48 NxB = − + = −

( ) ( ) ( ) 8 N 160 N 10 N∴ = − +A i j k

( ) ( ) ( )48 N 160 N 10 N= − + +B i j k

PROBLEM 3.118 While using a pencil sharpener, a student applies the forces and couple shown. (a) Determine the forces exerted at B and C knowing that these forces and the couple are equivalent to a force-couple system at A consisting of the force ( ) ( )3.9 lb 1.1 lbyR= + −R i j k and the couple

( ) ( )1.5 lb ft 1.1 lb ft .RA xM= + ⋅ − ⋅M i j k . (b) Find the corresponding

values of yR and .xM

SOLUTION

Have :Σ + =F B C R

: 3.9 lb or 3.9 lbx x x x xF B C B CΣ + = = − (1)

:y y yF C RΣ = (2)

: 1.1 lbz zF CΣ = − (3)

Have / /: RA B A C A B AΣ + + =M r B r C M M× ×

( ) ( ) ( )1 1 0 4.5 4 0 2.0 2 lb ft 1.5 lb ft 1.1 lb ft12 12

0 0 1.1x

x x y

x MB C C

∴ + + ⋅ = + ⋅ − ⋅−

i j k i j ki i j k

( ) ( ) ( )2 0.166667 0.375 0.166667 0.36667 0.33333y x x yC B C C− + + + +i j k

( ) ( )1.5 1.1xM= + −i j k

From - coefficient 2 0.166667 y xC M− =i (4)

- coefficient 0.375 0.166667 0.36667 1.5x xB C+ + =j (5)

- coefficient 0.33333 1.1 or 3.3 lby yC C= − = −k (6)

(a) From Equations (1) and (5):

( )0.375 3.9 0.166667 1.13333x xC C− + =

0.32917 1.58000 lb0.20833xC = =

From Equation (1): 3.9 1.58000 2.32 lbxB = − =

( ) 2.32 lb∴ =B i

( ) ( ) ( )1.580 lb 3.30 lb 1.1 lb= − −C i j k

(b) From Equation (2): 3.30 lby yR C= = − ( )or 3.30 lby = −R j

From Equation (4): ( )0.166667 3.30 2.0 2.5500 lb ftxM = − − + = ⋅

( )or 2.55 lb ftx = ⋅M i

PROBLEM 3.119 A portion of the flue for a furnace is attached to the ceiling at A. While supporting the free end of the flue at F, a worker pushes in at E and pulls out at F to align end E with the furnace. Knowing that the 10-lb force at F lies in a plane parallel to the yz plane, determine (a) the angle α the force at F should form with the horizontal if duct AB is not to tend to rotate about the vertical, (b) the force-couple system at B equivalent to the given force system when this condition is satisfied.

SOLUTION

(a) Duct AB will not have a tendency to rotate about the vertical or y-axis if:

( )/ / 0R RBy B F B F E B EM = Σ = + =j M j r F r F⋅ ⋅ × ×

where

( ) ( ) ( )/ 45 in. 23 in. 28 in.F B = − +r i j k

( ) ( ) ( )/ 54 in. 34 in. 28 in.E B = − +r i j k

( ) ( )10 lb sin cosF α α = + F j k

( )5 lbE = −F k

( ) ( )( ) 10 lb 45 in. 23 in. 28 in. 5 lb 2 in. 27 17 140 sin cos 0 0 1

RB

α α∴ Σ = − + −

−

i j k i j kM

( ) ( ) ( )230cos 280sin 170 450cos 270 450sin lb in.α α α α = − − + − − + ⋅ i j k

Thus, 450cos 270 0RByM α= − + =

cos 0.60α =

53.130α = °

or 53.1α = °

PROBLEM 3.119 CONTINUED

(b) E F= +R F F

where

( )5 lbE = −F k

( )( ) ( ) ( )10 lb sin 53.130 cos53.130 8 lb 6 lbF = ° + ° = +F j k j k

( ) ( ) 8 lb 1 lb∴ = +R j k

and ( ) ( ) ( ) ( )230 0.6 280 0.8 170 450 0.6 270 450 0.8RB = Σ = − + − − − + M M i j k

( ) ( ) ( )192 lb in. 0 360 lb in.= − ⋅ − + ⋅i j k

( ) ( )or 192 lb in. 360 lb in.= − ⋅ + ⋅M i k

PROBLEM 3.120 A portion of the flue for a furnace is attached to the ceiling at A. While supporting the free end of the flue at F, a worker pushes in at E and pulls out at F to align end E with the furnace. Knowing that the 10-lb force at F lies in a plane parallel to the yz plane and that α = 60°, (a) replace the given force system with an equivalent force-couple system at C, (b) determine whether duct CD will tend to rotate clockwise or counterclockwise relative to elbow C, as viewed from D to C.

SOLUTION

(a) Have F E= Σ = +R F F F

where ( ) ( ) ( ) ( )10 lb sin 60 cos60 8.6603 lb 5.0 lbF = ° + ° = + F j k j k

( )5 lbE = −F k

( ) 8.6603 lb∴ =R j ( )or 8.66 lb=R j

Have ( ) / /RC F C F E C E= Σ = +M r F r F r F× × ×

where ( ) ( )/ 9 in. 2 in.F C = −r i j

( ) ( )/ 18 in. 13 in.E C = −r i j

9 2 0 lb in. 18 13 0 lb in.0 8.6603 5.0 0 0 5

RC∴ = − ⋅ + − ⋅

−

i j k i j kM

( ) ( ) ( )55 lb in. 45 lb in. 77.942 lb in.= ⋅ + ⋅ + ⋅i j k

( ) ( ) ( )or 55.0 lb in. 45.0 lb in. 77.9 lb in.RC = ⋅ + ⋅ + ⋅M i j k

(b) To determine which direction duct section CD has a tendency to turn, have R RCD DC CM = Mλ ⋅

where

( ) ( ) ( )18 in. 4 in. 1 9 22 85 in. 85DC

− += = − +

i ji jλ

Then ( ) ( )1 9 2 55 45 77.942 lb in.85

RCDM = − + + + ⋅i j i j k⋅

( )53.690 9.7619 lb in.= − + ⋅

43.928 lb in.= − ⋅

Since 0,RDC C <Mλ ⋅ duct DC tends to rotate clockwise relative to elbow C as viewed from D to C.

PROBLEM 3.121 The head-and-motor assembly of a radial drill press was originally positioned with arm AB parallel to the z axis and the axis of the chuck and bit parallel to the y axis. The assembly was then rotated o25 about the y axis and o20 about the centerline of the horizontal arm AB, bringing it into the position shown. The drilling process was started by switching on the motor and rotating the handle to bring the bit into contact with the workpiece. Replace the force and couple exerted by the drill press with an equivalent force-couple system at the center O of the base of the vertical column.

SOLUTION

Have =R F

( ) ( ) ( ) ( )44 N sin 20 cos 25 cos 20 sin 20 sin 25 = ° ° − ° − ° ° i j k

( ) ( ) ( )13.6389 N 41.346 N 6.3599 N= − −i j k

( ) ( ) ( )or 13.64 N 41.3 N 6.36 N= − −R i j k

Have /O B O C= +M r F M×

where

( ) ( ) ( )/ 0.280 m sin 25 0.300 m 0.280 m cos 25B O = ° + + ° r i j k

( ) ( ) ( )0.118333 m 0.300 m 0.25377 m= + +i j k

( ) ( ) ( ) ( )7.2 N m sin 20 cos 25 cos 20 sin 20 sin 25C = ⋅ ° ° − ° − ° ° M i j k

( ) ( ) ( )2.2318 N m 6.7658 N m 1.04072 N m= ⋅ − ⋅ − ⋅i j k

0.118333 0.300 0.25377 N m13.6389 41.346 6.3599

O∴ = ⋅− −

i j kM

( )2.2318 6.7658 1.04072 N m+ − − ⋅i j k

( ) ( ) ( )10.8162 N m 2.5521 N m 10.0250 N m= ⋅ − ⋅ − ⋅i j k

( ) ( ) ( )or 10.82 N m 2.55 N m 10.03 N mO = ⋅ − ⋅ − ⋅M i j k

PROBLEM 3.122 While a sagging porch is leveled and repaired, a screw jack is used to support the front of the porch. As the jack is expanded, it exerts on the porch the force-couple system shown, where 300 NR = and

37.5 N m.M = ⋅ Replace this force-couple system with an equivalent force-couple system at C.

SOLUTION

From ( ) ( ) ( ) ( )0.2 m 1.4 m 0.5 m300 N 300 N

1.50 mC AB − + −

= = =

i j kR R λ

( ) ( ) ( )40.0 N 280 N 100 NC = − + −R i j k

From /C A C= +M r R M×

where

( ) ( )/ 2.6 m 0.5 mA C = +r i k

( ) ( ) ( ) ( ) ( )0.2 m 1.4 m 0.5 m37.5 N m 37.5 N m

1.50 mBA − +

= ⋅ = ⋅

i j kM λ

( ) ( ) ( )5.0 N m 35.0 N m 12.5 N m= ⋅ − ⋅ + ⋅i j k

( ) ( ) ( ) ( ) 10 N m 2.6 0 0.5 5.0 N m 35.0 N m 12.5 N m4 28 10

C∴ = ⋅ + ⋅ − ⋅ + ⋅− −

i j kM i j k

( ) ( ) ( )140 5 N m 20 260 35 N m 728 12.5 N m = − + ⋅ + − + − ⋅ + + ⋅ i j k

( ) ( ) ( )or 135.0 N m 205 N m 741 N mC = − ⋅ + ⋅ + ⋅M i j k

PROBLEM 3.123 Three children are standing on a 15 15-ft× raft. If the weights of the children at points A, B, and C are 85 lb, 60 lb, and 90 lb, respectively, determine the magnitude and the point of application of the resultant of the three weights.

SOLUTION

Have : A B CΣ + + =F F F F R

( ) ( ) ( )85 lb 60 lb 90 lb− − − =j j j R

( )235 lb− =j R or 235 lbR =

Have ( ) ( ) ( ) ( ):x A A B B C C DM F z F z F z R zΣ + + =

( )( ) ( )( ) ( )( ) ( )( )85 lb 9 ft 60 lb 1.5 ft 90 lb 14.25 ft 235 lb Dz+ + =

9.0957 ftDz∴ = or 9.10 ftDz =

Have ( ) ( ) ( ) ( ):z A A B B C C DM F x F x F x R xΣ + + =

( )( ) ( )( ) ( )( ) ( )( )85 lb 3 ft 60 lb 4.5 ft 90 lb 14.25 ft 235 lb Dx+ + =

7.6915 ftDx∴ = or 7.69 ftDx =

PROBLEM 3.124 Three children are standing on a 15 15-ft× raft. The weights of the children at points A, B, and C are 85 lb, 60 lb, and 90 lb, respectively. If a fourth child of weight 95 lb climbs onto the raft, determine where she should stand if the other children remain in the positions shown and the line of action of the resultant of the four weights is to pass through the center of the raft.

SOLUTION

Have : A B C DΣ + + + =F F F F F R

( ) ( ) ( ) ( )85 lb 60 lb 90 lb 95 lb− − − − =j j j j R

( ) 330 lb∴ = −R j

Have ( ) ( ) ( ) ( ) ( ):x A A B B C C D D HM F z F z F z F z R zΣ + + + =

( )( ) ( )( ) ( )( ) ( )( ) ( )( )85 lb 9 ft 60 lb 1.5 ft 90 lb 14.25 ft 95 lb 330 lb 7.5 ftDz+ + + =

3.5523 ftDz∴ = or 3.55 ftDz =

Have ( ) ( ) ( ) ( ) ( ):z A A B B C C D D HM F x F x F x F x R xΣ + + + =

( )( ) ( )( ) ( )( ) ( )( ) ( )( )85 lb 3 ft 60 lb 4.5 ft 90 lb 14.25 ft 95 lb 330 lb 7.5 ftDx+ + + =

7.0263 ftDx∴ = or 7.03 ftDx =

PROBLEM 3.125 The forces shown are the resultant downward loads on sections of the flat roof of a building because of accumulated snow. Determine the magnitude and the point of application of the resultant of these four loads.

SOLUTION

Have : A B C DΣ + + + =F F F F F R

( ) ( ) ( ) ( )580 kN 2350 kN 330 kN 140 kN− − − − =j j j j R

( ) 3400 kN∴ = −R j 3400 kNR =

Have ( ) ( ) ( ) ( ) ( ):x A A B B C C D D EM F z F z F z F z R zΣ + + + =

( )( ) ( )( ) ( )( ) ( )( ) ( )( )580 kN 8 m 2350 kN 16 m 330 kN 6 m 140 kN 33.5 m 3400 kN Ez+ + + =

14.3853 mEz∴ = or 14.39 mEz =

Have ( ) ( ) ( ) ( ) ( ):z A A B B C C D D EM F x F x F x F x R xΣ + + + =

( )( ) ( )( ) ( )( ) ( )( ) ( )( )580 kN 10 m 2350 kN 32 m 330 kN 54 m 140 kN 32 m 3400 kN Ex+ + + =

30.382 mEx∴ = or 30.4 mEx =

PROBLEM 3.126 The forces shown are the resultant downward loads on sections of the flat roof of a building because of accumulated snow. If the snow represented by the 580-kN force is shoveled so that the this load acts at E, determine a and b knowing that the point of application of the resultant of the four loads is then at B.

SOLUTION

Have : B C D EΣ + + + =F F F F F R

( ) ( ) ( ) ( )2350 kN 330 kN 140 kN 580 kN− − − − =j j j j R

( ) 3400 kN∴ = −R j

Have ( ) ( ) ( ) ( ) ( ):x B B C C D D E E BM F z F z F z F z R zΣ + + + =

( )( ) ( )( ) ( )( ) ( )( ) ( )( )2350 kN 16 m 330 kN 6 m 140 kN 33.5 m 580 kN 3400 kN 16 mb+ + + =

17.4655 mb∴ = or 17.47 mb =

Have ( ) ( ) ( ) ( ) ( ):z B B C C D D E E BM F x F x F x F x R xΣ + + + =

( )( ) ( )( ) ( )( ) ( )( ) ( )( )2350 kN 32 m 330 kN 54 m 140 kN 32 m 580 kN 3400 kN 32 ma+ + + =

19.4828 ma∴ = or 19.48 ma =

PROBLEM 3.127 A group of students loads a 2 4-m× flatbed trailer with two 0.6 0.6 0.6-m× × boxes and one 0.6 0.6 1.2-m× × box. Each of the boxes at the rear of the trailer is positioned so that it is aligned with both the back and a side of the trailer. Determine the smallest load the students should place in a second 0.6 0.6 1.2-m× × box and where on the trailer they should secure it, without any part of the box overhanging the sides of the trailer, if each box is uniformly loaded and the line of action of the resultant of the weights of the four boxes is to pass through the point of intersection of the centerlines of the trailer and the axle. (Hint: Keep in mind that the box may be placed either on its side or on its end.)

SOLUTION

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 ( ) ( ) ( ): 200 N 400 N 180 NΣ − − − =F j j j R

( ) 780 N∴ = −R j

Have ( )( ) ( )( ) ( )( ) ( )( ): 200 N 0.3 m 400 N 1.7 m 180 N 1.7 m 780 NzM xΣ + + =

1.34103 mx∴ =

Have ( )( ) ( )( ) ( )( ) ( )( ): 200 N 0.3 m 400 N 0.6 m 180 N 2.4 m 780 NxM zΣ + + =

0.93846 mz∴ =

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, 0.GΣ =M

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 1 m 1.8 m 3.7 mx z≤ ≤ ≤ ≤

PROBLEM 3.127 CONTINUED

Let 0.3 m,x = ( )( ) ( )( ) ( )( ) ( ): 200 N 0.7 m 400 N 0.7 m 180 N 0.7 m 0.7 m 0GzM WΣ − − + =

380 NW∴ =

( )( ) ( )( ) ( )( ) ( )( ): 200 N 1.5 m 400 N 1.2 m 180 N 0.6 m 380 N 1.8 m 0GxM zΣ − − + + − =

3.5684 m 3.7 m acceptablez∴ = < ∴

Let 3.7 m,z = ( )( ) ( )( ) ( )( ) ( ): 200 N 1.5 m 400 N 1.2 m 180 N 0.6 m 1.7 m 0GxM WΣ − − + + =

395.29 N 380 NW∴ = >

Since the weight W found for 0.3 mx = is less than W found for 3.7 m, 0.3 mz x= = results in the smallest weight W.

( )or 380 N at 0.3 m, 0, 3.57 mW =

PROBLEM 3.128 Solve Problem 3.127 if the students want to place as much weight as possible in the fourth box and that at least one side of the box must coincide with a side of the trailer.

Problem 3.127: A group of students loads a 2 4-m× flatbed trailer with two 0.6 0.6 0.6-m× × boxes and one 0.6 0.6 1.2-m× × box. Each of the boxes at the rear of the trailer is positioned so that it is aligned with both the back and a side of the trailer. Determine the smallest load the students should place in a second 0.6 0.6 1.2-m× × box and where on the trailer they should secure it, without any part of the box overhanging the sides of the trailer, if each box is uniformly loaded and the line of action of the resultant of the weights of the four boxes is to pass through the point of intersection of the centerlines of the trailer and the axle. (Hint: Keep in mind that the box may be placed either on its side or on its end.)

SOLUTION

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 0.6 m,x = with a small side of the box touching the z-axis, satisfies this condition.

Let 0.6 m,x = ( )( ) ( )( ) ( )( ) ( ): 200 N 0.7 m 400 N 0.7 m 180 N 0.7 m 0.4 m 0GzM WΣ − − + =

665 NW∴ =

and ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ): 200 N 1.5 m 400 N 1.2 m 180 N 0.6 m 665 N 1.8 m 0GXM zΣ − − + + − =

( ) 2.8105 m 2 m 4 m acceptablez z∴ = < < ∴

( )or 665 N at 0.6 m, 0, 2.81 mW =

PROBLEM 3.129 A block of wood is acted upon by three forces of the same magnitude P and having the directions shown. Replace the three forces with an equivalent wrench and determine (a) the magnitude and direction of the resultant R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xy plane.

SOLUTION

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

Have : P P PΣ − − =F i i k R

P∴ = −R k

Have ( ) ( ) ( ): 3 3 3 RO OP a P a P a aΣ − − + − + =M k j i j M

( ) 3RO Pa∴ = − −M i k

Then let vectors ( )1, R M represent the components of the wrench, where their directions are the same.

(a) P= −R k or Magnitude of P=R

Direction of : 90 , 90 , 180x y zθ θ θ= ° = ° = − °R

(b) Have 1R

R OM = Mλ ⋅

( )3Pa = − − − k i k⋅

3Pa=

and pitch 1 3 3M Pap aR P

= = = or 3p a=

PROBLEM 3.129 CONTINUED

(c) Have 1 2RO = +M M M

( ) ( )2 1 3 3RO Pa Pa Pa∴ = − = − − − − = −M M M i k k i

Require 2 /Q O=M r R×

( ) ( )Pa x y P Px Py− = + − = −i i j k j i×

From : or Pa Py y a− = − =i

: 0x =j

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

PROBLEM 3.130 A piece of sheet metal is bent into the shape shown and is acted upon by three forces. Replace the three forces with an equivalent wrench and determine (a) the magnitude and direction of the resultant R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the yz plane.

SOLUTION

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

Have ( ) ( ) ( ): 2P P PΣ − + =F i j j R

( ) 2P∴ =R i

Have ( ): RO O OΣ Σ =M r F M×

( )2 2 2.5 0 0 4 1.5 5 62 1 0 0 1 0

RO Pa Pa= + = − + −

−

i j k i j kM i j k

(a) 2P=R i or Magnitude of 2P=R

Direction of : 0 , 90 , 90x y zθ θ θ= ° = − ° = °R

(b) Have 1 RR O RM

R= =

RMλ ⋅ λ

( )1.5 5 6Pa Pa Pa= − + −i i j k⋅

1.5Pa= −

and pitch 1 1.5 0.752

M Pap aR P

−= = = − or 0.75p a= −

PROBLEM 3.130 CONTINUED

(c) Have 1 2RO = +M M M

( ) ( )2 1 5 6RO Pa Pa∴ = − = −M M M j k

Require 2 /Q O=M r R×

( ) ( ) ( ) ( ) ( ) ( )5 6 2 2 2Pa Pa y z P Py Pz− = + = − +j k j k i k j×

From : 5 2Pa Pz=i

2.5z a∴ =

From : 6 2Pa Py− = −k

3y a∴ =

∴ The axis of the wrench is parallel to the x-axis and intersects the yz-plane at 3 , 2.5y a z a= =

PROBLEM 3.131 The forces and couples shown are applied to two screws as a piece of sheet metal is fastened to a block of wood. Reduce the forces and the couples to an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xz plane.

SOLUTION

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

Have ( ) ( ): 10 N 11 NΣ − − =F j j R

( ) 21 N∴ = −R j

Have ( ): RO O C OΣ Σ + Σ =M r F M M×

( )0 0 0.5 N m 0 0 0.375 N m 12 N m0 10 0 0 11 0

RO = ⋅ + − ⋅ − ⋅

− −

i j k i j kM j

( ) ( )0.875 N m 12 N m= ⋅ − ⋅i j

(a) ( )21 N= −R j ( )or 21 N= −R j

(b) Have 1 RR O RM

R= =

RMλ ⋅ λ

( ) ( ) ( )0.875 N m 12 N m = − − j i j⋅ ⋅ ⋅

( )112 N m and 12 N m= ⋅ = − ⋅M j

and pitch 1 12 N m 0.57143 m21 N

MpR

⋅= = = or 0.571 mp =

PROBLEM 3.131 CONTINUED

(c) Have 1 2RO = +M M M

( )2 1 0.875 N mRO∴ = − = ⋅M M M i

Require 2 /Q O=M r R×

( ) ( ) ( ) 0.875 N m 21 Nx z ∴ ⋅ = + − i i k j×

( ) ( )0.875 21 21x z= − +i k i

From i: 0.875 21z=

0.041667 mz∴ =

From k: 0 21x= −

0z∴ =

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

PROBLEM 3.132 The forces and couples shown are applied to two screws as a piece of sheet metal is fastened to a block of wood. Reduce the forces and the couples to an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xz plane.

SOLUTION

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

Have ( ) ( ): 6 lb 4.5 lb 7.5 lbRΣ − − = =F i j R

Have ( ): RO O C OΣ ∑ + ∑ =M r F M M×

( ) ( ) ( )6 lb 8 in. 160 lb in. 72 lb in.RO = − − ⋅ − ⋅M j i j

( ) ( )160 lb in. 120 lb in.= − ⋅ − ⋅i j

200 lb in.ROM = ⋅

(a) ( ) ( )6 lb 4.5 lb= − −R i j

(b) Have 1 RR OM

R= =

RMλ ⋅ λ

( ) ( ) ( )0.8 0.6 160 lb in. 120 lb in. = − − − ⋅ − ⋅ i j i j⋅

200 lb in.= ⋅

and ( )1 200 lb in. 0.8 0.6= ⋅ − −M i j

Pitch 1 200 lb in. 26.667 in.7.50 lb

MpR

⋅= = =

or 26.7 in.p =

(c) From above note that

1RO=M M

Therefore, the axis of the wrench goes through the origin. The line of action of the wrench lies in the xy plane with a slope of

34

dydx

=

PROBLEM 3.133 Two bolts A and B are tightened by applying the forces and couple shown. Replace the two wrenches with a single equivalent wrench and determine (a) the resultant R, (b) the pitch of the single equivalent wrench, (c) the point where the axis of the wrench intersects the xz plane.

SOLUTION

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

Have ( ) ( ) ( ) ( ): 20 lb 21 lb 21 lb 20 lb 29 lbRΣ − − = − − = =F k j j k R

and ( ): RO O C OΣ ∑ + ∑ =M r F M M×

( ) ( ) ( )20 lb 4 in. 4 3 0 21 lb 4 in. 6 0 1 300 320 lb in.0 0 1 0 1 0

RO+ + − − ⋅ =

− −

i j k i j kj k M

( ) ( ) ( ) 156 lb in. 20 lb in. 824 lb in.RO∴ = − ⋅ + ⋅ − ⋅M i j k

(a) ( ) ( )21 lb 20 lb= − −R j k

(b) Have 1 RR O RM

R= =

RMλ ⋅ λ

( ) ( ) ( )21 20 156 lb in. 20 lb in. 824 lb in.29

− − = − − ⋅ + ⋅ − ⋅ j k i j k⋅

553.80 lb in.= ⋅

PROBLEM 3.133 CONTINUED

and ( ) ( )1 1 401.03 lb in. 381.93 lb in.RM= = − ⋅ − ⋅M j kλ

Then pitch 1 553.80 lb in. 19.0964 in.29 lb

MpR

⋅= = = or 19.10 in.p =

(c) Have 1 2RO = +M M M

( ) ( )2 1 156 20 824 401.03 381.93 lb in.RO ∴ = − = − + − − − − ⋅ M M M i j k j k

( ) ( ) ( )156.0 lb in. 421.03 lb in. 442.07 lb in.= − ⋅ + ⋅ − ⋅i j k

Require 2 /Q O=M r R×

( ) ( ) ( )156 421.03 442.07 21 20x z− + − = + − −i j k i k j k×

( ) ( ) ( )21 20 21z x x= + −i j k

From i: 156 21z− =

7.4286 in.z∴ = −

or 7.43 in.z = −

From k: 442.07 21x− = −

21.051 in.x∴ =

or 21.1 in.x =

∴ The axis of the wrench intersects the xz-plane at

21.1 in., 7.43 in.x z= = −

PROBLEM 3.134 Two bolts A and B are tightened by applying the forces and couple shown. Replace the two wrenches with a single equivalent wrench and determine (a) the resultant R, (b) the pitch of the single equivalent wrench, (c) the point where the axis of the wrench intersects the xz plane.

SOLUTION

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

(a) Have ( ) ( ) 8 15: 79.2 lb 51 lb17 17 Σ − − + =

F k i j R

( ) ( ) ( ) 24.0 lb 45.0 lb 79.2 lb∴ = − − −R i j k

and 94.2 lbR =

Have /: RB A B A A B BΣ + + =M r F M M M×

( )8 150 20 0 660 714 1584 660 42 8 1517 17

0 0 79.2

RB

= − − − + = − − + −

i j kM k i j i k i j

( ) ( ) ( ) 1248 lb in. 630 lb in. 660 lb in.RB∴ = ⋅ − ⋅ − ⋅M i j k

(b) Have 1 RR O RM

R= =

RMλ ⋅ λ

( ) ( ) ( )24.0 45.0 79.2 1248 lb in. 630 lb in. 660 lb in.94.2

− − − = ⋅ − ⋅ − ⋅ i j k i j k⋅

537.89 lb in.= ⋅

PROBLEM 3.134 CONTINUED

and 1 1 RM=M λ

( ) ( ) ( )137.044 lb in. 256.96 lb in. 452.24 lb in.= − ⋅ − ⋅ − ⋅i j k

Then pitch 1 537.89 lb in. 5.7101 in.94.2 lb

MpR

⋅= = = or 5.71 in.p =

(c) Have 1 2RB = +M M M

( ) ( )2 1 1248 630 660 137.044 256.96 452.24RB∴ = − = − − − − − −M M M i j k i j k

( ) ( ) ( )1385.04 lb in. 373.04 lb in. 207.76 lb in.= ⋅ − ⋅ − ⋅i j k

Require 2 /Q B=M r R×

1385.04 373.04 207.76 024 45 79.2x z− − =

− − −

i j ki j k

( ) ( ) ( ) ( )45 24 79.2 45z z x x= − + −i j j k

From i: 1385.04 45 30.779 in.z z= ∴ =

From k: 207.76 45 4.6169 in.x x− = − ∴ =

∴ The axis of the wrench intersects the xz-plane at

4.62 in., 30.8 in.x z= =

PROBLEM 3.135 A flagpole is guyed by three cables. If the tensions in the cables have the same magnitude P, replace the forces exerted on the pole with an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xz plane.

SOLUTION

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

Have : BA DC DEP P PΣ + + =F Rλ λ λ

4 3 3 4 9 4 125 5 5 5 25 5 25

P − = − + − + − +

R j k i j i j k

( )3 2 2025P

∴ = − −R i j k

( ) ( ) ( )2 2 23 27 52 20 125 25PR P= + + =

Have ( ): RO OPΣ Σ =M r M×

( ) ( ) ( )4 3 3 4 9 4 1224 20 205 5 5 5 25 5 25

RO

P P P P P P Pa a a− − − + − + − + =

j j k j i j j i j k M× × ×

( )24 5

RO

Pa∴ = − −M i k

(b) Have 1R

R OM = Mλ ⋅

where ( ) ( )3 25 12 20 2 2025 27 5 9 5RP

R P= = − − = − −

R i j k i j kλ

PROBLEM 3.135 CONTINUED

Then ( ) ( )11 24 82 20

59 5 15 5Pa PaM −

= − − − − =i j k i k⋅

and pitch 1 8 25 88115 5 27 5

M Pa apR P

− − = = =

or 0.0988p a= −

(c) ( ) ( )1 18 1 82 20 2 20

67515 5 9 5RPa PaM −

= = − − = − + +

M i j k i j kλ

Then ( ) ( ) ( )2 124 8 82 20 403 20 406

5 675 675RO

Pa Pa Pa= − = − − − − + + = − − −M M M i k i j k i j k

Require 2 /Q O=M r R×

( ) ( ) ( )8 3403 20 406 2 20675 25Pa Px z − − − = + − −

i j k i k i j k×

( )3 20 2 2025P z x z x = + + −

i j k

From i: ( ) 38 403 20675 25Pa Pz − =

1.99012z a∴ = −

From k: ( ) 38 406 20 2.0049675 25Pa Px x a − = − ∴ =

∴ The axis of the wrench intersects the xz-plane at

2.00 , 1.990x a z a= = −

PROBLEM 3.136 Determine whether the force-and-couple system shown can be reduced to a single equivalent force R. If it can, determine R and the point where the line of action of R intersects the yz plane. If it cannot be so reduced, replace the given system with an equivalent wrench and determine its resultant, its pitch, and the point where its axis intersects the yz plane.

SOLUTION

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

Have : DA ED DA DA ED EDF FΣ + = + =F F F Rλ λ

where ( ) ( ) ( )0.300 m 0.225 m 0.200 m136 N

0.425 mDA − + +

=

i j kF

( ) ( ) ( )96 N 72 N 64 N= − + +i j k

( ) ( ) ( ) ( )0.150 m 0.200 m120 N 72 N 96 N

0.250 mED − −

= = − −

i kF i k

( ) ( ) ( )168 N 72 N 32 N∴ = − + −R i j k

Have : RD A DΣ =M M M

or ( ) ( ) ( ) ( ) ( )0.150 m 0.150 m 0.450 m 16 N m16 N m 30.150 11 m 11

RD

− − + ⋅= ⋅ = − − +

i j kM i j k

PROBLEM 3.136 CONTINUED

The force-couple at D can be replaced by a single force if R is perpendicular to .RDM To be perpendicular,

0.RD =R M⋅

Have ( ) ( )16168 72 32 311

RD = − + − − − +R M i j k i j k⋅ ⋅

( )128 21 9 1211

= − −

0=

∴ Force-couple can be reduced to a single equivalent force.

To determine the coordinates where the equivalent single force intersects the yz-plane, /RD Q D=M r R×

where ( ) ( ) ( )/ 0 0.300 m 0.075 m 0 mQ D y z = − + − + − r i j k

( ) ( ) ( )16 N m 3 8 N 0.3 0.075 m11

21 9 4y z⋅

∴ − − + = − −− −

i j ki j k

or

( ) ( ) ( ) ( ) ( ){ }16 N m 3 8 N 4 0.075 9 21 1.2 2.7 21 0.075 m11

y z z y⋅ − − + = − − − + − − + − + − i j k i j k

From j: ( )16 8 21 1.211

z−= − − 0.028427 m 28.4 mmz∴ = − = −

From k: ( )48 8 2.7 21 0.075 0.28972 m 290 mm11

y y = − + − ∴ = =

∴ line of action of R intersects the yz-plane at

290 mm, 28.4 mmy z= = −

PROBLEM 3.137 Determine whether the force-and-couple system shown can be reduced to a single equivalent force R. If it can, determine R and the point where the line of action of R intersects the yz plane. If it cannot be so reduced, replace the given system with an equivalent wrench and determine its resultant, its pitch, and the point where its axis intersects the yz plane.

SOLUTION

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

Have : A GΣ + =F F F R

( ) ( ) ( ) ( ) ( ) ( ) ( )4 in. 6 in. 12 in. 10 lb 14 lb 4 lb 6 lb 2 lb

14 in. + −

∴ = + = + −

i j kR k i j k

and 56 lbR =

Have ( ): RO O C OΣ ∑ + ∑ =M r F M M×

( ) ( ) ( ) ( ) ( ) ( ){ }12 in. 10 lb 16 in. 4 lb 6 lb 12 lbRO = + + − M j k i i j k× ×

( ) ( ) ( ) ( ) ( ) ( ) ( )16 in. 12 in. 4 in. 12 in. 6 in.84 lb in. 120 lb in.

20 in. 14 in. − − +

+ ⋅ + ⋅

i j i j k

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

( ) ( ) ( )18.4572 lb ft 3.2286 lb ft 12.2858 lb ft= ⋅ + ⋅ + ⋅i j k

PROBLEM 3.137 CONTINUED

The force-couple at O can be replaced by a single force if the direction of R is perpendicular to .ROM

To be perpendicular 0RO =R M⋅

Have ( ) ( )4 6 2 18.4572 3.2286 12.2858 0?RO = + − + + =R M i j k i j k⋅ ⋅

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.

1 RR O RM

R= =

RMλ ⋅ λ

( ) ( )4 6 218.4572 3.2286 12.2858

56 + −

= + +

i j ki j k⋅

9.1709 lb ft= ⋅

and ( )( )1 1 9.1709 lb ft 0.53452 0.80178 0.26726RM= = ⋅ + −M i j kλ

And pitch 1 9.1709 lb ft 1.22551 ft56 lb

MpR

⋅= = =

or 1.226 ftp =

Have

( ) ( )( )2 1 18.4572 3.2286 12.2858 9.1709 0.53452 0.80178 0.26726RO= − = + + − + −M M M i j k i j k

( ) ( ) ( )13.5552 lb ft 4.1244 lb ft 14.7368 lb ft= ⋅ − ⋅ + ⋅i j k

Require 2 /Q O=M r R×

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

( ) ( ) ( )2 6 4 4y z z y= − + + −i j k

From j: 4.1244 4z− = or 1.0311 ftz = −

From k: 14.7368 4 or 3.6842 fty y= − = −

∴ line of action of the wrench intersects the yz plane at

3.68 ft, 1.031 fty z= − =

PROBLEM 3.138 Replace the wrench shown with an equivalent system consisting of two forces perpendicular to the y axis and applied respectively at A and B.

SOLUTION

Express the forces at A and B as

x zA A= +A i k

x zB B= +B i k

Then, for equivalence to the given force system

: 0x x xF A BΣ + = (1)

:z z zF A B RΣ + = (2)

( ) ( ): 0x z zM A a B a bΣ + + = (3)

( ) ( ):z x xM A a B a b MΣ − − + = (4)

From Equation (1), x xB A= −

Substitute into Equation (4)

( ) ( )x xA a A a b M− + + =

and x xM MA Bb b

∴ = = −

From Equation (2), z zB R A= −

and Equation (3), ( )( ) 0z zA a R A a b+ − + =

1zaA Rb

∴ = +

PROBLEM 3.138 CONTINUED

and 1zaB R Rb

= − +

zaB Rb

∴ = −

Then 1M aRb b

= + +

A i k

M a Rb b

= − −

B i k

PROBLEM 3.139 Show that, in general, a wrench can be replaced with two forces chosen in such a way that one force passes through a given point while the other force lies in a given plane.

SOLUTION

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

and x y z x y zR R R M M M M= + + = + +R i j k i j k

while the unknown forces A and B can be expressed as

and x y z x zA A A B B= + + = +A i j k B i k

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

Then, for equivalence of the two systems

:x x x xF R A BΣ = + (1)

:y y yF R AΣ = (2)

:z z z zF R A BΣ = + (3)

:x x z yM M yA zAΣ = − (4)

:y y x z zM M zA xA bBΣ = − − (5)

:z z y xM M xA yAΣ = − (6)

PROBLEM 3.139 CONTINUED

Based on the above six independent equations for the six unknowns ( ), , , , , ,x y z x zA A A B B b there exists a

unique solution for A and B.

From Equation (2) y yA R=

Equation (6) ( )1x y zA xR M

y

= −

Equation (1) ( )1x x y zB R xR M

y

= − −

Equation (4) ( )1z x yA M zR

y

= +

Equation (3) ( )1z z x yB R M zR

y

= − +

Equation (5) ( )( )

x y z

x z y

xM yM zMb

M yR zR

+ +=

− +

PROBLEM 3.140 Show that a wrench can be replaced with two perpendicular forces, one of which is applied at a given point.

SOLUTION

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 and and are known.R M= =R j M j

The unknown forces A and B can be expressed as

and x y z x y zA A A B B B= + + = + +A i j k B i j k

The distance a is known. It is assumed that force B intersects the xz plane at (x, 0, z). Then for equivalence

: 0x x xF A B∑ = + (1)

:y y yF R A B∑ = + (2)

: 0z z zF A B∑ = + (3)

: 0x yM zB∑ = − (4)

:y z z xM M aA xB zB∑ = − − + (5)

: 0z y yM aA xB∑ = + (6)

Since A and B are made perpendicular,

0 or 0x x y y z zA B A B A B= + + =A B⋅ (7)

There are eight unknowns: , , , , , , , x y z x y zA A A B B B x z

But only seven independent equations. Therefore, there exists an infinite number of solutions.

PROBLEM 3.140 CONTINUED

Next consider Equation (4): 0 yzB= −

If 0,yB = Equation (7) becomes 0x x z zA B A B+ =

Using Equations (1) and (3) this equation becomes 2 2 0x zA A+ =

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

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

(Note: Setting , ,y zA A or zB equal to zero results in unacceptable solutions.)

The defining equations then become.

0 xB= (1)′

y yR A B= + (2)

0 z zA B= + (3)

z zM aA xB= − − (5)′

0 y yaA xB= + (6)

0y y z zA B A B+ = (7)′

Then Equation (2) can be written y yA R B= −

Equation (3) can be written z zB A= −

Equation (6) can be written y

y

aAx

B= −

Substituting into Equation (5)′,

( )yz z

y

R BM aA a A

B

−= − − − −

or z yMA BaR

= − (8)

Substituting into Equation (7)′,

( ) 0y y y yM MR B B B BaR aR

− + − =

PROBLEM 3.140 CONTINUED

or 2 3

2 2 2ya RB

a R M=

+

Then from Equations (2), (8), and (3)

2 3 2

2 2 2 2 2 2ya R RMA R

a R M a R M= − =

+ +

2 3 2

2 2 2 2 2 2zM a R aR MAaR a R M a R M

= − = − + +

2

2 2 2zaR MB

a R M=

+

In summary

( )2 2 2RM M aR

a R M= −

+A j k

( )2

2 2 2aR aR M

a R M= +

+B j k

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 0 and 0,R M> > it follows from the equations found for A and B that 0 and 0.y yA B> >

From Equation (6), 0 (assuming 0).x a< > Then, as a consequence of letting 0,xA = 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.

PROBLEM 3.141 Show that a wrench can be replaced with two forces, one of which has a prescribed line of action.

SOLUTION

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 ( ), 0, .P x z Denoting the known direction of line AA′ by

A x y zλ λ λ= + +i j kλ

it follows that force A can be expressed as

( )A x y zA A λ λ λ= = + +i j kλA

Force B can be expressed as

x y zB B B= + +B i j k

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

: 0x x xF A BλΣ = + (1)

:y y yF R A BλΣ = + (2)

: 0z z zF A BλΣ = + (3)

: 0x yM zBΣ = − (4)

:y z x zM M aA zB xBλΣ = − + − (5)

: 0z y yM aA xBλΣ = + (6)

Since there are six unknowns ( ), , , , , x y zA B B B x z and six independent equations, it will be possible to

obtain a solution.

PROBLEM 3.141 CONTINUED

Case 1: Let 0z = to satisfy Equation (4)

Now Equation (2) y yA R Bλ = −

Equation (3) z zB Aλ= −

Equation (6) ( )yy

y y

aA ax R BB Bλ

= − = − −

Substitution into Equation (5)

( )( )z y zy

aM aA R B AB

λ λ = − − − − −

1 yz

MA BaRλ

∴ = −

Substitution into Equation (2)

1y y y

z

MR B BaR

λλ

= − +

2 z

yz y

aRBaR Mλ

λ λ∴ =

−

Then z y y z

MR RA aRaR MM

λ λ λ λ= − =

− −

xx x

z y

MRB AaR Mλλ

λ λ= − =

−

zz z

z y

MRB AaR Mλλ

λ λ= − =

−

In summary

A

y z

PaRM

λ λ=

−A λ

( )x z zz y

R M aR MaR M

λ λ λλ λ

= + +−

B i j k

and z21 1 y

y z

aR MRx a a RB aR

λ λλ

− = − = −

or y

z

MxR

λλ

=

Note that for this case, the lines of action of both A and B intersect the x axis.

PROBLEM 3.141 CONTINUED

Case 2: Let 0yB = to satisfy Equation (4)

Now Equation (2) y

RAλ

=

Equation (1) xx

yB R λ

λ

= −

Equation (3) zz

yB R λ

λ

= −

Equation (6) 0 which requires 0yaA aλ = =

Substitution into Equation (5)

or x zz x y

y y

MM z R x R x zR

λ λ λ λ λλ λ

= − − − − =

This last expression is the equation for the line of action of force B.

In summary

Ay

Rλ

=

A λ

( )x zy

R λ λλ

= − −

B i k

Assuming that , , 0,x y zλ λ λ > the equivalent force system is as shown below.

Note that the component of A in the xz plane is parallel to B.

PROBLEM 3.142 A worker tries to move a rock by applying a 360-N force to a steel bar as shown. (a) Replace that force with an equivalent force-couple system at D. (b) Two workers attempt to move the same rock by applying a vertical force at A and another force at D. Determine these two forces if they are to be equivalent to the single force of part a.

SOLUTION

(a) Have ( ) ( ) ( ): 360 N sin 40 cos40 231.40 N 275.78 NΣ − ° − ° = − − =F i j i j F

or 360 N=F 50°

Have /:D B DΣ =M r R M×

where ( ) ( )/ 0.65 m cos30 0.65 m sin 30B D = − ° + ° r i j

( ) ( )0.56292 m 0.32500 m= − +i j

( ) 0.56292 0.32500 0 N m 155.240 75.206 N m231.40 275.78 0

∴ = − ⋅ = + ⋅ − −

i j kM k

( )230.45 N m= ⋅ k or 230 N m= ⋅M

(b) Have /:D A D AΣ =M M r F×

where ( ) ( )/ 1.05 m cos30 1.05 m sin 30A D = − ° + ° r i j

( ) ( )0.90933 m 0.52500 m= − +i j

PROBLEM 3.142 CONTINUED

[ ] 0.90933 0.52500 0 N m 230.45 N m0 1 0

AF∴ − ⋅ = ⋅−

i j kk

or ( )0.90933 230.45AF =k k

253.42 NAF∴ = or 253 NA =F

Have : A DΣ = +F F F F

( ) ( ) ( ) ( )231.40 N 275.78 N 253.42 N cos sinDF θ θ− − = − + − −i j j i j

From : 231.40 N cosDF θ=i (1)

: 22.36 N sinDF θ=j (2)

Equation (2) divided by Equation (1)

tan 0.096629θ =

5.5193θ∴ = ° or 5.52θ = °

Substitution into Equation (1)

231.40 232.48 Ncos5.5193DF = =

°

or 232 ND =F 5.52°

PROBLEM 3.143 A worker tries to move a rock by applying a 360-N force to a steel bar as shown. If two workers attempt to move the same rock by applying a force at A and a parallel force at C, determine these two forces so that they will be equivalent to the single 360-N force shown in the figure.

SOLUTION

Have : A CΣ = +F R F F

( ) ( ) ( ) ( )360 N sin 40 360 N cos 40 sin cosA C A CF F F Fθ θ − ° − ° = − + − + i j i j

From ( ) ( ): 360 N sin 40 sinA CF F θ° = +i (1)

( ) ( ): 360 N cos 40 cosA CF F θ° = +j (2)

Dividing Equation (1) by Equation (2),

tan 40 tanθ° =

40θ∴ = °

Substituting 40θ = ° into Equation (1),

360 NA CF F+ = (3)

Have / /:C B C A C AΣ =M r R r F× ×

where ( )( ) ( ) ( )/ 0.35 m cos30 sin 30 0.30311 m 0.175 mB C = − ° + ° = − +r i j i j

PROBLEM 3.143 CONTINUED

( )( ) ( ) ( )360 N sin40 cos 40 231.40 N 275.78 N= − ° − ° = − −R i j i j

( )( ) ( ) ( )/ 0.75 m cos30 sin 30 0.64952 m 0.375 mA C = − ° + = − +r i j i j

( ) ( )sin 40 cos40 0.64279 0.76604A A AF F= − ° − ° = − −F i j i j

0.30311 0.175 0 N m 0.64952 0.375 0 N m231.40 275.78 0 0.64279 0.76604 0

AF∴ − ⋅ = − ⋅− − − −

i j k i j k

( )83.592 40.495 0.49756 0.24105 AF+ = +

168.002 NAF∴ = or 168.0 NAF =

Substituting into Equation (3),

360 168.002 191.998 NCF = − = or 192.0 NCF =

or 168.0 NA =F 50°

192.0 NC =F 50°

PROBLEM 3.144 A force and a couple are applied as shown to the end of a cantilever beam. (a) Replace this system with a single force F applied at point C, and determine the distance d from C to a line drawn through points D and E. (b) Solve part a if the directions of the two 360-N forces are reversed.

SOLUTION

(a)

(b)

(a) Have ( ) ( ) ( ): 360 N 360 N 600 NΣ = − −F F j j k

or ( )600 N= −F k

and ( )( ) ( )( ): 360 N 0.15 m 600 NDM dΣ =

0.09 md∴ =

or 90.0 mm below d ED=

(b) Have from part a ( )600 N= −F k

and ( )( ) ( )( ): 360 N 0.15 m 600 NDM dΣ − = −

0.09 md∴ =

or 90.0 mm above d ED=

PROBLEM 3.145 A crate of mass 80 kg is held in the position shown. Determine (a) the moment produced by the weight W of the crate about E, (b) the smallest force applied at B which creates a moment of equal magnitude and opposite sense about E.

SOLUTION

(a) By definition ( )280 kg 9.81 m/s 784.8 NW mg= = =

Have ( )( ): 784.8 N 0.25 mE EM MΣ =

196.2 N mE∴ = ⋅M

(b) For the force at B to be the smallest, resulting in a moment ( )EM about E, the line of action of force BF must be perpendicular to the line connecting E to B. The sense of BF must be such that the force produces a counterclockwise moment about E.

Note: ( ) ( )2 20.85 m 0.5 m 0.98615 md = + =

Have ( ): 196.2 N m 0.98615 mE BM FΣ ⋅ =

198.954 NBF∴ =

and 1 0.85 mtan 59.5340.5 m

θ − = = °

or 199.0 NB =F 59.5°

PROBLEM 3.146 A crate of mass 80 kg is held in the position shown. Determine (a) the moment produced by the weight W of the crate about E, (b) the smallest force applied at A which creates a moment of equal magnitude and opposite sense about E, (c) the magnitude, sense, and point of application on the bottom of the crate of the smallest vertical force which creates a moment of equal magnitude and opposite sense about E.

SOLUTION

(a) By definition ( )280 kg 9.81 m/s 784.8 NW mg= = =

Have ( )( ): 784.8 N 0.25 mE EM MΣ =

196.2 N mE∴ = ⋅M

(b) For the force at A to be the smallest, resulting in a moment about E, the line of action of force AF must be perpendicular to the line connecting E to A. The sense of AF must be such that the force produces a counterclockwise moment about E.

Note: ( ) ( )2 20.35 m 0.5 m 0.61033 md = + =

Have ( ): 196.2 N m 0.61033 mE AM FΣ ⋅ =

321.47 NAF∴ =

and 1 0.35 mtan 34.9920.5 m

θ − = = °

or 321 NA =F 35.0°

(c) The smallest force acting on the bottom of the crate resulting in a moment about E will be located at the point on the bottom of the crate farthest from E and acting perpendicular to line CED. The sense of the force will be such as to produce a counterclockwise moment about E. A force acting vertically upward at D satisfies these conditions.

PROBLEM 3.146 CONTINUED

Have /:E E D E DΣ =M M r F×

( ) ( ) ( )196.2 N m 0.85 m DF⋅ =k i j×

( ) ( )196.2 N m 0.85 DF⋅ =k k

230.82 NDF∴ =

or 231 ND =F

PROBLEM 3.147 A farmer uses cables and winch pullers B and E to plumb one side of a small barn. Knowing that the sum of the moments about the x axis of the forces exerted by the cables on the barn at points A and D is equal to 4728 lb ft,⋅ determine the magnitude of DET when 255 lb.AB =T

SOLUTION

The moment about the x-axis due to the two cable forces can be found using the z-components of each force acting at their intersection with the xy-plane (A and D). The x-components of the forces are parallel to the x-axis, and the y-components of the forces intersect the x-axis. Therefore, neither the x or y components produce a moment about the x-axis.

Have ( ) ( ) ( ) ( ):x AB A DE D xz zM T y T y MΣ + =

where ( ) ( )AB AB AB ABzT T= =k T k⋅ ⋅ λ

12 12255 lb 180 lb17

− − + = =

i j kk ⋅

( ) ( )DE DE DE DEzT T= =k T k⋅ ⋅ λ

1.5 14 12 0.6486518.5DE DET T

− + = =

i j kk ⋅

12 ftAy =

14 ftDy =

4728 lb ftxM = ⋅

( )( ) ( )( ) 180 lb 12 ft 0.64865 14 ft 4728 lb ftDET∴ + = ⋅

and 282.79 lbDET =

or 283 lbDET =

PROBLEM 3.148 Solve Problem 3.147 when the tension in cable AB is 306 lb.

Problem 3.147: A farmer uses cables and winch pullers B and E to plumb one side of a small barn. Knowing that the sum of the moments about the x axis of the forces exerted by the cables on the barn at points A and D is equal to 4728 lb ft,⋅ determine the magnitude of DET when

255 lb.AB =T

SOLUTION

The moment about the x-axis due to the two cable forces can be found using the z components of each force acting at the intersection with the xy plane (A and D). The x components of the forces are parallel to the x axis, and the y components of the forces intersect the x axis. Therefore, neither the x or y components produce a moment about the x axis.

Have ( ) ( ) ( ) ( ):x AB A DE D xz zM T y T y MΣ + =

where ( ) ( )AB AB AB ABzT T= =k T k⋅ ⋅ λ

12 12306 lb 216 lb17

− − + = =

i j kk ⋅

( ) ( )DE DE DE DEzT T= =k T k⋅ ⋅ λ

1.5 14 12 0.6486518.5DE DET T

− + = =

i j kk ⋅

12 ftAy =

14 ftDy =

4728 lb ftxM = ⋅

( )( ) ( )( ) 216 lb 12 ft 0.64865 14 ft 4728 lb ftDET∴ + = ⋅

and 235.21 lbDET =

or 235 lbDET =

PROBLEM 3.149 As an adjustable brace BC is used to bring a wall into plumb, the force-couple system shown is exerted on the wall. Replace this force-couple system with an equivalent force-couple system at A knowing that

21.2 lbR = and 13.25 lb ft.M = ⋅

SOLUTION

Have : A BCΣ = =F R R Rλ

where ( ) ( ) ( )42 in. 96 in. 16 in.106 in.BC

− −=

i j kλ

( )21.2 lb 42 96 16106A∴ = − −R i j k

( ) ( ) ( )or 8.40 lb 19.20 lb 3.20 lbA = − −R i j k

Have /:A C A AΣ + =M r R M M×

where ( ) ( ) ( )/142 in. 48 in. 42 48 ft

12C A = + = +r i k i k

( ) ( )3.5 ft 4.0 ft= +i k

( ) ( ) ( )8.40 lb 19.20 lb 3.20 lb= − −R i j k

BCM= −M λ

( )42 96 16 13.25 lb ft106

− + += ⋅

i j k

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

PROBLEM 3.149 CONTINUED

Then ( )3.5 0 4.0 lb ft 5.25 12 2 lb ft8.40 19.20 3.20

A⋅ + − + + ⋅ =− −

i j ki j k M

( ) ( ) ( ) 71.55 lb ft 56.80 lb ft 65.20 lb ftA∴ = ⋅ + ⋅ − ⋅M i j k

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

PROBLEM 3.150 Two parallel 60-N forces are applied to a lever as shown. Determine the moment of the couple formed by the two forces (a) by resolving each force into horizontal and vertical components and adding the moments of the two resulting couples, (b) by using the perpendicular distance between the two forces, (c) by summing the moments of the two forces about point A.

SOLUTION

(a) Have 1 2:B x yd C d CΣ − + =M M

where ( )1 0.360 m sin 55 0.29489 md = ° =

( )2 0.360 m cos55 0.20649 md = ° =

( )60 N cos 20 56.382 NxC = ° =

( )60 N sin 20 20.521 NyC = ° =

( )( ) ( )( ) ( ) 0.29489 m 56.382 N 0.20649 m 20.521 N 12.3893 N m∴ = − + = − ⋅M k k k

or 12.39 N m= ⋅M

(b) Have ( ) ( ) ( ) ( )60 N 0.360 m sin 55 20Fd = − = ° − ° − M k k

( )12.3893 N m= − ⋅ k

or 12.39 N m= ⋅M

PROBLEM 3.150 CONTINUED

(c) Have ( ) / /:A A B A B C A CΣ Σ = + =M r F r F r F M× × ×

( )( ) ( )( ) 0.520 m 60 N cos55 sin 55 0 0.880 m 60 N cos55 sin 55 0cos 20 sin 20 0 cos 20 sin 20 0

M∴ = ° ° + ° °− ° − ° ° °

i j k i j k

( ) ( )17.8956 N m 30.285 N m 12.3892 N m= ⋅ − ⋅ = − ⋅k k

or 12.39 N m= ⋅M

PROBLEM 3.151 A 32-lb motor is mounted on the floor. Find the resultant of the weight and the forces exerted on the belt, and determine where the line of action of the resultant intersects the floor.

SOLUTION

Have ( ) ( ) ( )( ): 60 lb 32 lb 140 lb cos30 sin 30Σ − + ° + ° =F i j i j R

( ) ( ) 181.244 lb 38.0 lb∴ = +R i j

or 185.2 lb=R 11.84°

Have :O O yM M xRΣ Σ =

( ) ( ) ( ) ( ) 140 lb cos 30 4 2 cos 30 in. 140 lb sin 30 2 in. sin 30 ∴ − ° + ° − ° °

( )( ) ( )60 lb 2 in. 38.0 lbx− =

( )1 694.97 70.0 120 in.38.0

x = − − −

and 23.289 in.x = −

Or, resultant intersects the base (x axis) 23.3 in. to the left of the vertical centerline (y axis) of the motor.

PROBLEM 3.152 To loosen a frozen valve, a force F of magnitude 70 lb is applied to the handle of the valve. Knowing that 25 ,θ = ° 61 lb ft,xM = − ⋅ and

43 lb ft,zM = − ⋅ determine θ and d.

SOLUTION

Have /:O A O OΣ =M r F M×

where ( ) ( ) ( )/ 4 in. 11 in.A O d= − + −r i j k

( )cos cos sin cos sinF θ φ θ θ φ= − +F i j k

For 70 lb, 25F θ= = °

( ) ( ) ( )70 lb 0.90631cos 0.42262 0.90631sinφ φ = − + F i j k

( ) 70 lb 4 11 in.0.90631cos 0.42262 0.90631sin

O dφ φ

∴ = − −− −

i j kM

( ) ( ) ( )70 lb 9.9694sin 0.42262 0.90631 cos 3.6252sind dφ φ φ= − + − + i j

( )1.69048 9.9694cos in.φ + − k

and ( )( ) ( )( )70 lb 9.9694sin 0.42262 in. 61 lb ft 12 in./ftxM dφ= − = − ⋅ (1)

( )( )70 lb 0.90631 cos 3.6252sin in.yM d φ φ= − + (2)

( )( ) ( )70 lb 1.69048 9.9694cos in. 43 lb ft 12 in./ftzM φ= − = − ⋅ (3)

PROBLEM 3.152 CONTINUED

From Equation (3)

1 634.33cos 24.636697.86

φ − = = °

or 24.6φ = °

From Equation (1)

1022.90 34.577 in.29.583

d = =

or 34.6 in.d =

PROBLEM 3.153 When a force F is applied to the handle of the valve shown, its moments about the x and z axes are, respectively, 77 lb ftxM = − ⋅ and

81 lb ft.zM = − ⋅ For 27 in.,d = determine the moment yM of F about the y axis.

SOLUTION

Have /:O A O OΣ =M r F M×

where ( ) ( ) ( )/ 4 in. 11 in. 27 in.A O = − + −r i j k

( )cos cos sin cos sinF θ φ θ θ φ= − +F i j k

4 11 27 lb in.cos cos sin cos sin

O Fθ φ θ θ φ

∴ = − − ⋅−

i j kM

( ) ( )11cos sin 27sin 27cos cos 4cos sinF θ φ θ θ φ θ φ= − + − + i j

( ) ( )4sin 11cos cos lb in.θ θ φ + − ⋅k

and ( )( )11cos sin 27sin lb in.xM F θ φ θ= − ⋅ (1)

( )( )27cos cos 4cos sin lb in.yM F θ φ θ φ= − + ⋅ (2)

( )( )4sin 11cos cos lb in.zM F θ θ φ= − ⋅ (3)

Now, Equation (1) 1cos sin 27sin11

xMF

θ φ θ = +

(4)

and Equation (3) 1cos cos 4sin11

zMF

θ φ θ = −

(5)

Substituting Equations (4) and (5) into Equation (2),

1 127 4sin 4 27sin11 11

z xy

M MM FF F

θ θ = − − + +

or ( )1 27 411y z xM M M= +

PROBLEM 3.153 CONTINUED

Noting that the ratios 2711

and 411

are the ratios of lengths, have

( ) ( )27 481 lb ft 77 lb ft 226.82 lb ft11 11yM = − ⋅ + − ⋅ = ⋅

or 227 lb ftyM = − ⋅