Beer & Johnston 9Ed Resolução do Capítulo 3 Corpos Rígidos

Prévia do material em texto

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
 
W 
 
 
 
 
 
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 54.819
86 mm
α θ − −   = = = = °      
y
x
 
Then min=A ABM r P 
or min = A
AB
MP
r
 
 
2.20 N m 1000 mm
149.265 mm 1 m
⋅  =    
 14.7389 N= 
min 14.74 N∴ =P 54.8° 
 or min 14.74 N=P 35.2°W 
 
 
 
 
 
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 54.819
86 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∴ = ° °W 
 
 
 
 
 
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
 
W 
 
 
 
 
 
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
 
W 
 
 
 
 
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
 
W 
(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 =
 
W 
(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°W 
 
 
 
 
 
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
 
W 
(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 34.095
6.5 ft
φ −  = = °   
90 90 34.095 55.905θ φ= ° − = ° − ° = ° 
 or ( )
min 36.0 lbC =F 55.9°W 
 
 
 
 
 
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
 
W 
(b) Have /A C A C=M r F× 
 or =A CM xF 
282.56 lb ft 43.471 lb
6.5 ft
A
C
MF
x
⋅∴ = = = 
 or 43.5 lbC =F
 
W 
(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 = + − = 
( )
min
282.56 lb ft
 42.627 lb
6.6287 ft
⋅∴ = = =AB MF d 
 and 1 6.5 fttan 78.690
4.4 ft 3.1 ft
θ −  = = ° −  
 or ( )
min 42.6 lb=BF 78.7°W 
 
 
 
 
 
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 mmcos
244.48 mm
θ = 
 
46.6 mm
sin
244.48 mm
θ = 
and cos sinCB CB CBF Fθ θ= −F i j 
 ( ) ( )125 N 240 mm 46.6 mm
244.48 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 46.6
244.48A
 = − − M i j i j× 
 ( ) ( )27878 N mm 27.878 N m= ⋅ = ⋅k k 
 or 27.9 N mA = ⋅M
 
W 
 
 
 
 
 
 
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 mmcos
376.25 mm
θ = 152.4 mmsin
376.25 mm
θ = 
and ( ) ( )cos sinCB CB CBF Fθ θ= −F i j 
 ( ) ( )125 N 344 mm 152.4 mm
376.25 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 344 152.4
376.25A
 = − − M i j i j× 
 ( )30770 N mm= ⋅ k 
 ( )30.770 N m= ⋅ k 
 or 30.8 N mA = ⋅M
 
WPROBLEM 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. 5
76 in. 8 in. 12
EC = =+ 
 Then ( )12
13ABx AB
T T= 
 ( )12 260 lb 240 lb
13
= = 
 and ( )5 260 lb 100 lb
13ABy
T = = 
 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
 
W 
(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
 
W 
 
 
 
 
 
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. 7
112 in. 8 in. 24
EC = =+ 
Then 24
25ABx AB
T T= 
and 7
25ABy AB
T T= 
Have ( ) ( )D ABx AByM T y T x= + 
( ) ( )24 77840 lb in. 0 112 in.
25 25AB AB
T T∴ ⋅ = + 
250 lbABT = 
 or 250 lbABT = W 
 
 
 
 
 
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 m
sin
0.24 1.05 md
θ =
+ +
 
 ( ) ( ) ( )2 2
1.051152 N m 2880 N
0.24 1.05
d
d
  ∴ ⋅ =  + +  
 
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 = W 
 
 
 
 
 
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 N
78 86
BxF = =+
 
 ( ) ( ) ( )2 2
86 580 N 429.62 N
78 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 
 
W 
 
 
 
 
 
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 sin
2 2
α β α β α β∴ = + + − W 
 
 
 
 
 
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 A
AB
B A
= r
r
λ 
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 33
650 mm 65AB
− +∴ = = − +i j i 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 mm
65
d  ∴ = − − − + = i j i j× 
 84.0 mmd = W 
 
 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 B
A 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 10
2 6 5
= − = − + − + + − = − + +
− −
i j k
A 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 kλ or ( )1 4 7 10
165
= − + +i j kλ W 
(b) Have = A B
A 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 3
6 3 2
= − = − + − + − + = − + −
− −
i j k
A 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 kλ or ( )1 2 2 3
17
= − + −i j kλ W 
 
 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 in
3 4 2
 = − = + + − + + 
−
i j k
P 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 = W 
(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 in
2 5 3
 = − = − − + − + − − 
−
i j k
P 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 = W 
 
 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 m
5 2 3
O  ∴ = − − ⋅ = − − + − + − − ⋅ 
− −
i j k
M 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 W 
(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 m
5 2 3
O  ∴ = − ⋅ = + + − + + + ⋅ 
− −
i j k
M 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 W 
(c) Have O =Mr 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 m
5 2 3
O  ∴ = − ⋅ = − + − + − + ⋅ 
− −
i j k
M i j k 
or 0O =M W 
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 ft
1.5 3 2
O  = − ⋅ = − + − + + − ⋅ 
− −
i j k
M i j k 
 or ( ) ( ) ( )4 lb ft 2 lb ft 6 lb ftO = − ⋅ + ⋅ + ⋅M i j kW 
(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 ft
1.5 3 2
O  = − ⋅ = − + − + + − ⋅ 
− −
i j k
M i j k 
 or 0O =M W 
 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 ft
1.5 3 2
O  = − ⋅ = − + − + + − ⋅ 
− −
i j k
M i j k 
 or ( ) ( ) ( )19 lb ft 2.5 lb ft 10.5 lb ftO = − ⋅ − ⋅ + ⋅M i j kW 
 
 
 
 
 
 
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 m
777 N
0.9 8.4 7.2 m
AB BA ABT
− − += =
+ +
i j k
T λ 
( ) ( ) ( )
( ) ( ) ( ) ( )2 2 2
5.1 m 8.4 m 1.2 m
990 N
5.1 8.4 1.2 m
BC BC BCT
− += =
+ +
i j k
T 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 m
447 1428 624
O = ⋅ = ⋅ − ⋅
−
i j k
M i k 
or ( ) ( )5.24 kN m 3.75 kN mO = ⋅ − ⋅M i k W 
 
 
 
 
 
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 A O=M r R× 
where ( ) ( ) ( )/ 0 m 9 m 1 mA O = + +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 m
124 N
1.5 m 9 m 1.8 m
T
 − + = =  + +  
i j k
T 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 m
102.116 141.532 24
O = ⋅
− −
i j k
M 
 ( ) ( ) ( )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 kW 
 
 
 
 
 
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 k
M 
 ( ) ( ) ( )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 W 
 
 
 
 
 
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 A C BA=M r F× 
where ( ) ( ) ( )/ 0.96 m 0.12 m 0.72 mA C = − +r i j k 
and BA BA BAF=F λ 
 
( ) ( ) ( )
( ) ( ) ( ) ( )2 2 2
0.1 m 1.8 m 0.6 m
228 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 m
12.0 216 72
C∴ = − ⋅
− −
i j k
M 
 ( ) ( ) ( )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 kW 
 
 
 
 
 
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 in
48 264 240
A∴ = ⋅ = − ⋅ − ⋅
−
i j k
M i k 
 or ( ) ( )1840 lb ft 368 lb ftA = − ⋅ − ⋅M i k W 
(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 k
T λ 
 ( ) ( ) ( )48 lb 264 lb 240 lb= − + −i j k 
 108 92 0 lb in.
48 264 240
A∴ = ⋅
− −
i j k
M 
 ( ) ( ) ( )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 kW 
 
 
 
 
 
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 xz
x
r 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 m
1.93476 5.6569 5.3157
O∴ = ⋅
− −
i j k
M 
 ( ) ( ) ( )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 kW 
 
 
 
 
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. Knowingthat 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 m
777 N
0.9 8.4 7.2 m
BA BA ABT
− − += =
+ +
i j k
T λ 
 ( ) ( ) ( )63.0 N 588 N 504 N= − − +i j k 
( ) ( )0 8.4 0 N m 4233.6 N m 529.2 N m
63.0 588 504
O∴ = ⋅ = ⋅ + ⋅
− −
i j k
M 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 = W 
 
 
 
 
 
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 m
990 N
5.1 8.4 1.2 m
BC BC BCT
− += =
+ +
i j k
T λ 
 ( ) ( ) ( )510 N 840 N 120 N= − +i j k 
( ) ( ) 0 8.4 0 1008 N m 4284 N m
510 840 120
O∴ = = ⋅ − ⋅
−
i j k
M 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 = W 
 
 
 
 
 
 
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 m
228 N
0.1 1.8 0.6 m
BA BA BAF
− + −= =
+ +
i j k
F λ 
 ( ) ( ) ( )12.0 N 216 N 72 N= − + −i j k 
 0 0.12 0.72 N m
12.0 216 72
D∴ = − ⋅
− −
i j k
M 
 ( ) ( ) ( )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 = W 
 
 
 
 
 
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 A C BA=M r F× 
( ) ( ) ( )/ 0.96 m 0.12 m 0.72 mA C = − +r i j k 
( ) ( ) ( )( )
( ) ( ) ( ) ( )2 2 2
0.1 m 1.8 m 0.6
228 N
0.1 1.8 0.6 m
BA BA BAF
− + −= =
+ +
i j k
F λ 
 ( ) ( ) ( )12.0 N 216 N 72 N= − + −i j k 
 0.96 0.12 0.72 N m
12.0 216 72
C∴ = − ⋅
− −
i j k
M 
 ( ) ( ) ( )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 = W 
 
 
 
 
 
 
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 k
T λ 
 ( ) ( ) ( )48 lb 264 lb 240 lb= + −i j k 
 0 92 0 N m
48 264 240
A∴ = ⋅
−
i j k
M 
 ( ) ( )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 = W 
 
 
 
 
 
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 k
M 
 ( ) ( ) ( )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 = W 
 
 
 
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 k
T λ 
 ( ) ( ) ( )48 lb 264 lb 240 lb= + −i j k 
 108 92 0 lb in.
48 264 240
B∴ = − ⋅
−
i j k
M 
 ( ) ( ) ( )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 = W 
 
 
 
 
 
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 A C 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 m
ABF F
+ −= =
+ +
i j k
F λ 
 ( ) ( ) ( )0.57437 0.50257 0.64616F= + −i j k 
( ) ( ) ( )/ 1 m 2.8 m 3 mA C a= − − −r i j k 
 1 2.8 3
0.57437 0.50257 0.64616
C a F∴ = − −
−
i j k
M 
 ( ) ( )0.30154 0.50257 2.3693 0.57437a a= + + − i j 
 ]2.1108 F+ k 
Since ( )22 2/ /or C A C A C 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 = W 
 
 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⋅ W 
( ) ( )7 2 5 8 9= − + + −P S i j k i j k⋅ ⋅ 
 ( )( ) ( )( ) ( )( )7 8 2 1 5 9= + − + − 
 9= 
 or 9=P S⋅ W( ) ( )3 4 6 8 9= − − + + −Q S i j k i j k⋅ ⋅ 
 ( )( ) ( )( ) ( )( )3 8 4 1 6 9= − + − + − 
 82= − 
 or 82= −Q S⋅ W 
 
 
 
 
 
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 cos
2 2
α β α β α β= + + − W 
 
 
 
 
 
 
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 = °W 
 
 
 
 
 
 
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 0.77778
8.1
θ = = 
 38.942θ = ° 
 or 38.9θ = °W 
 
 
 
 
 
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 12
20.516 4.5 12 m
BC
− −= = − −
+ +
i j k
i j kλ 
 
( ) ( ) ( )
( ) ( ) ( ) ( )2 2 2
7 m 6 m 6 m 1 7 6 6
11.07 6 6 m
EF
− − += = − − +
+ +
i j k
i j kλ 
( ) ( )16 4.5 12 7 6 6
cos
20.5 11.0
θ− − − − +∴ =i j k i j k⋅ 
( )( ) ( )( ) ( )( ) ( )( )16 7 4.5 6 12 6 20.5 11.0 cosθ− + − − + − = 
and 1 157cos 134.125
225.5
θ − − = = °   
 or 134.1θ = °W 
(b) By definition ( ) cosEF EFBCT T θ= 
 ( )330 N cos134.125= ° 
 229.26 N= − 
 or ( ) 230 NEF BCT = − W 
 
 
 
 
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 12
20.516 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 4.875
11.1258 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 4.875
cos
20.5 11.25BC EG
θ+ − − + −∴ = =λ ⋅ λ 
and 1 96.5cos 64.967
228.06
θ −  = = °   
 or 65.0θ = °W 
(b) By definition ( ) cosEG EGBCT T θ= 
 ( )445 N cos64.967= ° 
 188.295 N= 
 or ( ) 188.3 NEG BCT = W 
 
 
 
 
 
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 m
OA
+ −=
+ +
i j kλ 
 
2 2 1
3 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 1
3
 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 m
PC
+ +=
+ +
i j kλ 
 0.27037 0.59481 0.75703= + +i j k 
( )2 2 1 0.27037 0.59481 0.75703 cos
3 3 3
θ ∴ + − + + =  i j k i j k⋅ 
 and ( )1cos 0.32445 71.068θ −= = ° 
 or 71.1θ = °W 
(b) ( ) ( )cos 30 N cos71.068PC PCOAT T θ= = ° 
 ( ) 9.7334 NPC OAT = 
 or ( ) 9.73 NPC OAT = W 
 
 
 
 
 
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 m
OA
+ −=
+ +
i j kλ 
 
2 2 1
3 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 0
3 3 3
x y z   ∴ + − − + − + − =    i j k i j k⋅ (1) 
Since ( )/ 2 2 ,3OPP O OA OP
dd= = + −r i j kλ 
Then 
,
2 2 1
,
3 3 3OP OP OP
x 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 0
3 3 3 3OP OP OP
d d d      + − − + − + + =            i j k i j k⋅ 
or 3 0.36 0.60 0.24 0.72OPd = + − = 
 0.24 mOPd∴ = 
 or 240 mmOPd = W 
 
 
 
 
 
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 3
7 1 2
3 2 4 in 14 168 20 3 36 20 in
5 6 1
− −
= − = − + + − + −
− −
 
 
3187 in= 
 
3or Volume 187 in= W 
(b) ( )Vol = P Q S⋅ × 
 ( )3 3
1 2 1
8 1 9 in 1 27 36 16 24 2 in
2 3 1
−
= − − = − − + + + − 
 
346 in= 
 or 3Volume 46 in= W 
 
 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 2
1 3 5 12 40 60 2 2 18
6 2 1
z
z
P
O P
−
∴ = − = − + − − + +
− −
 
so that 34 1.70
20z
P = = 
 or 1.700zP = WPROBLEM 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 ( )/OEDE E D
DE
T
d
=T r 
 ( )54 N 0.360 0.720 0.720
1.08
= + −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 m
18.0 36.0 36.0
A = ⋅
−
i j k
M 
 
 
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 = − ⋅
 
W 
 
 
 
 
 
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
CE
T
d
= rT 
 
( ) ( ) ( ) ( )0.840 m 0.720 m 0.720 m 54 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 m
34.363 29.454 29.454
A = ⋅
− −
i j k
M 
 ( ) ( ) ( )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∴ = − ⋅ = ⋅ = ⋅ W 
 
 
 
 
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 m 56 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 m
3.0 m CD
T
− −= i j k 
 ( )1 2 2
3 CD
T= − −i j k 
 ( ) ( ) ( ) ( ) ( ){ }66 N m 1 m 24 N 16 N 48 N ∴ − ⋅ = − + k j i j k⋅ × 
 ( ) ( )11 m 2 2
3 CD
T
  + − −    k j i j k⋅ × 
or 
266 24
3 CD
T− = − − 
( )3 66 24 N
2CD
T∴ = − 
 or 63.0 NCDT = W 
 
 
 
 
 
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 m
3.5 m BA
T
− −= i j k 
 ( )1.5 3
3.5
BAT= − +i j k 
 CD CD CDT=T λ 
 
( ) ( ) ( ) ( )2 m 1 m 2 m 33 N
3.0 m
− −= i j k 
 ( )22 11 22 N= − −i j k 
 ( ) ( ) ( )212 N m 8 m 1.5 3
3.5
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 BA
T= + 
 
168
18.6667BA
T∴ = 
 or 49.0 NBAT = W 
 
 
 
 
 
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 ft
BA
BA
a
T
d
− −= i j k 
 100 400 6 4 0
6 12
BA
x
BA
TM
d
a
∴ + − =
− −
i j k
i j k 
 ( ) ( ) ( )4 6 96BA
BA
T
a a
d
 = − + − i j k 
From -coefficient:j 100 6AB BAd aT= 100or 6BA BAT da= (1) 
From -coefficient:k 400 96AB BAd T− = − 400or 96BA BAT d= (2) 
Equating Equations (1) and (2) yields ( )( )
100 96
6 400
a = 
 or 4.00 fta = W 
 
 
 
 
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 B
BA BA BA BA
BA
T T
d
= = rT λ 
 
( ) ( ) ( ) ( )6 ft 12 ft 200 lb
BA
a
d
− −= i j k 
 ( )200 6 12
BA
a
d
= − −i j k 
 
0 1 0
200175 lb ft 6 4 0
6 12 BA
d
a
∴ ⋅ =
− −
 
 ( ) 200175 0 6
BA
a
d
 = − −  
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 = W 
 
 
 
 
 
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
PM
M P
φ
φ
−= − 
or 
35
tan 0.175 9.9262
200
φ φ−= = = °− 
 or 9.93φ = °W 
Substituting into Equation (2)φ 
( )200 lb in. cos9.9262 (5 in.)P− ⋅ = − ° 
 40.608 lbP = 
 or 40.6 lbP = W 
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 = °W 
 
 
 
 
 
 
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
PM
M P
φ
φ
−= − 
or 
30
tan
180
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 = ⋅ W 
 
 
 
 
 
PROBLEM 3.53 
The triangular plate ABC is supported by ball-and-socket joints at B and 
D and is held in theposition 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 lb
44 in.AE AE AE
T
 − + = = i j kT 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 = ⋅ W 
 
 
 
 
 
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 CF
T
− −= = i j kT 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 = − ⋅ W 
 
 
 
 
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 4
171.6 0.4 m
DI
−= = −
+
i j
i 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 m 66 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 06 N 5.4 m 0 0 1
17 2 6 9
DIM
−
∴ =
−
 
 ( ) ( )7.8582 0 24 2 0 = + + − −  
 172.879 N m= ⋅ 
 or 172.9 N mDIM = ⋅ W 
 
 
 
 
 
 
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
0.4 17 mDI
−= i jλ 
 ( )1 4
17
= −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 m 61.5 N
12.3 m
− −= i j k 
 ( ) ( ) ( )5 1.2 N 3.6 N 11.7 N = − − i j k 
( ) 4 1 05 N 11.7 m
 0 0 1
17 1.2 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 = − ⋅ W 
 
 
 
 
 
 
 
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 
 ( )
2x
aOA = 
( ) ( ) 1tan30
2 3 2 3z x
a aOA OA  = ° = =   
Since ( ) ( ) ( ) ( )2 2 2 2zx yOA OA OA OA= + + 
or ( ) 22 22
2 2 3y
a a
a OA   = + +       
( ) 2 22 2 
4 12 3y
a aOA a a∴ = − − = 
Then /
2
2 3 2 3A O
a a
a= + +r i j k 
and 1 2 1
2 3 2 3OA
= + +i j kλ 
 BCP=P λ 
 
( ) ( ) ( )sin30 cos30a a P
a
° − °= i k 
 ( )32P= −i k 
/C O a=r i 
 
 
 
 
 
PROBLEM 3.57 CONTINUED 
( )
1 2 1
2 3 2 3
 1 0 0 2
1 0 3
OA
PM a  ∴ =   
−
 
 ( )( )2 1 32 3aP  = − −    
 
2
aP= 
2OA
aPM = W 
 
 
 
 
 
 
 
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 =JJJG JJJG⋅ 
 where 
 From triangle OBC 
 ( )
2x
aOA = 
( ) ( ) 1tan30
2 3 2 3z x
a aOA OA  = ° = =   
( ) 
2 2 3y
a aOA OA   ∴ = + +      i j k
JJJG
 
 and ( ) ( )sin 30 cos30BC a a= ° − °i kJJJG 
 
3
2 2
a a= −i k 
 ( )32a= −i k 
 Then ( ) ( )3 02 22 3ya a aOA  + + − =    i j k i k⋅ 
 or ( ) ( )2 20 0
4 4y
a aOA+ − = 
 0OA BC∴ =JJJG JJJG⋅ 
 so that OA
JJJG
 is perpendicular to .BC
JJJG W 
 
 
 
 
 
 
PROBLEM 3.58 CONTINUED 
(b) Have ,OAM Pd= with P acting along BC and d the 
 perpendicular distance from to .OA BC
JJJG JJJG
 
 From the results of Problem 3.57, 
2OA
PaM = 
 
2
Pa Pd∴ = 
 or 
2
ad = W 
 
 
 
 
 
 
 
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 8
6524 3 ft
AD
+= = +
+
i j
i jλ 
( ) ( )/ 7 ft 3 ftE A = −r i j 
( ) ( )( ) ( )
( ) ( ) ( ) ( )
8
24
2 2 2
8 ft 7 ft 3 ft 3 ft 8 ft
5400 lb
1 4 8 ft
EF EF EFT
 − + + + = =
+ +
i j k
T λ 
 ( ) ( ) ( )600 1 lb 4 lb 8 lb = + + i j k 
( )
8 1 0
600 600
 7 3 0 lb ft 192 56 lb ft
65 651 4 8
ADM∴ = − ⋅ = − − ⋅ 
 18,456.4 lb ft= − ⋅ or 18.46 kip ftADM = − ⋅ W 
 
 
 
 
 
 
 
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 8
6524 3 ft
AD
+= = +
+
i j
i jλ 
( ) ( ) ( ) ( )/ 20 ft 6 ft 2 10 ft 3 ftG A  = − = − r i j i j 
( ) ( )( ) ( )
( ) ( ) ( ) ( )
16
24
2 2 2
16 ft 20 ft 6 ft 3 ft 8 ft
4800 lb
4 8 8 ft
GH GH GHT
 − + + + = =
+ +
i j k
T 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 20
65 651 2 2
ADM
⋅∴ = − = − −
−
 
 26,989 lb ft= − ⋅ or 27.0 kip ftADM = − ⋅ W 
 
 
 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∴ = W 
 
 
 
 
 
PROBLEM 3.62 
In Problem 3.53, determine theperpendicular 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 40
DBM
−
∴ = − ⋅
−
 
 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 = W 
 
 
 
 
 
 
 
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 jλ 
 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 24
DBM
−
∴ = − ⋅
− −
 
 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 = W 
 
 
 
 
 
 
 
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 4
0.4 17 m 17DI
−= = −i j i jλ 
( )/ 5.4 mF I =r k 
( ) ( ) ( ) ( )1.2 m 3.6 m 5.4 m 66 N
6.6 mEF EF EF
T
− += = i j kT λ 
 ( ) ( ) ( )6 2 N 6 N 9 N = − + i j k 
( )( ) 4 1 06 N 5.4 m
 0 0 1 172.879 N m
17 2 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 N
17
 = − − + i j i j k⋅ 
 ( )1 48 36 N
17
= + 
 
84 N
17
= 
 
 
 
 
 
 
PROBLEM 3.64 CONTINUED 
Since ( ) ( )perp. parallelEF EF EF= +T T T 
 ( ) ( ) ( )2 2perp. parallel EF EF EFT T T∴ = − 
 ( ) 22 8466
17
 = −    
 62.777 N= 
Then ( ) ( )perp.DI EFM T d= 
( )( )172.879 N m 62.777 N d⋅ = 
2.7539 md = 
 or 2.75 md = W 
 
 
 
 
 
 
 
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 4
0.4 17 m 17DI
−= = −i j i jλ 
( ) ( )/ 10.9 m 0.8 m 11.7 mG I = − + = −r k k 
( ) ( ) ( ) ( )1.2 m 3.6 m 11.7 m 61.5 N
12.3 mEG EG EG
T
− −= = i j kT λ 
 ( ) ( ) ( )5 1.2 N 3.6 N 11.7 N = − − i j k 
( )( ) 4 1 05 N 11.7 m
 0 0 1
17 1.2 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 N
17
 = − − − i j i j k⋅ 
 ( )5 4.8 3.6 N
17
= + 
 
42 N
17
= 
 
 
 
 
 
PROBLEM 3.65 CONTINUED 
Since ( ) ( )perp. parallelEF EG EG= +T T T 
 ( ) ( ) ( )2 2perp. parallel EG EG EGT T T∴ = − 
 ( ) 22 4261.5
17
 = −    
 60.651 N= 
Then ( ) ( )perp.DI EGM T d= 
( )( )187.286 N m 60.651 N d⋅ = 
3.0880 md = 
 or 3.09 md = W 
 
 
 
 
 
 
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 1
0.36 m 3 3 3OA
+ −= = + −i j k i 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 0.22
3 3
0 1 0
BC
OA 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. 
( )parallel
2 2 1
3 3 3OA BC BCBC
F F = = + −  F i j k jλ ⋅ ⋅ 
 
2
3 BC
F= 
Since ( ) ( )parallel perp.BC BC BC= +F F F 
( ) ( ) ( ) ( ) 22 2 2perp. parallel 2 3BCBC BC BC BC
FF 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 = W 
 
 
 
 
 
 
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 0.132mz = − 
 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 m 54 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 36
yM∴ = = ⋅
−
 
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 = W 
 
 
 
 
 
 
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
 
W 
(b) Have 1 2 0+ =M M 
 or ( )28.40 N m 12 N 0d⋅ − = 
 2 0.700 md∴ = W 
(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 = °W 
 
 
 
 
 
 
 
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 = W 
(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 = W 
(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 = W 
 
 
 
 
 
 
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
 
W 
(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 m
tan 1.33333
0.285 m
θ = = 
 and 53.130θ = ° 
 or 53.1θ = °W 
(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 = W 
 
 
 
 
 
 
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
 
W 
(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
 N
2x
F −= 
45 N
2y
F −= 
( ) 54.0 N m 45 N 0.25 m 1.125 N m 1.125 N ma∴ − ⋅ = − + − ⋅ − ⋅ 
 ( ) ( )45 N 45 N0.2 m 2
2 2
a a
− + − 
0.20 21.20 0.25 0.025 0.025
2 2 2
a a
a= + + + + + + 
 3.1213 0.75858a = 
0.24303 ma = 
 or 243 mma = W 
 
 
 
 
 
 
 
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 = ⋅ W 
( ) ( )8 N m 6 N m 0.8 0.6
10 N m
− ⋅ − ⋅= = = − −⋅
j kM j k
M
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= ° = ° = °W 
 
 
 
 
 
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 m
0 0 100
∴ = − = ⋅ + ⋅
−
i j k
M 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 m
146 N
0.48 0.55 m
− +=
+
i k
 
 ( ) ( )96 N 110 N= − +i k 
 2 0 0.20 0.55 N m
96 0 110
∴ = − ⋅
−
i j k
M 
 ( ) ( ) ( )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 = ⋅ W 
 
62.0 148.8 19.2
162.339
+ += =M i j k
M
λ 
 0.38192 0.91660 0.118271= + +i j k 
cos 0.38192 67.547x xθ θ= ∴ = ° 
 or 67.5xθ = °W 
cos 0.91660 23.566y yθ θ= ∴ = ° 
 or 23.6yθ = °W 
cos 0.118271 83.208z zθ θ= ∴ = ° 
 or 83.2zθ = °W 
 
 
 
 
 
 
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 7
83
= − + +i j k 
( )7 7 lb in. 7 lb in. 5 3 0 21 35 083 835 3 7
⋅ ⋅∴ = − = + +
−
i j k
M i j k 
 ( )0.76835 21 35 lb in.= + ⋅i j 
 
 
 
 
 
 
PROBLEM3.74 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 = ⋅ W 
 
( ) ( )16.1353 lb in. 66.892 lb in.
68.811 lb in.
⋅ + ⋅= = ⋅
i jM
M
λ 
 0.23449 0.97212= +i j 
cos 0.23449 76.438x xθ θ= ∴ = ° 
 or 76.4xθ = °W 
 cos 0.97212 13.5615y yθ θ= ∴ = ° 
 or 13.56yθ = °W 
 cos 0.0zθ = 90zθ∴ = ° 
 or 90.0zθ = °W 
 
 
 
 
 
 
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 / 7 75 3 0 lb in. 0.76835 21 35 lb in.835 3 7D F D
 = = − ⋅ = + ⋅  −
i j k
M 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 = ⋅ W 
18.8647 66.892 50 0.22034 0.78129 0.58399
85.618
− + += = = − + +M i j k i j k
M
λ 
 cos 0.22034xθ = − 102.729xθ∴ = ° or 102.7xθ = °W 
 cos 0.78129 38.621y yθ θ= ∴ = ° or 38.6yθ = °W 
 cos 0.58399 54.268z zθ θ= ∴ = ° or 54.3zθ = °W 
 
 
 
 
 
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 m
0 0 100
C B C= = − = ⋅ + ⋅
−
i j k
M r P i j× 
( ) ( ) ( )2 / 2 0 0.20 0.55 22.0 N m 52.8 N m 19.2 N m
96 0 110
D A E= = − = ⋅ + ⋅ + ⋅
−
i j k
M r P i j k× 
(See Solution to Problem 3.73.) 
( ) ( )/ 0.48 0.20 1.10 231 N m 100.8 N m
0 210 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 = ⋅ W 
293 148.8 120 0.83752 0.42533 0.34301
349.84
+ += = = + +M i j k i j k
M
λ 
 cos 0.83752 33.121x xθ θ= ∴ = ° or 33.1xθ = °W 
 cos 0.42533 64.828y yθ θ= ∴ = ° or 64.8yθ = °W 
 cos 0.34301 69.940z zθ θ= ∴ = ° or 69.9zθ = °W 
 
 
 
 
 
 
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 = ⋅ W 
( )0 3.3185 0.020254
3.3186
− −= = i j kM
M
λ 
 0.99997 0.0061032= − −j k 
 cos 0xθ = 90xθ∴ = ° or 90.0xθ = °W 
 cos 0.99997yθ = − 179.555yθ∴ = ° or 179.6yθ = °W 
 cos 0.0061032 90.349z zθ θ= − ∴ = ° or 90.3zθ = °W 
 
 
 
 
 
 
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°W 
 ( )( ): sin 50A A AM M T dΣ = ° 
 ( ) ( )1000 N sin 50 2.25 m= ° 
 1723.60 N m= ⋅ 
 or 1724 N mA = ⋅M
 
W 
(b) Based on : 1000 NBF F TΣ = = 
 or 1000 NB =F 20°W 
 ( )( ): sin 50B B BM M T dΣ = ° 
 ( ) ( )1000 N sin 50 1.25 m= ° 
 957.56 N m= ⋅ 
 or 958 N mB = ⋅M
 
W 
 
 
 
 
 
 
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 W 
 : B BM M PdΣ = 
 ( )20 lb 5 in.= 
 100 lb in.= ⋅ 
 or 100 lb in.B = ⋅M
 
W 
(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
 
W 
: 0y D CF P PΣ = − 
 25 lbDP∴ = 
 or 25 lbD =P
 
W 
 
 
 
 
 
 
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°W 
: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
 
W 
(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
 
W 
 
 
 
 
 
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 55.991
350
Dy
Dx
P
P
θ − −   = = = °      
 or 626 NDP = 56.0°W 
 
 
 
 
 
 
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 30
Equation 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°W 
From Equation (1), ( )100 N cos30 240cos30CF− + ° = − ° 
140 NCF = 
 or 140.0 NC =F 30°W 
 
 
 
 
 
 
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°W 
( ) ( ): 240 N cos10 0.25 mC A C AM