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3.1
A PROBLEM 13.2N force P is applied to the lever which controls the auger of a snowblower. Determine the moment of P about A when is equal to 30.
SOLUTION
First note Px = P sin = (13.2 N ) sin 30 = 6.60 N
Py = P cos = (13.2 N ) cos 30 = 11.4315 N Noting that the direction of the moment of each force component about A is counterclockwise, M A = xB/ A Py + yB/ A Px = ( 0.086 m )(11.4315 N ) + ( 0.122 m )( 6.60 N ) = 1.78831 N m or M A = 1.788 N 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.
rAB =
(86 mm )2 + (122 mm )2
y
= 149.265 mm
= = tan 1 = tan 1 = 54.819 x 86 mm
Then or
122 mm
M A = rAB Pmin Pmin =
=
MA rAB
2.20 N m 1000 mm 149.265 mm 1 m
= 14.7389 N Pmin = 14.74 N 54.8 or Pmin = 14.74 N 35.2
PROBLEM 3.3
A 13.1N force P is applied to the lever which controls the auger of a snowblower. Determine the value of knowing that the moment of P about A is counterclockwise and has a magnitude of 1.95 N m.
SOLUTION
By definition where and
M A = rB/ A P sin
= + ( 90  ) = tan 1 = 54.819 86 mm
rB/ A =
122 mm
Also Then or or and
(86 mm )2 + (122 mm )2
= 149.265 mm
1.95 N m = ( 0.149265 m )(13.1 N ) sin ( 54.819 + 90  ) sin (144.819  ) = 0.99725 144.819  = 85.752 144.819  = 94.248 = 50.6, 59.1
PROBLEM 3.4
A foot valve for a pneumatic system is hinged at B. Knowing that = 28, determine the moment of the 4lb force about point B by resolving the force into horizontal and vertical components.
SOLUTION
Note that and
=  20 = 28  20 = 8
Fx = ( 4 lb ) cos8 = 3.9611 lb Fy = ( 4 lb ) sin 8 = 0.55669 lb
Also
x = ( 6.5 in.) cos 20 = 6.1080 in. y = ( 6.5 in.) sin 20 = 2.2231 in.
Noting that the direction of the moment of each force component about B is counterclockwise,
M B = xFy + yFx
= ( 6.1080 in.)( 0.55669 lb ) + ( 2.2231 in.)( 3.9611 lb ) = 12.2062 lb in. or M B = 12.21 lb in.
PROBLEM 3.5
A foot valve for a pneumatic system is hinged at B. Knowing that = 28, determine the moment of the 4lb force about point B by resolving the force into components along ABC and in a direction perpendicular to ABC.
SOLUTION
First resolve the 4lb force into components P and Q, where
Q = ( 4.0 lb ) sin 28 = 1.87787 lb
Then
M B = rA/BQ
= ( 6.5 in.)(1.87787 lb ) = 12.2063 lb in. or M B = 12.21 lb in.
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 = 10, (c) the smallest force P which creates the same moment about B.
SOLUTION
(a) Have
M B = rC/B FN
= ( 0.1 m )( 800 N ) = 80.0 N m or M B = 80.0 N m (b) By definition
M B = rA/B P sin
where
= 90  ( 90  70 ) 
= 90  20  10 = 60 80.0 N m = ( 0.45 m ) P sin 60 P = 205.28 N or P = 205 N
(c) For P to be minimum, it must be perpendicular to the line joining points A and B. Thus, P must be directed as shown. Thus or M B = dPmin = rA/B Pmin 80.0 N m = ( 0.45 m ) Pmin Pmin = 177.778 N or Pmin = 177.8 N 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
M A = rB/ A TBF
Noting that the direction of the moment of each force component about A is counterclockwise, M A = xTBFy + yTBFx = ( 6.5 ft )( 45 lb ) sin 60 + ( 4.4 ft  3.1 ft )( 45 lb ) cos 60 = 282.56 lb ft or M A = 283 lb ft (b) Have
M A = rC/ A ( FC )min
For FC to be minimum, it must be perpendicular to the line joining points A and C. M A = d ( FC )min where d = rC/ A =
( 6.5 ft )2 + ( 4.4 ft )2
= 7.8492 ft
282.56 lb ft = ( 7.8492 ft ) ( FC )min
( FC )min
= 35.999 lb
= tan 1 = 34.095 6.5 ft = 90  = 90  34.095 = 55.905
or
4.4 ft
( FC )min
= 36.0 lb
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
M A = rB/ A TBF
Noting that the direction of the moment of each force component about A is counterclockwise, M A = xTBFy + yTBFx = ( 6.5 ft )( 45 lb ) sin 60 + ( 4.4 ft  3.1 ft )( 45 lb ) cos 60 = 282.56 lb ft or M A = 283 lb ft (b) Have or
FC =
M A = rC/ A FC
M A = xFC
MA 282.56 lb ft = = 43.471 lb x 6.5 ft or FC = 43.5 lb
(c) Have
M A = rB/ A ( FB )min
For FB to be minimum, it must be perpendicular to the line joining points A and B. M A = d ( FB )min where and d =
( 6.5 ft )2 + ( 4.4 ft
=
 3.1 ft ) = 6.6287 ft
2
( FB )min
MA 282.56 lb ft = = 42.627 lb d 6.6287 ft 6.5 ft or
= tan 1 = 78.690 4.4 ft  3.1 ft
( FB )min
= 42.6 lb
78.7
PROBLEM 3.9
The tailgate of a car is supported by the hydraulic lift BC. If the lift exerts a 125N force directed along its center line on the ball and socket at B, determine the moment of the force about A.
SOLUTION
First note dCB =
( 240 mm )2 + ( 46.6 mm )2
= 244.48 mm Then cos = sin = and 240 mm 244.48 mm 46.6 mm 244.48 mm
FCB = FCB cos i  FCB sin j
= 125 N ( 240 mm ) i  ( 46.6 mm ) j 244.48 mm M A = rB/ A FCB rB/ A = ( 306 mm ) i  ( 240 mm + 46.6 mm ) j = ( 306 mm ) i  ( 286.6 mm ) j
Now where
Then
125 N M A = ( 306 mm ) i  ( 286.6 mm ) j 244.48 ( 240i  46.6 j) = ( 27878 N mm ) k = ( 27.878 N m ) k or M A = 27.9 N m
PROBLEM 3.10
The tailgate of a car is supported by the hydraulic lift BC. If the lift exerts a 125N force directed along its center line on the ball and socket at B, determine the moment of the force about A.
SOLUTION
First note Then and dCB =
( 344 mm )2 + (152.4 mm )2
344 mm 376.25 mm sin =
= 376.25 mm 152.4 mm 376.25 mm
cos =
FCB = ( FCB cos ) i  ( FCB sin ) j = 125 N ( 344 mm ) i + (152.4 mm ) j 376.25 mm M A = rB/ A FCB rB/ A = ( 410 mm ) i  ( 87.6 mm ) j 125 N M A = ( 410 mm ) i  ( 87.6 mm ) j 376.25 ( 344i  152.4 j) = ( 30770 N mm ) k = ( 30.770 N m ) k or M A = 30.8 N m
Now where Then
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 EC =
35 in. 5 = 76 in. + 8 in. 12
Then
TABx = = TABy =
12 (TAB ) 13 12 ( 260 lb ) = 240 lb 13 5 ( 260 lb ) = 100 lb 13
and Then
M D = TABx ( 35 in.)  TABy ( 8 in.)
= ( 240 lb )( 35 in.)  (100 lb )( 8 in.) = 7600 lb in.
or M D = 7600 lb in.
(b) Have
M D = TABx ( y ) + TABy ( x )
= ( 240 lb )( 0 ) + (100 lb )( 76 in.) = 7600 lb in. or M D = 7600 lb in.
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 a = 8 in., b = 35 in., and d = 112 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 EC =
35 in. 7 = 112 in. + 8 in. 24 24 TAB 25 7 TAB 25
Then
TABx = TABy =
and Have
M D = TABx ( y ) + TABy ( x ) 7840 lb in. = 24 7 TAB ( 0 ) + TAB (112 in.) 25 25
TAB = 250 lb
or TAB = 250 lb
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 a = 0.24 m and b = 1.05 m.
SOLUTION
The minimum value of d can be found based on the equation relating the moment of the force TAB about D:
M D = (TAB max ) y ( d )
where
M D = 1152 N m
(TAB max ) y
Now
= TAB max sin = ( 2880 N ) sin
sin = 1.05 m
(d
+ 0.24 ) + (1.05 ) m
2 2
1152 N m = 2880 N
1.05
( d + 0.24 )2 + (1.05)2
= 2.625d
(d )
or or or
( d + 0.24 )2 + (1.05)2
(d
2 2
+ 0.24 ) + (1.05 ) = 6.8906d 2
5.8906d 2  0.48d  1.1601 = 0
Using the quadratic equation, the minimum values of d are 0.48639 m and 0.40490 m. Since only the positive value applies here, d = 0.48639 m or d = 486 mm
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
M C = rB/C FB
Noting the direction of the moment of each force component about C is clockwise, M C = xFBy + yFBx where x = 144 mm  78 mm = 66 mm y = 86 mm + 108 mm = 194 mm and FBx = 78
( 78)
2
+ ( 86 )
2
( 580 N ) = 389.65 N ( 580 N ) = 429.62 N
FBy =
86
( 78) + (86 )
2
2
M C = ( 66 mm )( 429.62 N ) + (194 mm )( 389.65 N ) = 103947 N mm = 103.947 N m or M C = 103.9 N m
PROBLEM 3.15
Form the vector products B C and B C, where B = B, and use the results obtained to prove the identity sin cos =
1 sin 2
( + ) +
1 sin 2
(  ) .
SOLUTION
First note
B = B ( cos i + sin j) B = B ( cos i  sin j) C = C ( cos i + sin j)
By definition
B C = BC sin (  ) B C = BC sin ( + )
(1) (2)
Now
B C = B ( cos i + sin j) C ( cos i + sin j)
= BC ( cos sin  sin cos ) k (3)
B C = B ( cos i  sin j) C ( cos i + sin j)
= BC ( cos sin + sin cos ) k Equating magnitudes of B C from Equations (1) and (3), sin (  ) = cos sin  sin cos Similarly, equating magnitudes of B C from Equations (2) and (4), sin ( + ) = cos sin + sin cos Adding Equations (5) and (6) sin (  ) + sin ( + ) = 2cos sin sin cos = 1 1 sin ( + ) + sin (  ) 2 2 (6) (4) (5)
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 where d = AB rO/ A AB =
rB/ A rB/ A
and
rB/ A = ( 140 mm  420 mm ) i + 180 mm  ( 150 mm ) j
=  ( 560 mm ) i + ( 330 mm ) j
rB/ A =
( 560 )2 + ( 330 )2 mm = 650 mm
 ( 560 mm ) i + ( 330 mm ) j 1 = ( 56i + 33j) 650 mm 65
AB =
rO/ A = ( 0  x A ) i + ( 0  y A ) j =  ( 420 mm ) i + (150 mm ) j
d = 1 ( 56i  33j)  ( 420 mm ) i + (150 mm ) j = 84.0 mm 65 d = 84.0 mm
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 where
= AB AB
A = 4i  2 j + 3k B = 2i + 6 j  5k
i j k A B = 4 2 3 = (10  18 ) i + ( 6 + 20 ) j + ( 24  4 ) k = 2 ( 4i + 7 j + 10k ) 2 6 5 AB = 2 =
Then
and
( 4 )2 + ( 7 )2 + (10 )2
= 2 165 or =
1 ( 4i + 7 j + 10k ) 165
2 ( 4i + 7 j + 10k ) 2 165
= AB AB
(b) Have where
A = 7i + j  4k B = 6i  3j + 2k i j k A B = 7 1 4 = ( 2  12 ) i + ( 24  14 ) j + ( 21 + 6 ) k = 5 ( 2i + 2 j  3k ) 6 3 2 AB = 5 =
Then
and
( 2 )2 + ( 2 )2 + ( 3)2
= 5 17 or =
1 ( 2i + 2 j  3k ) 17
5 ( 2i + 2 j  3k ) 5 17
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 where
A = PQ
P = ( 8 in.) i + ( 2 in.) j  (1 in.) k Q =  ( 3 in.) i + ( 4 in.) j + ( 2 in.) k i j k P Q = 8 2 1 in 2 = ( 4 + 4 ) i + ( 3  16 ) j + ( 32 + 6 ) k in 2 3 4 2
Then
= 8 in 2 i  13 in 2 j + 38 in 2 k = (b) Have where
(
) (
) (
)
(8)2 + ( 13)2 + ( 38)2 in 2
= 40.951 in 2
A = PQ
or A = 41.0 in 2
P =  ( 3 in.) i + ( 6 in.) j + ( 4 in.) k Q = ( 2 in.) i + ( 5 in.) j  ( 3 in.) k i j k P Q = 3 6 4 in 2 = ( 18  20 ) i + ( 8  9 ) j + ( 15  12 ) k in 2 2 5 3 =  38 in 2 i  1 in 2 j  27 in 2 k =
Then
(
) (
) (
)
( 38)2 + ( 1)2 + ( 27 )2 in 2
= 46.626 in 2
or A = 46.6 in 2
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 where MO = r F F =  (5 N ) i  ( 2 N ) j + (3 N ) k r = ( 4 m ) i  ( 2 m ) j  (1 m ) k i j k = 4 2 1 N m = ( 6  2 ) i + ( 5  12 ) j + ( 8  10 ) k N m 5 2 3 = ( 8i  7 j  18k ) N m or M O =  ( 8 N m ) i  ( 7 N m ) j  (18 N m ) k (b) Have where MO = r F F =  (5 N ) i  ( 2 N ) j + (3 N ) k r =  (8 m ) i + ( 3 m ) j  ( 4 m ) k i j k = 8 3 4 N m = ( 9 + 8 ) i + ( 20 + 24 ) j + (16 + 15 ) k N m 5 2 3 = (17i + 4 j + 31k ) N m or M O = (17 N m ) i + ( 4 N m ) j + ( 31 N m ) k (c) Have where MO = r F F =  (5 N ) i  ( 2 N ) j + (3 N ) k r = ( 7.5 m ) i + ( 3 m ) j  ( 4.5 m ) k
MO
MO
PROBLEM 3.19 CONTINUED
MO i j k = 7.5 3 4.5 N m = ( 9  9 ) i + ( 22.5  22.5) j + ( 15 + 15 ) k N m 5 2 3 or M O = 0 2 r . Therefore, vector F has a line of action This answer is expected since r and F are proportional F = 3 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 where MO = r F F =  (1.5 lb ) i + ( 3 lb ) j + ( 2 lb ) k r = ( 2.5 ft ) i  (1 ft ) j + ( 2 ft ) k Then MO i j k = 2.5 1 2 lb ft = ( 2  6 ) i + ( 3 + 5 ) j + ( 7.5  1.5 ) k lb ft 1.5 3 2 or M O =  ( 4 lb ft ) i + ( 2 lb ft ) j + ( 6 lb ft ) k (b) Have where MO = r F F =  (1.5 lb ) i + ( 3 lb ) j  ( 2 lb ) k r = ( 4.5 ft ) i  ( 9 ft ) j + ( 6 ft ) k Then MO i j k = 4.5 9 6 lb ft = (18  18 ) i + ( 9 + 9 ) j + (13.5  13.5 ) k lb ft 1.5 3 2 or M O = 0 1 r . This answer is expected since r and F are proportional F = 3 Therefore, vector F has a line of action passing through the origin at O. (c) Have where MO = r F F =  (1.5 lb ) i  ( 3 lb ) j  ( 2 lb ) k r = ( 4 ft ) i  (1 ft ) j + ( 7 ft ) k Then MO i j k = 4 1 7 lb ft = ( 2  21) i + ( 10.5 + 8 ) j + (12  1.5 ) k lb ft 1.5 3 2 or M O =  (19 lb ft ) i  ( 2.5 lb ft ) j + (10.5 lb ft ) 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 where
M O = rB/O FB
rB/O = ( 8.4 m ) j
FB = TAB + TBC TAB = BATAB =  ( 0.9 m ) i  ( 8.4 m ) j + ( 7.2 m ) k
( 0.9 )2 + (8.4 )2 + ( 7.2 )2
m
( 777 N )
TBC = BCTBC =
( 5.1 m ) i  (8.4 m ) j + (1.2 m ) k 990 N ( ) ( 5.1)2 + (8.4 )2 + (1.2 )2 m
PROBLEM 3.21 CONTINUED
FB =  ( 63.0 N ) i  ( 588 N ) j + ( 504 N ) k + ( 510 N ) i  ( 840 N ) j + (120 N ) k
= ( 447 N ) i  (1428 N ) j + ( 624 N ) k
and
MO i j k = 0 8.4 0 N m = ( 5241.6 N m ) i  ( 3754.8 N m ) k 447 1428 624 or M O = ( 5.24 kN m ) i  ( 3.75 kN m ) 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 where M O = rA/O R rA/O = ( 0 m ) i + ( 9 m ) j + (1 m ) k R = T1 + T2
T1 =  (124 N ) cos10 i  (124 N ) sin10 j
=  (122.116 N ) i  ( 21.532 N ) j
(1.5 m ) i  ( 9 m ) j + (1.8 m ) k T2 = T2 = 2 2 2 (1.5 m ) + ( 9 m ) + (1.8 m ) (124 N )
= ( 20 N ) i  (120 N ) j + ( 24 N ) k R =  (102.116 N ) i  (141.532 N ) j + ( 24 N ) k i j k = 0 9 1 Nm 102.116 141.532 24 = ( 357.523 N m ) i  (102.116 N m ) j + ( 919.044 N m ) k or M O = ( 358 N m ) i  (102.1 N m ) j + ( 919 N m ) k
MO
PROBLEM 3.23
An 8lb 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 where M A = rC/ A F rC/ A = ( 8.5 in.) i  ( 2.0 in.) j + ( 5.5 in.) k
Fx =  ( 8cos 45 sin12 ) lb
Fy =  ( 8sin 45 ) lb Fz =  ( 8cos 45 cos12 ) lb
F =  (1.17613 lb ) i  ( 5.6569 lb ) j  ( 5.5332 lb ) k i j k MA = 8.5 2.0 5.5 lb in. 1.17613 5.6569 5.5332 = ( 42.179 lb in.) i + ( 40.563 lb in.) j  ( 50.436 lb in.) k or M A = ( 42.2 lb in.) i + ( 40.6 lb in.) j  ( 50.4 lb in.) k
and
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 where and M C = rA/C FBA rA/C = ( 0.96 m ) i  ( 0.12 m ) j + ( 0.72 m ) k FBA = BA FBA
 ( 0.1 m ) i + (1.8 m ) j  ( 0.6 m ) k = 228 N ) ( 2 2 2 ( 0.1) + (1.8) + ( 0.6 ) m
=  (12.0 N ) i + ( 216 N ) j  ( 72 N ) k i j k = 0.96 0.12 0.72 N m 12.0 216 72 =  (146.88 N m ) i + ( 60.480 N m ) j + ( 205.92 N m ) k or M C =  (146.9 N m ) i + ( 60.5 N m ) j + ( 206 N m ) k
MC
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 where TDE = DETDE = M A = rE/ A TDE rE/ A = ( 92 in.) j
( 24 in.) i + (132 in.) j  (120 in.) k 360 lb ( ) ( 24 )2 + (132 )2 + (120 )2 in.
= ( 48 lb ) i + ( 264 lb ) j  ( 240 lb ) k i j k M A = 0 92 0 lb in. =  ( 22, 080 lb in.) i  ( 4416 lb in ) k 48 264 240 or M A =  (1840 lb ft ) i  ( 368 lb ft ) k (b) Have where TCG = CGTCG = M A = rG/ A TCG rG/ A = (108 in.) i + ( 92 in.) j  ( 24 in.) i + (132 in.) j  (120 in.) k
( 24 )2 + (132 )2 + (120 )2 in.
( 360 lb )
=  ( 48 lb ) i + ( 264 lb ) j  ( 240 lb ) k i j k M A = 108 92 0 lb in. 48 264 240 =  ( 22, 080 lb in.) i + ( 25,920 lb in.) j + ( 32,928 lb in.) k or M A =  (1840 lb ft ) i + ( 2160 lb ft ) j + ( 2740 lb ft ) k
PROBLEM 3.26
The arms AB and BC of a desk lamp lie in a vertical plane that forms an angle of 30 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 AB = 450 mm, BC = 325 mm, and line CD is parallel to the z axis.
o
SOLUTION
Have where M O = rC/O FC
( rC/O ) x = ( ABxz + BCxz ) cos 30
ABxz = ( 0.450 m ) sin 45 = 0.31820 m BC xz = ( 0.325 m ) sin 50 = 0.24896 m
( rC/O ) y = (OAy + ABy  BC y ) = 0.150 m + ( 0.450 m ) cos 45
 ( 0.325 m ) cos 50 = 0.25929 m
( rC/O ) z = ( ABxz + BCxz ) sin 30
= ( 0.31820 m + 0.24896 m ) sin 30 = 0.28358 m or
rC/O = ( 0.49118 m ) i + ( 0.25929 m ) j + ( 0.28358 m ) k
( FC ) x
( FC ) y
=  ( 8 N ) cos 45 sin 20 = 1.93476 N =  ( 8 N ) sin 45 = 5.6569 N = ( 8 N ) cos 45 cos 20 = 5.3157 N
( FC ) z
or
FC =  (1.93476 N ) i  ( 5.6569 N ) j + ( 5.3157 N ) k
MO
i j k = 0.49118 0.25929 0.28358 N m 1.93476 5.6569 5.3157
= ( 2.9825 N m ) i  ( 3.1596 N m ) j  ( 2.2769 N m ) k or M O = ( 2.98 N m ) i  ( 3.16 N m ) j  ( 2.28 N m ) 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
 M O  = TBAd
d = perpendicular distance from O to line AB.
O = rB/O TBA rB/O = ( 8.4 m ) j TBA = BATAB =
where Now and
 ( 0.9 m ) i  ( 8.4 m ) j + ( 7.2 m ) k
( 0.9 ) + (8.4 ) + ( 7.2 ) m
2 2 2
( 777 N )
=  ( 63.0 N ) i  ( 588 N ) j + ( 504 N ) k MO =
i j k 0 8.4 0 N m = ( 4233.6 N m ) i + ( 529.2 N m ) k 63.0 588 504
and
 MO  =
( 4233.6 )2 + ( 529.2 )2
d = 5.4911 m
= 4266.5 N m
4266.5 N m = ( 777 N ) d
or
or d = 5.49 m
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 where  M O  = TBC d
d = perpendicular distance from O to line BC.
M O = rB/O TBC rB/O = 8.4 m j TBC = BCTBC =
( 5.1 m ) i  (8.4 m ) j + (1.2 m ) k 990 N ( ) ( 5.1)2 + (8.4 )2 + (1.2 )2 m
= ( 510 N ) i  ( 840 N ) j + (120 N ) k
i j k = 0 8.4 0 = (1008 N m ) i  ( 4284 N m ) k 510 840 120
MO
and
 MO  =
(1008)2 + ( 4284 )2
= 4401.0 N m
4401.0 N m = ( 990 N ) d d = 4.4454 m
or d = 4.45 m
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 where  M D  = FBAd
d = perpendicular distance from D to line AB.
M D = rA/D FBA rA/D =  ( 0.12 m ) j + ( 0.72 m ) k
FBA = BA FBA =
(  ( 0.1 m ) i + (1.8 m ) j  ( 0.6 m ) k ) ( 228 N )
( 0.1)2 + (1.8)2 + ( 0.6 )2
m
=  (12.0 N ) i + ( 216 N ) j  ( 72 N ) k MD
i j k = 0 0.12 0.72 N m 12.0 216 72
=  (146.88 N m ) i  ( 8.64 N m ) j  (1.44 N m ) k
and
MD  =
(146.88)2 + (8.64 )2 + (1.44 )2
147.141 N m = ( 228 N ) d d = 0.64536 m
= 147.141 N m
or d = 0.645 m
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 where  M C  = FBAd
d = perpendicular distance from C to line AB.
M C = rA/C FBA rA/C = ( 0.96 m ) i  ( 0.12 m ) j + ( 0.72 m ) k
FBA = BA FBA =
(  ( 0.1 m ) i + (1.8 m ) j  ( 0.6 ) k ) ( 228 N )
( 0.1)2 + (1.8)2 + ( 0.6 )2
m
=  (12.0 N ) i + ( 216 N ) j  ( 72 N ) k
i j k M C = 0.96 0.12 0.72 N m 12.0 216 72
=  (146.88 N m ) i  ( 60.48 N m ) j + ( 205.92 N m ) k
and
 MC  =
(146.88)2 + ( 60.48)2 + ( 205.92 )2
260.07 N m = ( 228 N ) d d = 1.14064 m
= 260.07 N m
or d = 1.141 m
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 where
M A = TDE d d = perpendicular distance from A to line DE. M A = rE/ A TDE
rE/ A = ( 92 in.) j TDE = DETDE =
( 24 in.) i + (132 in.) j  (120 in.) k 360 lb ( ) ( 24 )2 + (132 )2 + (120 )2 in.
= ( 48 lb ) i + ( 264 lb ) j  ( 240 lb ) k
i j k M A = 0 92 0 Nm 48 264 240 =  ( 22, 080 lb in.) i  ( 4416 lb in.) k
PROBLEM 3.31 CONTINUED
and MA =
( 22, 080 )2 + ( 4416 )2
= 22,517 lb in.
22,517 lb in. = ( 360 lb ) d
d = 62.548 in.
or d = 5.21 ft
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 where
M A = TCG d
d = perpendicular distance from A to line CG.
M A = rG/ A TCG rG/ A = (108 in.) i + ( 92 in.) j
TCG = CGTCG =  ( 24 in.) i + (132 in.) j  (120 in.) k
( 24 )
2
+ (132 ) + (120 ) in.
2 2
( 360 lb )
=  ( 48 lb ) i + ( 264 lb ) j  ( 240 lb ) k i j k M A = 108 92 0 lb in.  48 264 240 =  ( 22, 080 lb in.) i + ( 25,920 lb in.) j + ( 32,928 lb in.) k
and
MA =
( 22, 080 )2 + ( 25,920 )2 + ( 32,928)2
47,367 lb in. = ( 360 lb ) d d = 131.575 in.
= 47,367 lb in.
or d = 10.96 ft
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 where
M B = TDE d d = perpendicular distance from B to line DE. M B = rE/B TDE rE/B =  (108 in.) i + ( 92 in.) j TDE = DETDE =
( 24 in.) i + (132 in.) j  (120 in.) k 360 lb ( ) 2 2 2 24 ) + (132 ) + (120 ) in. (
= ( 48 lb ) i + ( 264 lb ) j  ( 240 lb ) k i j k M B = 108 92 0 lb in. 48 264 240 =  ( 22, 080 lb in.) i  ( 25,920 lb in.) j  ( 32,928 lb in.) k
and
MB =
( 22, 080 )2 + ( 25,920 )2 + ( 32,928)2
47,367 lb in. = ( 360 lb ) d d = 131.575 in.
= 47,367 lb in.
or d = 10.96 ft
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,
M C = rA/C F = F ( d )
where
d = perpendicular distance from C to line AB F = AB F =
(8 m ) i + ( 7 m ) j  ( 9 m ) k F ( 8 )2 + ( 7 )2 + ( 9 )2 m
= F ( 0.57437 ) i + ( 0.50257 ) j  ( 0.64616 ) k rA/C = (1 m ) i  ( 2.8 m ) j  ( a  3 m ) k i j k MC = 1 2.8 3a F 0.57437 0.50257 0.64616
= ( 0.30154 + 0.50257a ) i + ( 2.3693  0.57437a ) j
+ 2.1108k ] F
Since
MC = rA/C F 2
2
or
rA/C F 2 = ( dF )
2
2
( 0.30154 + 0.50257a ) + ( 2.3693  0.57437a ) + ( 2.1108 ) = d 2
2
Setting
d d 2 = 0 to find a to minimize d da 2 ( 0.50257 )( 0.30154 + 0.50257a ) + 2 ( 0.57437 )( 2.3693  0.57437a ) = 0
( )
Solving
a = 2.0761 m
or a = 2.08 m
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
P Q = ( 7i  2 j + 5k ) ( 3i  4 j + 6k )
= ( 7 )( 3) + ( 2 )( 4 ) + ( 5 )( 6 )
= 17
or P Q = 17
P S = ( 7i  2 j + 5k ) ( 8i + j  9k )
= ( 7 )( 8 ) + ( 2 )(1) + ( 5 )( 9 ) =9 or P S = 9 Q S = ( 3i  4 j + 6k ) ( 8i + j  9k ) = ( 3)( 8 ) + ( 4 )(1) + ( 6 )( 9 ) = 82 or Q S = 82
PROBLEM 3.36
Form the scalar products B C and B C, where B = B, and use the results obtained to prove the identity cos cos =
1 2
cos ( + ) + 1 cos (  ) . 2
SOLUTION
By definition B C = BC cos (  ) where
B = B ( cos ) i + ( sin ) j C = C ( cos ) i + ( sin ) j
( B cos )( C cos ) + ( B sin )( C sin ) = BC cos (  ) or By definition B C = BC cos ( + ) where
B = ( cos ) i  ( sin ) j
cos cos + sin sin = cos (  )
(1)
( B cos )( C cos ) + (  B sin )( C sin ) = BC cos ( + ) or cos cos  sin sin = cos ( + ) 2 cos cos = cos (  ) + cos ( + ) or cos cos = 1 1 cos ( + ) + cos (  ) 2 2 (2)
Adding Equations (1) and (2),
PROBLEM 3.37
Consider the volleyball net shown. Determine the angle formed by guy wires AB and AC.
SOLUTION
First note
AB = rB/ A =
( 1.95 m )2 + ( 2.4 m )2 + ( 0.6 m )2
= 3.15 m
AC = rC/ A =
( 0 m )2 + ( 2.4 m )2 + (1.8 m )2
= 3.0 m and rB/ A =  (1.95 m ) i  ( 2.40 m ) j + ( 0.6 m ) k rC/ A =  ( 2.40 m ) j + (1.80 m ) k By definition rB/ A rC/ A = rB/ A rC/ A cos or
( 1.95i  2.40 j + 0.6k ) ( 2.40 j + 1.80k ) = ( 3.15)( 3.0 ) cos ( 1.95)( 0 ) + ( 2.40 )( 2.40 ) + ( 0.6 )(1.8) = 9.45cos
cos = 0.72381
and
= 43.630
or = 43.6
PROBLEM 3.38
Consider the volleyball net shown. Determine the angle formed by guy wires AC and AD.
SOLUTION
First note
AC = rC/ A = AD = rD/ A =
( 2.4 )2 + (1.8)2
m = 3m m = 2.7 m
(1.2 )2 + ( 2.4 )2 + ( 0.3)2
and
rC/ A =  ( 2.4 m ) j + (1.8 m ) k rD/ A = (1.2 m ) i  ( 2.4 m ) j + ( 0.3 m ) k
By definition rC/ A rD/ A = rC/ A rD/ A cos or
( 2.4 j + 1.8k ) (1.2i  2.4 j + 0.3k ) = ( 3)( 2.7 ) cos ( 0 )(1.2 ) + ( 2.4 )( 2.4 ) + (1.8)( 0.3) = 8.1cos
and
cos =
6.3 = 0.77778 8.1
= 38.942
or = 38.9
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 where BC =
BC EF = (1)(1) cos
(16 m ) i  ( 4.5 m ) j  (12 m ) k (16 )2 + ( 4.5)2 + (12 )2 m
 (7 m) i  (6 m) j + (6 m)k
=
1 (16i  4.5j  12k ) 20.5
EF =
( 7 ) 2 + ( 6 )2 + ( 6 ) 2 m
20.5 11.0
=
1 ( 7i  6 j + 6k ) 11.0 = cos
(16i  4.5j  12k ) ( 7i  6 j + 6k )
(16 )( 7 ) + ( 4.5)( 6 ) + ( 12 )( 6 ) = ( 20.5)(11.0 ) cos
and
= cos 1 = 134.125 225.5
or = 134.1
157
(b) By definition
(TEF )BC
= TEF cos = ( 330 N ) cos134.125 = 229.26 N or (TEF ) BC = 230 N
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 where BC =
BC EG = (1)(1) cos
(16 m ) i  ( 4.5 m ) j  (12 m ) k (16 m )2 + ( 4.5)2 + (12 )2 m
=
16i  4.5j  12k 20.5
= 0.78049i  0.21951j  0.58537k EG =
(8 m ) i  ( 6 m ) j + ( 4.875 m ) k (8)2 + ( 6 )2 + ( 4.875)2 m
=
8i  6 j + 4.875k 11.125
= 0.71910i  0.53933j + 0.43820k BC EG = and 16 ( 8 ) + ( 4.5 )( 6 ) + ( 12 )( 4.875 ) = cos ( 20.5)(11.25)
= cos 1 = 64.967 228.06
or = 65.0
96.5
(b) By definition
(TEG )BC
= TEG cos = ( 445 N ) cos 64.967 = 188.295 N or (TEG ) BC = 188.3 N
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 where OA = OA PC = (1)(1) cos
( 0.24 m ) i + ( 0.24 m ) j  ( 0.12 m ) k ( 0.24 )2 + ( 0.24 )2 + ( 0.12 )2 m
2 2 1 i+ j k 3 3 3
=
Knowing that  rA/O  = LOA = 0.36 m and that P is located 0.12 m from O, it follows that the coordinates of P are 1 the coordinates of A. 3 P ( 0.08 m, 0.08 m,  0.040 m ) Then PC =
( 0.10 m ) i + ( 0.22 m ) j + ( 0.28 m ) k ( 0.10 )2 + ( 0.22 )2 + ( 0.28)2 m
= 0.27037i + 0.59481j + 0.75703k
2 1 2 i + j  k ( 0.27037i + 0.59481j + 0.75703k ) = cos 3 3 3 and
= cos 1 ( 0.32445) = 71.068
or = 71.1
(b)
(TPC )OA
= TPC cos = ( 30 N ) cos 71.068
(TPC )OA
= 9.7334 N or (TPC )OA = 9.73 N
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
OA PC = 0 where OA = or OA rC/P = 0
( 0.24 m ) i + ( 0.24 m ) j  ( 0.12 m ) k ( 0.24 )2 + ( 0.24 )2 + ( 0.12 )2 m
2 2 1 i+ j k 3 3 3
=
Letting the coordinates of P be P ( x, y, z ) , we have
rC/P = ( 0.18  x ) i + ( 0.30  y ) j + ( 0.24  z ) k m
2 1 2 i + j  k ( 0.18  x ) i + ( 0.30  y ) j + ( 0.24  z ) k = 0 3 3 3 Since Then
rP/O = OAdOP =
(1)
dOP ( 2i + 2 j  k ) , 3
x=
2 dOP , 3
y =
2 dOP, 3
z =
1 dOP 3
(2)
Substituting the expressions for x, y, and z from Equation (2) into Equation (1), 1 2 2 1 ( 2i + 2 j  k ) 0.18  dOP i + 0.30  dOP j + 0.24 + dOP k = 0 3 3 3 3 or 3dOP = 0.36 + 0.60  0.24 = 0.72 dOP = 0.24 m or dOP = 240 mm
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 )
7 1 2 = 3 2 4 in 3 = ( 14 + 168 + 20  3 + 36  20 ) in 3 5 6 1 = 187 in 3 or Volume = 187 in 3 (b) Vol = P ( Q S ) 1 2 1 = 8 1 9 in 3 = ( 1  27 + 36 + 16 + 24  2 ) in 3 2 3 1 = 46 in 3 or Volume = 46 in 3
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.
P (Q S ) = 0
4 2 Pz O = 1 3 5 = 12 + 40  60  2 + Pz ( 2 + 18 ) 6 2 1 so that Pz = 34 = 1.70 20 or Pz = 1.700
PROBLEM 3.45
The 0.732 1.2m 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 z =
( 0.732 )2  ( 0.132 )2
m
= 0.720 m Then d DE =
( 0.360 )2 + ( 0.720 )2 + ( 0.720 )2
m
= 1.08 m and Have
rE/D = ( 0.360 m ) i + ( 0.720 m ) j  ( 0.720 m ) k TDE =
= TOE rE/D d DE
(
)
54 N ( 0.360i + 0.720 j  0.720k ) 1.08
= (18.0 N ) i + ( 36.0 N ) j  ( 36.0 N ) k Now where
M A = rD/ A TDE rD/ A = ( 0.132 m ) j + ( 0.720 m ) k
i j k M A = 0 0.132 0.720 N m 18.0 36.0 36.0
Then
PROBLEM 3.45 CONTINUED
M A = ( 0.132 )( 36.0 )  ( 0.720 )( 36.0 ) i + ( 0.720 )(18.0 )  0 j + 0  ( 0.132 )(18.0 ) k N m or M A =  ( 30.7 N m ) i + (12.96 N m ) j  ( 2.38 N m ) k M x = 30.7 N m, M y = 12.96 N m, M z = 2.38 N m
{
}
PROBLEM 3.46
The 0.732 1.2m 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
z =
( 0.732 )2  ( 0.132 )2
m
= 0.720 m Then
dCE =
( 0.840 )2 + ( 0.720 )2 + ( 0.720 )2
m
= 1.32 m and TCE = rE/C
dCE
(TCE )
=
 ( 0.840 m ) i + ( 0.720 m ) j  ( 0.720 m ) k ( 54 N ) 1.32 m
=  ( 36.363 N ) i + ( 29.454 N ) j  ( 29.454 N ) k Now where M A = rE/ A TCE rE/ A = ( 0.360 m ) i + ( 0.852 m ) j i j k M A = 0.360 0.852 0 Nm 34.363 29.454 29.454 =  ( 25.095 N m ) i + (10.6034 N m ) j + ( 39.881 N m ) k M x = 25.1 N m, M y = 10.60 N m, M z = 39.9 N m
Then
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 TCD when TBA = 56 N.
SOLUTION
Based on where
 M z  = k ( rB ) y TBA + k ( rC ) y TCD
M z =  ( 66 N m ) k
( rB ) y
= ( rC ) y = (1 m ) j
TBA = BATBA
=
(1.5 m ) i  (1 m ) j + ( 3 m ) k
3.5 m
( 56 N )
= ( 24 N ) i  (16 N ) j + ( 48 N ) k
TCD = CDTCD
=
( 2 m ) i  (1 m ) j  ( 2 m ) k T
3.0 m
CD
=
1 TCD ( 2i  j  2k ) 3
 ( 66 N m ) = k (1 m ) j ( 24 N ) i  (16 N ) j + ( 48 N ) k 1 + k (1 m ) j TCD ( 2i  j  2k ) 3 or 66 = 24  TCD = 2 TCD 3
{
}
3 ( 66  24 ) N 2 or TCD = 63.0 N
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 TBA when
TCD = 33 N.
SOLUTION
Based on where
 M y  = j ( rB ) z TBA + ( rC ) z TCD
M y = ( 212 N m ) j
( rB ) z
= (8 m ) k = (2 m) k
( rC ) z
TBA = BATBA
=
(1.5 m ) i  (1 m ) j  ( 3 m ) k T
3.5 m
BA
=
TBA (1.5i  j + 3k ) 3.5
TCD = CDTCD
=
( 2 m ) i  (1 m ) j  ( 2 m ) k
3.0 m
( 33 N )
= ( 22i  11j  22k ) N T ( 212 N m ) = j ( 8 m ) k BA (1.5i  j + 3k ) 3.5
+ j ( 2 m ) k ( 22 i  11j  22k ) N
or
212 =
8 (1.5 ) TBA + 2 ( 22 ) 3.5 168 18.6667 or TBA = 49.0 N
TBA =
PROBLEM 3.49
To lift a heavy crate, a man uses a block and tackle attached to the bottom of an Ibeam 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 where
M O = rA/O TBA
M O = M xi + M y j + M zk
= M xi + (100 lb ft ) j  ( 400 lb ft ) k
rA/O = ( 6 ft ) i + ( 4 ft ) j TBA = BATBA
=
( 6 ft ) i  (12 ft ) j  ( a ) k T
d BA
BA
i j k T M xi + 100 j  400k = 6 4 0 BA d 6 12 a BA
= From jcoefficient: From k coefficient:
TBA  ( 4a ) i + ( 6a ) j  ( 96 ) k d BA
100d AB = 6aTBA or TBA =
100 d BA 6a 400 d BA 96
(1) (2)
400d AB = 96TBA or TBA =
a=
Equating Equations (1) and (2) yields
100 ( 96 ) 6 ( 400 ) or a = 4.00 ft
PROBLEM 3.50
To lift a heavy crate, a man uses a block and tackle attached to the bottom of an Ibeam at hook B. Knowing that the man applies a 200lb 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 where
 M y  = j rA/O TBA
(
)
rA/O = ( 6 ft ) i + ( 4 ft ) j
TBA = BATBA = = rA/B
d BA d BA
TBA
( 6 ft ) i  (12 ft ) j  ( a ) k
200 ( 6i  12 j  ak ) d BA
( 200 lb )
=
0 1 0 200 4 0 175 lb ft = 6 d 6 12 a BA 200 175 = 0  6 ( a ) d
BA
where
d BA =
( 6 )2 + (12 )2 + ( a )2
ft
= 180 + a 2 ft
175 180 + a 2 = 1200a
or Squaring each side
180 + a 2 = 6.8571a
180 + a 2 = 47.020a 2 Solving
a = 1.97771 ft
or a = 1.978 ft
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 M x = 230 lb in.,
M y = 200 lb in., and M z = 35 lb in., determine the magnitude of P and the values of and .
SOLUTION
Based on
M x = ( P cos ) ( 9 in.) sin  ( P sin ) ( 9 in.) cos
(1)
(2) (3)
M y =  ( P cos )( 5 in.) M z =  ( P sin )( 5 in.) Then
 ( P sin ) (5) Equation (3) M z : = Equation (2) M y  ( P cos ) (5)
or
tan =
35 = 0.175 200
= 9.9262
or = 9.93
Substituting into Equation (2) 200 lb in. =  ( P cos 9.9262 ) (5 in.)
P = 40.608 lb
or P = 40.6 lb Then, from Equation (1)
230 lb in. = ( 40.608 lb ) cos 9.9262 ( 9 in.) sin  ( 40.608 lb ) sin 9.9262 ( 9 in.) cos
or
0.98503sin  0.172380cos = 0.62932
Solving numerically,
= 48.9
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 M y = 180 lb in. and
M z = 30 lb in., determine the moment M x of P about the x axis when = 60.
SOLUTION
Based on
M x = ( P cos ) ( 9 in.) sin  ( P sin ) ( 9 in.) cos
(1) (2) (3)
M y =  ( P cos )( 5 in.) M z =  ( P sin )( 5 in.)
Then
 ( P sin )( 5 ) Equation (3) M z : = Equation (2) M y  ( P cos )( 5 ) 30 = tan 180 = 9.4623
or
From Equation (3), 30 lb in. =  ( P sin 9.4623 )( 5 in.) P = 36.497 lb From Equation (1),
M x = ( 36.497 lb )( 9 in.)( cos 9.4623 sin 60  sin 9.4623 cos 60 )
= 253.60 lb in. or M x = 254 lb in.
PROBLEM 3.53
The triangular plate ABC is supported by ballandsocket 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 where
DB =
M DB = DB rA/D TAE
(
)
( 48 in.) i  (14 in.) j
50 in.
= 0.96i  0.28 j
rA/D =  ( 4 in.) j + ( 8 in.) k
( 36 in.) i  ( 24 in.) j + ( 8 in.) k 220 lb TAE = AETAE = ( ) 44 in. = (180 lb ) i  (120 lb ) j + ( 40 lb ) k
0.960 0.280 0 8 lb in. = 0 4 180 120 40
= ( 0.960 ) ( 4 )( 40 )  ( 8 )( 120 ) + ( 0.280 ) 8 (180 )  0
M DB
= 364.8 lb in. or M DB = 365 lb in.
PROBLEM 3.54
The triangular plate ABC is supported by ballandsocket 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 where
DB =
M DB = DB rC/D TCF
(
)
( 48 in.) i  (14 in.) j
50 in.
= 0.96i  0.28 j
rC/D = ( 8 in.) j  (16 in.) k
TCF = CFTCF =
( 24 in.) i  ( 36 in.) j  (8 in.) k
44 in.
(132 lb )
= ( 72 lb ) i  (108 lb ) j  ( 24 lb ) k 0.96 0.28 0 8 = 0 16 lb in. 72 108 24
= 0.96 ( 8 )( 24 )  ( 16 )( 108 ) + ( 0.28 ) ( 16 )( 72 )  0
M DB
= 1520.64 lb in. or M DB = 1521 lb in.
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 where DI = M DI = DI rF /I TEF
(1.6 m ) i  ( 0.4 m ) j (1.6 )2 + ( 0.4 )2 m
=
1 ( 4i  j) 17
rF /I = ( 4.6 m + 0.8 m ) k = ( 5.4 m ) k
TEF = EF TEF
=
(1.2 m ) i  ( 3.6 m ) j + ( 5.4 m ) k
6.6 m
( 66 N )
= (12 N ) i  ( 36 N ) j + ( 54 N ) k
= 6 ( 2 N ) i  ( 6 N ) j + ( 9 N ) k
M DI =
( 6 N )( 5.4 m ) 0
17
1 0 0 1 2 6 9 4
= 7.8582 ( 0 + 24 ) + ( 2  0 )
= 172.879 N m or M DI = 172.9 N m
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 where M DI = DI rG/I TEG
DI =
=
(1.6 m ) i  ( 0.4 m ) j
0.4 17 m
1 ( 4i  j) 17
rG/I =  (10.9 m + 0.8 m ) k =  (11.7 m ) k
TEG = EGTEG =
(1.2 m ) i  ( 3.6 m ) j  (11.7 m ) k
12.3 m
( 61.5 N )
= 5 (1.2 N ) i  ( 3.6 N ) j  (11.7 N ) k
M DI =
5 N (11.7 m ) 17
4 1 0 1 0 0 1.2 3.6 11.7
= (14.1883 N m ) 0  ( 4 )( 1)( 3.6 ) + ( 1)( 1)(1.2 )  0 = 187.286 N m
{
}
or M DI = 187.3 N m
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 where From triangle OBC
a 2
M OA = OA rC/O P
(
)
( OA) x
( OA) z
Since or
=
= ( OA ) x tan 30 =
a 1 a = 2 3 2 3
2 2
( OA)2
= ( OA) x + ( OA) y + ( OAz )
2
a 2 a a 2 = + ( OA) y + 2 2 3
2
2
( OA) y
=
a2 
a2 a2 2  =a 4 12 3
Then and
rA/O =
OA =
a 2 a i +a j+ k 2 3 2 3
1 i+ 2 2 1 j+ k 3 2 3
P = BC P
= =
( a sin 30 ) i  ( a cos30 ) k
a P i  3k 2
( P)
(
)
rC/O = ai
PROBLEM 3.57 CONTINUED
1 2 = 1 1 2 1 3 2 3 P (a) 0 0 2 0  3
M OA
= =
aP 2  (1)  3 2 3
(
)
M OA = aP 2
aP 2
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,
OA BC = 0
where From triangle OBC a 2
a 1 a = 2 3 2 3
( OA) x
( OA) z
=
= ( OA ) x tan 30 =
a a OA = i + ( OA)y j + k 2 2 3
and
BC = ( a sin 30 ) i  ( a cos 30 ) k = = a a 3 i k 2 2 a i  3k 2
(
)
( )
Then
a a a =0 i + ( OA ) y j + k i  3k 2 2 2 3 a2 a2 + ( OA)y ( 0 )  =0 4 4
OA BC = 0
or
so that
OA is perpendicular to BC.
PROBLEM 3.58 CONTINUED
(b) Have M OA = Pd , with P acting along BC and d the perpendicular distance from OA to BC. From the results of Problem 3.57,
M OA = Pa 2
Pa = Pd 2 d = a 2
or
PROBLEM 3.59
The 8ftwide 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 where AD =
M AD = AD rE/ A TEF
(
)
( 24 ft ) i + ( 3 ft ) j ( 24 )2 + ( 3)2 ft
=
1 (8i + j) 65
rE/ A = ( 7 ft ) i  ( 3 ft ) j
TEF = EFTEF =
8 (8 ft  7 ft ) i + 3 ft + ( 24 ) ( 3 ft ) j + (8 ft ) k
(1)2 + ( 4 )2 + (8)2
ft
( 5400 lb )
= 600 (1 lb ) i + ( 4 lb ) j + ( 8 lb ) k
M AD =
8 1 0 600 600 7 3 0 lb ft = ( 192  56 ) lb ft 65 65 1 4 8 or M AD = 18.46 kip ft
= 18,456.4 lb ft
PROBLEM 3.60
The 8ftwide 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 where AD =
M AD = AD rG/ A TGH
(
)
( 24 ft ) i + ( 3 ft ) j ( 24 )2 + ( 3)2 ft
=
1 (8i + j) 65
rG/ A = ( 20 ft ) i  ( 6 ft ) j = 2 (10 ft ) i  ( 3 ft ) j TGH = GH TGH =
(16 ft  20 ft ) i + 6 ft + ( 16 ) ( 3 ft ) j + (8 ft ) k 24 ( 4 )2 + ( 8 )2 + ( 8 )2
ft
( 4800 lb )
= 1600  (1 lb ) i + ( 2 lb ) j + ( 2 lb ) k
M AD =
(1600 lb )( 2 ft ) 10
65
1 0 3200 lb ft 3 0 = ( 48  20 ) 65 1 2 2 or M AD = 27.0 kip ft
8
= 26,989 lb ft
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
F1 = F11
and
F2 = F2 2
Let M1 = moment of F2 about the line of action of M1 and M 2 = moment of F1 about the line of action of M 2 Now, by definition M1 = 1 rB/ A F2 = 1 rB/ A 2 F2 M 2 = 2 rA/B F1 = 2 rA/B 1 F1 Since F1 = F2 = F and
(
)
(
)
(
)
(
)
rA/B = rB/ A
M1 = 1 rB/ A 2 F M 2 = 2 rB/ A 1 F Using Equation (3.39) 1 rB/ A 2 = 2 rB/ A 1 so that
(
)
(
)
(
)
(
) )
M12 = M 21
M 2 = 1 rB/ A 2 F
(
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 ballandsocket 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 where DB = M DB = DB rA/D TAE
(
)
( 48 in.) i  (14 in.) j
50 in.
= 0.96i  0.28 j
rA/D =  ( 4 in.) j + ( 8 in.) k TAE = AETAE
=
( 36 in.) i  ( 24 in.) j + (8 in.) k
44 in.
( 220 lb )
= (180 lb ) i  (120 lb ) j + ( 40 lb ) k 0.96 0.28 0 8 lb in. = 0 4 180 120 40 = 364.8 lb in. Only the perpendicular component of TAE contributes to the moment of
M DB
TAE about line DB. The parallel component of TAE will be used to find the perpendicular component.
PROBLEM 3.62 CONTINUED
Have
(TAE )parallel
= DB TAE
= ( 0.96i  0.28 j) (180 lb ) i  (120 lb ) j + ( 40 lb ) k = ( 0.96 )(180 ) + ( 0.28 )( 120 ) + ( 0 )( 40 ) lb
= (172.8 + 33.6 ) lb
= 206.4 lb
Since TAE = ( TAE )perpendicular + ( TAE )parallel
(TAE )perpendicular
= =
(TAE )2  (TAE )2 parallel ( 220 )2  ( 206.41)2
= 76.151 lb Then
M DB = (TAE )perpendicular ( d )
364.8 lb in. = ( 76.151 lb ) d d = 4.7905 in. or d = 4.79 in.
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 ballandsocket 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 where
( M DB ) = DB ( rC/D TCF )
DB =
( 48 in.) i  (14 in.) j
50 in.
= 0.96i  0.28j
rC/D = ( 8 in.) j  (16 in.) k TCF = CF TCF
=
( 24 in.) i  ( 36 in.) j  (8 in.) k
44 in.
(132 lb )
= ( 72 lb ) i  (108 lb ) j  ( 24 lb ) k 0.96 0.28 0 = 0 16 lb in 8 72 108 24 = 1520.64 lb in. Only the perpendicular component of TCF contributes to the moment of
TCF about line DB. The parallel component of TCF will be used to obtain the perpendicular component.
M DB
PROBLEM 3.63 CONTINUED
Have
(TCF )parallel
= DB TCF
= ( 0.96i  0.28 j) ( 72 lb ) i  (108 lb ) j  ( 24 lb ) k = ( 0.96 )( 72 ) + ( 0.28 )( 108 ) + ( 0 )( 24 ) lb
= 99.36 lb Since TCF = ( TCF )perp. + ( TCF )parallel
(TCF )perp.
= =
(TCF )2  (TCF )2 parallel
(132 )2  ( 99.36 )2
M DB = (TCF )perp. ( d )
= 86.900 lb Then
1520.64 lb in. = ( 86.900 lb ) d d = 17.4988 in. or d = 17.50 in.
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 where DI = M DI = DI rF /I TEF
(
)
(1.6 m ) i  ( 0.4 m ) j
0.4 17 m
rF /I = ( 5.4 m ) k
=
1 ( 4i  j) 17
TEF = EFTEF =
(1.2 m ) i  ( 3.6 m ) j + ( 5.4 m ) k
6.6 m
( 66 N )
= 6 ( 2 N ) i  ( 6 N ) j + ( 9 N ) k
M DI =
( 6 N )( 5.4 m ) 0
17
4 1 0 0 1 = 172.879 N m 2 6 9
Only the perpendicular component of TEF contributes to the moment of
TEF about line DI. The parallel component of TEF will be used to find the perpendicular component.
Have
(TEF )parallel
= DI TEF
= = = 1 ( 4i  j) (12 N ) i  ( 36 N ) j + ( 54 N ) k 17 1 ( 48 + 36 ) N 17 84 N 17
PROBLEM 3.64 CONTINUED
Since TEF = ( TEF )perp. + ( TEF )parallel
(TEF )perp.
=
(TEF )2  (TEF )2 parallel
( 66 )2 
84 17
2
=
= 62.777 N Then M DI = (TEF )perp. ( d ) 172.879 N m = ( 62.777 N )( d ) d = 2.7539 m or d = 2.75 m
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 where DI = M DI = DI rG/I TEG
(1.6 m ) i  ( 0.4 m ) j
0.4 17 m
=
1 ( 4i  j) 17
rG/I =  (10.9 m + 0.8 m ) k =  (11.7 m ) k TEG = EGTEG =
(1.2 m ) i  ( 3.6 m ) j  (11.7 m ) k
12.3 m
( 61.5 N )
= 5 (1.2 N ) i  ( 3.6 N ) j  (11.7 N ) k
M DI =
( 5 N )(11.7 m )
17
4 1 0 1 0 0 1.2 3.6 11.7
= 187.286 N m Only the perpendicular component of TEG contributes to the moment of
TEG about line DI. The parallel component of TEG will be used to find the perpendicular component.
Have
TEG ( parallel ) = DI TEG
= = =
1 ( 4i  j) 5 (1.2 N ) i  ( 3.6 N ) j  (11.7 N ) k 17 5 ( 4.8 + 3.6 ) N 17 42 N 17
PROBLEM 3.65 CONTINUED
Since TEF = ( TEG )perp. + ( TEG )parallel
(TEG )perp.
=
2 (TEG )2  (TEG )parallel 2
=
( 61.5)
2
42  17
= 60.651 N Then M DI = (TEG )perp. ( d ) 187.286 N m = ( 60.651 N )( d ) d = 3.0880 m or d = 3.09 m
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 FBC where Have where OA =
FBC = FBC j M OA = OA rB/O FBC
(
)
= 2 2 1 i+ j k 3 3 3
( 0.24 m ) i + ( 0.24 m ) j  ( 0.12 m ) k
0.36 m
rB/O = ( 0.18 m ) i + ( 0.24 m ) k
M OA 2 2 1 1 F = FBC 0.18 0 0.24 = BC ( 0.48  0.18 ) = 0.22 FBC 3 3 0 1 0
Only the perpendicular component of FBC contributes to the moment of FBC about line OA. The parallel component will be found first so that the perpendicular component of FBC can be determined.
2 1 2 FBC ( parallel ) = OA FBC = i + j  k FBC j 3 3 3
= Since
2 FBC 3
FBC = ( FBC )parallel + ( FBC )perp.
( FBC )perp.
=
( FBC )2  ( FBC )2 parallel
=
( FBC )2 
2 FBC 3
2
= 0.74536 FBC
Then
M OA = ( FBC )perp. ( d )
0.22 FBC = ( 0.74536 FBC ) d
d = 0.29516 m or d = 295 mm
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 z =
( 0.732 )2  ( 0.132 )2
m
= 0.720 m Have where
TDE = DETDE
M y = j rD/ A TDE
(
)
rD/ A = ( 0.132 j + 0.720k ) m
=
( 0.360 m ) i + ( 0.732 m ) j  ( 0.720 m ) k
1.08 m
( 54 N )
= (18 N ) i + ( 36 N ) j  ( 36 N ) k 0 1 0 M y = 0 0.132 0.720 = 12.96 N m 36 18 36 Only the perpendicular component of TDE contributes to the moment of
TDE about the yaxis. The parallel component will be found first so that
the perpendicular component of TDE can be determined.
TDE ( parallel ) = j TDE = 36 N
PROBLEM 3.67 CONTINUED
Since
( TDE ) = ( TDE )parallel + ( TDE )perp. (TDE )perp.
= =
2 (TDE )2  (TDE )parallel
( 54 )2  ( 36 )2
= 40.249 N
Then
M y = (TDE )perp. (d ) 12.96 N m = ( 40.249 N )( d ) d = 0.32199 m or d = 322 mm
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 21N forces, (b) the perpendicular distance between the 12N forces if the resultant of the two couples is zero, (c) the value of if the resultant couple is 1.8 N m clockwise and d is 1.05 m.
SOLUTION
(a) Have where M1 = d1F1 d1 = 0.4 m F1 = 21 N M1 = ( 0.4 m )( 21 N ) = 8.4 N m or M1 = 8.40 N m (b) Have or
M1 + M 2 = 0
8.40 N m  d 2 (12 N ) = 0 d 2 = 0.700 m
(c) Have or
M total = M1 + M 2
1.8 N m = 8.40 N m  (1.05 m )( sin )(12 N ) sin = 0.52381
and
= 31.588
or = 31.6
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 or M = Pd 1 ft 10 lb ft = P (10 in.) 12 in. P = 12 lb (b) d BC = = Have or Pmin = 12.00 lb
( BE )2 + ( EC )2 (10 in.)2 + ( 6 in.)2
M = Pd = 11.6619 in.
1 ft 10 lb ft = P (11.6619 in.) 12 in. P = 10.2899 lb (c) d AC = = Have or P = 10.29 lb
( AD )2 + ( DC )2 (10 in.)2 + (16 in.)2
M = Pd AC = 2 89 in.
1 ft 10 lb ft = P 2 89 in. 12 in. P = 6.3600 lb or P = 6.36 lb
(
)
PROBLEM 3.70
Two 60mmdiameter 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 10N 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 ) = ( 36 N )( 0.345 m )  (10 N )( 0.380 m ) = 8.62 N m
M = 8.62 N m
(b)
Have For T to be minimum, d must be maximum.
M = Td = 8.62 N m
Tmin must be perpendicular to line AC tan = and 0.380 m = 1.33333 0.285 m
= 53.130
or = 53.1 M = Tmin d max M = 8.62 N m d max =
(c) Have where
( 0.380 )2 + ( 0.285)2
+ 2 ( 0.030 ) m = 0.535 m
8.62 N m = Tmin ( 0.535 m ) Tmin = 16.1121 N or Tmin = 16.11 N
PROBLEM 3.71
The steel plate shown will support six 50mmdiameter 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 a = 0.2 m, rC/F is perpendicular to the inclined 45 N forces. Have M = ( Fd )
=  ( 45 N ) a + 0.2 m + 2 ( 0.025 m )
 ( 45 N ) 2a 2 + 2 ( 0.025 m )
For a = 0.2 m, M =  ( 45 N )( 0.450 m + 0.61569 m ) = 47.956 N m or M = 48.0 N m (b)
M = 54.0 N m
M = Moment of couple due to horizontal forces at A and D + Moment of forcecouple systems at C and F about C.
54.0 N m = 45 N a + 0.2 m + 2 ( 0.025 m )
+ M C + M F + Fx ( a + 0.2 m ) + Fy ( 2a ) where M C =  ( 45 N )( 0.025 m ) = 1.125 N m M F = M C = 1.125 N m
PROBLEM 3.71 CONTINUED
Fx = Fy = 45 N 2 45 N 2
 54.0 N m = 45 N ( a + 0.25 m )  1.125 N m  1.125 N m
45 N 45 N ( a + 0.2 m )  ( 2a ) 2 2 1.20 = a + 0.25 + 0.025 + 0.025 + a 0.20 2a + + 2 2 2
3.1213a = 0.75858 a = 0.24303 m or a = 243 mm
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 where
M1 =  ( 8 N m ) j M2 =  (6 Nm)k M = M1 + M 2
M =  (8 N m ) j  ( 6 N m ) k and
M =
( 8 )2 + ( 6 ) 2
= 10 N m or M = 10.00 N m
= or
 (8 N m ) j  ( 6 N m ) k M = = 0.8j  0.6k 10 N m M
M = M = (10 N m )( 0.8j  0.6k )
cos x = 0 cos y = 0.8 cos z = 0.6
x = 90 y = 143.130 z = 126.870
or x = 90.0, y = 143.1, z = 126.9
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 where
M = M1 + M 2 M1 = rC/B P1C rC/B = ( 0.96 m ) i  ( 0.40 m ) j P1C =  (100 N ) k
i j k M1 = 0.96 0.40 0 = ( 40 N m ) i + ( 96 N m ) j 0 0 100
Also,
M 2 = rD/ A P2 E rD/ A = ( 0.20 m ) j  ( 0.55 m ) k P2 E = ED P2 E
=
 ( 0.48 m ) i + ( 0.55 m ) k
( 0.48)2 + ( 0.55)2
m
(146 N )
=  ( 96 N ) i + (110 N ) k
i j k M 2 = 0 0.20 0.55 N m 96 0 110
= ( 22.0 N m ) i + ( 52.8 N m ) j + (19.2 N m ) k
PROBLEM 3.73 CONTINUED
and
M = ( 40 N m ) i + ( 96 N m ) j + ( 22.0 N m ) i
+ ( 52.8 N m ) j + (19.2 N m ) k
= ( 62.0 N m ) i + (148.8 N m ) j + (19.2 N m ) k
M =
2 2 2 Mx + My + Mz =
( 62.0 )2 + (148.8)2 + (19.2 )2
= 162.339 N m or M = 162.3 N m =
M 62.0i + 148.8j + 19.2k = M 162.339
= 0.38192i + 0.91660 j + 0.118271k cos x = 0.38192 x = 67.547 or x = 67.5 cos y = 0.91660 y = 23.566 or y = 23.6 cos z = 0.118271 z = 83.208 or z = 83.2
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 where
M = M4 + M7 M 4 = rG/C F4G rG/C =  (10 in.) i F4G = ( 4 lb ) k
M 4 =  (10 in.) i ( 4 lb ) k = ( 40 lb in.) j Also,
M 7 = rD/F F7 D rD/F =  ( 5 in.) i + ( 3 in.) j F7 D = ED F7 D
=
 ( 5 in.) i + ( 3 in.) j + ( 7 in.) k
( 5 )2 + ( 3)2 + ( 7 )2
7 lb ( 5i + 3j + 7k ) 83
in.
( 7 lb )
=
i 7 lb in. M7 = 5 83 5
j k 7 lb in. 3 0 = ( 21i + 35j + 0k ) 83 3 7
= 0.76835 ( 21i + 35 j) lb in.
PROBLEM 3.74 CONTINUED
and
M = ( 40 lb in.) j + 0.76835 ( 21i + 35j) lb in.
= (16.1353 lb in.) i + ( 66.892 lb in.) j
M =
( M x )2 + ( M y )
2
=
(16.1353)2 + ( 66.892 )2
= 68.811 lb in. or M = 68.8 lb in. =
(16.1353 lb in.) i + ( 66.892 lb in.) j M = M 68.811 lb in.
= 0.23449i + 0.97212 j cos x = 0.23449 x = 76.438 or x = 76.4 cos y = 0.97212 y = 13.5615 or y = 13.56 cos z = 0.0 z = 90 or z = 90.0
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 where
M 4 = rG/C F4G i j k = 10 0 0 lb in. = ( 40 lb in.) j 0 0 4 M = M 4 + M 7 + M5
M 7 = rD/F F7 D
i j k 7 = 5 3 0 lb in. = 0.76835 ( 21i + 35 j) lb in. 83 5 3 7
(See Solution to Problem 3.74.)
M 5 = rC/ A F5C i j k = 10 6 7 lb in. =  ( 35 lb in.) i + ( 50 lb in.) k 0 5 0
M = (16.1353  35 ) i + ( 40 + 26.892 ) j + ( 50 ) k lb in.
=  (18.8647 lb in.) i + ( 66.892 lb in.) j + ( 50 lb in.) k
M =
2 2 2 Mx + My + Mz =
(18.8647 )2 + ( 66.892 )2 + ( 50 )2
= 85.618 lb in.
or M = 85.6 lb in.
= M 18.8647i + 66.892 j + 50k = = 0.22034i + 0.78129 j + 0.58399k 85.618 M
cos x = 0.22034 cos y = 0.78129 cos z = 0.58399
x = 102.729 y = 38.621 z = 54.268
or x = 102.7 or y = 38.6 or z = 54.3
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
M = M1 + M 2 + M P
i j k = 0.96 0.40 0 = ( 40 N m ) i + ( 96 N m ) j 0 0 100
where
M1 = rC/B P1C
M 2 = rD/ A P2 E
i j k = 0 0.20 0.55 = ( 22.0 N m ) i + ( 52.8 N m ) j + (19.2 N m ) k 96 0 110 (See Solution to Problem 3.73.) i j k PE = 0.48 0.20 1.10 = ( 231 N m ) i + (100.8 N m ) k 0 210 0
M P = rE/ A
M = ( 40 + 22 + 231) i + ( 96 + 52.8 ) j + (19.2 + 100.8 ) k N m
= ( 293 N m ) i + (148.8 N m ) j + (120 N m ) k
M =
2 2 2 Mx + My + Mz =
( 293)2 + (148.8)2 + (120 )2
= 349.84 N m
or M = 350 N m
= 293i + 148.8 j + 120k M = = 0.83752i + 0.42533j + 0.34301k 349.84 M
cos x = 0.83752 cos y = 0.42533 cos z = 0.34301
x = 33.121 y = 64.828 z = 69.940
or x = 33.1 or y = 64.8 or z = 69.9
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 where
M = M1 + M 2 + M 3 M1 =  (1.1 lb ft )( cos 25 j + sin 25k ) M 2 =  (1.1 lb ft ) j M 3 =  (1.3 lb ft )( cos 20 j  sin 20k )
M = ( 0.99694  1.1  1.22160 ) j + ( 0.46488 + 0.44463) k =  ( 3.3185 lb ft ) j  ( 0.020254 lb ft ) k
and
M =
2 2 2 Mx + My + Mz =
( 0 )2 + ( 3.3185)2 + ( 0.020254 )2
= 3.3186 lb ft
or M = 3.32 lb ft
=
( 0 ) i  3.3185j  0.020254k M = M 3.3186
= 0.99997 j  0.0061032k
cos x = 0 cos y = 0.99997 cos z = 0.0061032
x = 90 y = 179.555 z = 90.349
or x = 90.0 or y = 179.6 or z = 90.3
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 forcecouple system (a) at A, (b) at B.
SOLUTION
(a) Based on
F : FA = T = 1000 N
or FA = 1000 N
20
M A : M A = (T sin 50 )( dA ) = (1000 N ) sin 50 ( 2.25 m ) = 1723.60 N m or M A = 1724 N m (b) Based on F : FB = T = 1000 N or FB = 1000 N MB : M B = (T sin 50 )( d B ) = (1000 N ) sin 50 (1.25 m ) = 957.56 N m or M B = 958 N m 20
PROBLEM 3.79
The 20lb horizontal force P acts on a bell crank as shown. (a) Replace P with an equivalent forcecouple 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 F : PB = P = 20 lb or PB = 20 lb M : MB = Pd B = 20 lb ( 5 in.) = 100 lb in. or M B = 100 lb in. (b) If the two vertical forces are to be equivalent to MB , they must be a couple. Further, the sense of the moment of this couple must be counterclockwise. Then, with PC and PD acting as shown, M : M D = PC d 100 lb in. = PC ( 4 in.) PC = 25 lb or PC = 25 lb Fy : 0 = PD  PC PD = 25 lb or PD = 25 lb
PROBLEM 3.80
A 700N force P is applied at point A of a structural member. Replace P with (a) an equivalent forcecouple system at C, (b) an equivalent system consisting of a vertical force at B and a second force at D.
SOLUTION
(a) Based on F : PC = P = 700 N or PC = 700 N M C : M C =  Px dCy + Py dCx where
Px = ( 700 N ) cos60 = 350 N
Py = ( 700 N ) sin 60 = 606.22 N dCx = 1.6 m dCy = 1.1 m M C =  ( 350 N )(1.1 m ) + ( 606.22 N )(1.6 m ) = 385 N m + 969.95 N m = 584.95 N m or M C = 585 N m (b) Based on Fx : PDx = P cos 60 = ( 700 N ) cos 60 = 350 N M D :
60
( P cos 60 )( d DA ) =
PB ( d DB )
( 700 N ) cos 60 ( 0.6 m ) = PB ( 2.4 m )
PB = 87.5 N or PB = 87.5 N
PROBLEM 3.80 CONTINUED
Fy : P sin 60 = PB + PDy
( 700 N ) sin 60 = 87.5 N + PDy
PDy = 518.72 N PD = =
( PDx )2 + ( PDy )
2
( 350 )2 + ( 518.72 )2
= 625.76 N
= tan 1
PDy 1 518.72 = tan = 55.991 350 PDx or PD = 626 N 56.0
PROBLEM 3.81
A landscaper tries to plumb a tree by applying a 240N 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 240N force shown in the figure.
SOLUTION
Based on Fx :  ( 240 N ) cos30 =  FB cos  FC cos or  ( FB + FC ) cos =  ( 240 N ) cos 30 Fy : or From (1)
( 240 N ) sin 30 =
FB sin + FC sin (2)
( FB + FC ) sin
= ( 240 N ) sin 30
Equation (2) : tan = tan 30 Equation (1) = 30 Based on
M C : ( 240 N ) cos ( 30  20 ) ( 0.25 m ) = ( FB cos10 )( 0.60 m )
FB = 100 N or FB = 100.0 N From Equation (1),  (100 N + FC ) cos30 = 240cos30
FC = 140 N
30
or FC = 140.0 N
30
PROBLEM 3.82
A landscaper tries to plumb a tree by applying a 240N force as shown. (a) Replace that force with an equivalent forcecouple 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
Fx :  ( 240 N ) cos30 =  FC cos30 FC = 240 N or FC = 240 N 30
M C : ( 240 N ) cos10 ( d A ) = M C
d A = 0.25 m
M C = 59.088 N m or M C = 59.1 N m (b) Based on or Fy :
( 240 N ) sin 30 =
FB sin
FB sin = 120
(1)
M B : 59.088 N m  ( 240 N ) cos10 ( dC ) =  FC ( dC cos 20 ) 59.088 N m  ( 240 N ) cos10 ( 0.60 m ) =  FC ( 0.60 m ) cos 20
0.56382 FC = 82.724
FC = 146.722 N
or FC = 146.7 N and Fx :  ( 240 N ) cos30 = 146.722 N  FB cos
FB cos = 61.124
(2)
From Equation (1) : Equation (2)
tan =
120 = 1.96323 61.124 or = 63.0 120 = 134.670 N sin 63.007 or FB = 134.7 N 63.0
= 63.007
From Equation (1), FB =
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 of with the vertical. Then for equivalence, Fx :
( 250 lb ) sin 30 =
FA sin + FB sin
(1) (2)
Fy :  ( 250 lb ) cos 30 =  FA cos  FB cos Dividing Equation (1) by Equation (2),
( 250 lb ) sin 30  ( 250 lb ) cos 30
Simplifying yields = 30 Based on
=
( FA + FB ) sin  ( FA + FB ) cos
M C : ( 250 lb ) cos 30 (12 ft ) = ( FA cos 30 )( 32 ft )
FA = 93.75 lb or FA = 93.8 lb Based on
M A :  ( 250 lb ) cos 30 ( 20 ft ) = ( FC cos 30 ) ( 32 ft )
60
FC = 156.25 lb or FC = 156.3 lb 60
PROBLEM 3.84
Three workers trying to move a 3 3 4ft crate apply to the crate the three horizontal forces shown. (a) If P = 60 lb, replace the three forces with an equivalent forcecouple system at A. (b) Replace the forcecouple 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) Based on Fz :  50 lb + 50 lb + 60 lb = FA FA = 60 lb or FA = ( 60.0 lb ) k Based on M A : (a)
( 50 lb )( 2 ft )  ( 50 lb )( 0.6 ft ) = M A
M A = 70 lb ft or M A = ( 70.0 lb ft ) j
(b) Based on Fz :  50 lb + 50 lb + 60 lb = F
F = 60 lb
or F = ( 60.0 lb ) k Based on (b) M A : 70 lb ft = 60 lb (x )
x = 1.16667 ft
or x = 1.167 ft from A along AB (c) Based on MB :  ( 50 lb ) (1 ft ) + ( 50 lb ) (2.4 ft )  P (3 ft ) = R (0 )
P=
70 = 23.333 lb 3 or P = 23.3 lb
(c)
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 600N forces are reversed.
SOLUTION
(a)
Have
Fy : FC + FD + FE = F
F = 800 N + 600 N  600 N F = 800 N
Have M G : FC ( d  1.5 m )  FD ( 2 m ) = 0 or F = 800 N
(800 N )( d
 1.5 m )  ( 600 N )( 2 m ) = 0
d =
1200 + 1200 800 or d = 3.00 m
d = 3m
(b)
Changing directions of the two 600 N forces only changes sign of the couple. F = 800 N and M G : FC ( d  1.5 m ) + FD ( 2 m ) = 0 or F = 800 N
(800 N )( d
d =
 1.5 m ) + ( 600 N )( 2 m ) 1200  1200 =0 800 or d = 0
PROBLEM 3.86
Three cables attached to a disk exert on it the forces shown. (a) Replace the three forces with an equivalent forcecouple system at A. (b) Determine the single force which is equivalent to the forcecouple system obtained in part a, and specify its point of application on a line drawn through points A and D.
SOLUTION
(a) Have
F : FB + FC + FD = FA FB = FD FA = FC = 110 N
Since
20 or FA = 110.0 N 20.0
Have
M A :  FBT ( r )  FCT ( r ) + FDT ( r ) = M A
 (140 N ) sin15 ( 0.2 m )  (110 N ) sin 25 ( 0.2 m ) + (140 N ) sin 45 ( 0.2 m ) = M A
M A = 3.2545 N m
or M A = 3.25 N m (b) Have
F : FA = FE M : M A = [ FE cos 20] ( a )
3.2545 N m = (110 N ) cos 20 ( a )
or FE = 110.0 N
20.0
a = 0.031485 m or a = 31.5 mm below 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 forcecouple. Have FB = 26.5 N + 2.5 N, where the 26.5 N force be part of the couple. Combining the two parallel forces,
M couple = ( 26.5 N ) ( 0.080 m + 0.070 m ) cos 25
= 3.60 N m
and, M couple = 3.60 N m
A single equivalent force will be located in the negative zdirection. Based on
MB :  3.60 N m = (2.5 N )cos 25 ( a )
a = 1.590 m
F = ( 2.5 N )( cos 25i + sin 25 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, specify the magnitude and the line of action of the equivalent force. (b) Specify the value of if the line of action of the equivalent force is to intersect line CD 12 in. to the right of D.
SOLUTION
(a) Have
Fx :  ( 3 lb ) sin 40 + ( 3 lb ) sin 40 = Fx Fx = 0
Have
Fy :  ( 3 lb ) cos 40  10 lb + ( 3 lb ) cos 40 = Fy Fy = 10 lb or F = 10.00 lb
Note: The two 3lb forces form a couple and M A : rC/ A PC + rB/ A PB = rX / A F
i j k i j k i j k 3 16 10 0 + 160 1 0 0 = 10 d 0 0 sin 40 cos 40 0 0 1 0 0 1 0 k : 3 (16 ) cos 40  ( 10 ) 3sin 40  160 = 10d 36.770 + 19.2836  160 = 10d d = 10.3946 in. or F = 10.00 lb (b) From part (a), Have at 10.39 in. right of A or at 5.61 in. left of B
F = 10.00 lb
M A : rC/ A PC + rB/ A PB = (12 in.) i F
i j k i j k i j k 3 16 10 0 + 160 1 0 0 = 120 1 0 0 sin cos 0 0 1 0 0 1 0 k : 48cos + 30sin  160 = 120
24cos = 20  15sin
PROBLEM 3.88 CONTINUED
Squaring both sides of the equation, and using the identity cos 2 = 1  sin 2 , results in sin 2  0.74906sin  0.21973 = 0 Using quadratic formula sin = 0.97453 so that sin =  0.22547
= 77.0
and
= 13.03
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 Pmin is perpendicular to a line connecting B and E, = 30 Then, M E : rB/E Pmin + rD/ A PD = 0 where
rB/E =  ( 0.30 m ) i + 2 ( 0.30 m ) cos 30 j
=  ( 0.30 m ) i + ( 0.51962 m ) j
rD/ A = 0.30 m + 2 ( 0.3 m ) sin 30 i
= ( 0.60 m ) i
PD = ( 450 N ) j
Pmin = Pmin ( cos 30 ) i + ( sin 30 ) j
Pmin
i j k i j k 0.30 0.51962 0 + 0.60 0 0 N m = 0 0.86603 0.50 0 0 450 0
Pmin ( 0.15 m  0.45 m ) k + ( 270 N m ) k = 0 Pmin = 450 N or Pmin = 450 N 30
PROBLEM 3.90
An eccentric, compressive 270lb force P is applied to the end of a cantilever beam. Replace P with an equivalent forcecouple system at G.
SOLUTION
Have F :  ( 270 lb ) i = F F =  ( 270 lb ) i Also, have M G : rA/G P = M
i j k 270 0 4 2.4 lb in. = M 1 0 0
M = ( 270 lb in.) ( 2.4 )( 1) j  ( 4 )( 1) k
or M = ( 648 lb in.) j  (1080 lb in.) k
PROBLEM 3.91
Two workers use blocks and tackles attached to the bottom of an Ibeam to lift a large cylindrical tank. Knowing that the tension in rope AB is 324 N, replace force the exerted at A by rope AB with an equivalent forcecouple system at E.
SOLUTION
Have where F : TAB = F
TAB = ABTAB
=
( 0.75 m ) i  ( 6.0 m ) j + ( 3.0 m ) k
6.75 m TAB = 36 N ( i  8j + 4k )
( 324 N )
so that Have
F = ( 36.0 N ) i  ( 288 N ) j + (144.0 N ) k
M E : rA/E TAB = M
or
i j k ( 7.5 m )( 36 N ) 0 1 0 = M 1 8 4
M = ( 270 N m )( 4i  k ) or M = (1080 N m ) i  ( 270 N m ) k
PROBLEM 3.92
Two workers use blocks and tackles attached to the bottom of an Ibeam 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 forcecouple system at O.
SOLUTION
Have where F : TCD = F
TCD = CDTCD
=  ( 0.3 m ) i  ( 5.6 m ) j + ( 2.4 m ) k ( 366 N ) 6.1 m
TCD = ( 6.0 N )( 3i  56 j + 24k ) so that Have M O : rC/O TCD = M
F =  (18.00 N ) i  ( 336 N ) j + (144.0 N ) k
or
i j k ( 7.5 m )( 6 N ) 0 1 0 = M 3 56 24
M = ( 45 N m )( 24i + 3k ) or M = (1080 N m ) i + (135.0 N m ) 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 45lb force directed along line AB. Replace that force with an equivalent forcecouple system at C.
SOLUTION
Have F : PAB = FC where
PAB = AB PAB
=
( 2.0 in.) i + ( 38 in.) j  ( 24 in.) k
44.989 in.
( 45 lb )
or FC = ( 2.00 lb ) i + ( 38.0 lb ) j  ( 24.0 lb ) k Have M C : rB/C PAB = M C
MC
i j k = 2 29.5 33 0 lb in. 1 19 12
= ( 2 lb in.) {( 33)( 12 ) i  ( 29.5 )( 12 ) j
+ ( 29.5 )(19 )  ( 33)(1) k
}
or M C = ( 792 lb in.) i + ( 708 lb in.) j + (1187 lb in.) k
PROBLEM 3.94
A 25lb force acting in a vertical plane parallel to the yz plane is applied to the 8in.long horizontal handle AB of a socket wrench. Replace the force with an equivalent forcecouple system at the origin O of the coordinate system.
SOLUTION
Have F : PB = F where
PB = 25 lb  ( sin 20 ) j + ( cos 20 ) k
=  ( 8.5505 lb ) j + ( 23.492 lb ) k or F =  ( 8.55 lb ) j + ( 23.5 lb ) k Have M O : rB/O PB = M O where
rB/O = ( 8cos 30 ) i + (15 ) j  ( 8sin 30 ) k in.
= ( 6.9282 in.) i + (15 in.) j  ( 4 in.) k i j k 6.9282 15 4 lb in. = M O 0 8.5505 23.492
M O = ( 318.18 ) i  (162.757 ) j  ( 59.240 ) k lb in.
or M O = ( 318 lb in.) i  (162.8 lb in.) j  ( 59.2 lb in.) k
PROBLEM 3.95
A 315N force F and 70N m couple M are applied to corner A of the block shown. Replace the given forcecouple system with an equivalent forcecouple system at corner D.
SOLUTION
Have F : F = FD = AI F =
( 0.360 m ) i  ( 0.120 m ) j + ( 0.180 m ) k
0.420 m
( 315 N )
= ( 750 N )( 0.360i  0.120 j + 0.180k ) or FD = ( 270 N ) i  ( 90.0 N ) j + (135.0 N ) k Have M D : M + rI /D F = M D where M = AC M =
( 0.240 m ) i  ( 0.180 m ) k
0.300 m
( 70.0 N m )
= ( 70.0 N m )( 0.800i  0.600k ) rI /D = ( 0.360 m ) k MD i j k = ( 70.0 N m )( 0.8i  0.6k ) + 0 0 0.36 ( 750 N m ) 0.36 0.12 0.18 = ( 56.0 N m ) i  ( 42.0 N m ) k + ( 32.4 N m ) i + ( 97.2 N m ) j or M D = ( 88.4 N m ) i + ( 97.2 N m ) j  ( 42.0 N m ) 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 M1 and M 2 can be replaced with a single equivalent force.
Further assuming that M1 = 0.068 N m and M 2 = 0.065 N 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 M1 and M 2 act at the origin. Now and
W = Wj M = M1 + M 2 =  ( M 2 cos 25 ) i + ( M1  M 2 sin 25 ) k
Note that since W and M are perpendicular, it follows that they can be replaced with a single equivalent force. (a) Have
F =W or F = Wj =  ( 2.4 N ) j or F =  ( 2.40 N ) j (b) Assume that the line of action of F passes through point P (x, 0, z). Then for equivalence M = rP/O F where rP/O = xi + zk  ( M 2 cos 25 ) i + ( M1  M 2 sin 25 ) k
i j k = x 0 z = (Wz ) i  (Wx ) k 0 W 0
PROBLEM 3.96 CONTINUED
Equating the i and k coefficients,
z = (b) For
M z cos 25 W
and
M  M 2 sin 25 x =  1 W
W = 2.4 N, M1 = 0.068 N m, M 2 = 0.065 N m
x=
0.068  0.065sin 25 = 0.0168874 m 2.4 or x = 16.89 mm
z =
0.065cos 25 = 0.024546 m 2.4 or z = 24.5 mm
PROBLEM 3.97
A 20lb force F1 and a 40 lb ft couple M1 are applied to corner E of the bent plate shown. If F1 and M1 are to be replaced with an equivalent forcecouple system ( F2 , M 2 ) at corner B and if ( M 2 ) z = 0, determine (a) the distance d, (b) F2 and M 2.
SOLUTION
(a) Have
M Bz : M 2 z = 0
k rH /B F1 + M1z = 0
(
)
(1)
where
rH /B = ( 31 in.) i  ( 2 in.) j F1 = EH F1
= =
( 6 in.) i + ( 6 in.) j  ( 7 in.) k
11.0 in. 20 lb ( 6i + 6 j  7k ) 11.0
( 20 lb )
M1z = k M1
M1 = EJ M1
= Then from Equation (1), 0 0 1 20 lb in. ( 7 )( 480 lb in.) 31 2 0 + =0 11.0 d 2 + 58 6 6 7 di + ( 3 in.) j  ( 7 in.) k d 2 + 58 in.
( 480 lb in.)
PROBLEM 3.97 CONTINUED
Solving for d, Equation (1) reduces to
20 lb in. 3360 lb in. =0 (186 + 12 )  2 11.0 d + 58
From which
d = 5.3955 in. or d = 5.40 in.
(b)
F2 = F1 =
20 lb ( 6i + 6 j  7k ) 11.0
= (10.9091i + 10.9091j  12.7273k ) lb or F2 = (10.91 lb ) i + (10.91 lb ) j  (12.73 lb ) k M 2 = rH /B F1 + M1 i j k 20 lb in. ( 5.3955 ) i + 3j  7k = 31 2 0 + ( 480 lb in.) 11.0 9.3333 6 6 7 = ( 25.455i + 394.55 j + 360k ) lb in. + ( 277.48i + 154.285 j  360k ) lb in. M 2 =  ( 252.03 lb in.) i + ( 548.84 lb in.) j or M 2 =  ( 21.0 lb ft ) i + ( 45.7 lb ft ) j
PROBLEM 3.98
A 4ftlong beam is subjected to a variety of loadings. (a) Replace each loading with an equivalent forcecouple system at end A of the beam. (b) Which of the loadings are equivalent?
SOLUTION
(a) (a) Have Fy :  200 lb  100 lb = Ra or R a = 300 lb and M A : 900 lb ft  (100 lb )( 4 ft ) = M a or M a = 500 lb ft (b) Have Fy :  300 lb = Rb or R b = 300 lb and M A :  450 lb ft = M b or M b = 450 lb ft (c) Have Fy : 150 lb  450 lb = Rc or R c = 300 lb and M A : 2250 lb ft  ( 450 lb )( 4 ft ) = M c or M c = 450 lb ft (d) Have Fy :  200 lb + 400 lb = Rd or R d = 200 lb and M A :
( 400 lb )( 4 ft )  1150 lb ft
= Md
or M d = 450 lb ft (e) Have Fy :  200 lb  100 lb = Re or R e = 300 lb and M A : 100 lb ft + 200 lb ft  (100 lb )( 4 ft ) = M e or M e = 100 lb ft
PROBLEM 3.98 CONTINUED
(f) Have Fy :  400 lb + 100 lb = R f or R f = 300 lb and M A : 150 lb ft + 150 lb ft + (100 lb )( 4 ft ) = M f or M f = 400 lb ft (g) Have Fy : 100 lb  400 lb = Rg or R g = 500 lb and M A : 100 lb ft + 2000 lb ft  ( 400 lb )( 4 ft ) = M g or M g = 500 lb ft (h) Have Fy : 150 lb  150 lb = Rh or R h = 300 lb and M A : 1200 lb ft  150 lb ft  (150 lb )(4 ft ) = M h or M h = 450 lb ft (b) Therefore, loadings (c) and (h) are equivalent
PROBLEM 3.99
A 4ftlong beam is loaded as shown. Determine the loading of Problem 3.98 which is equivalent to this loading.
SOLUTION
Have Fy : 100 lb  200 lb = R or R = 300 lb and M A :  200 lb ft + 1400 lb ft  ( 200 lb )( 4 ft ) = M or M = 400 lb ft Equivalent to case (f) of Problem 3.98 Problem 3.98 Equivalent forcecouples at A
case
(a) (b) (c) (d ) (e) (f )
R
M
500 lb ft 450 lb ft 450 lb ft 450 lb ft 100 lb ft 400 lb ft
300 lb 300 lb 300 lb 200 lb 300 lb 300 lb 500 lb 300 lb
(g ) (h)
500 lb ft 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 4ftlong beam is subjected to a variety of loadings. (a) Replace each loading with an equivalent forcecouple system at end A of the beam. (b) Which of the loadings are equivalent?
SOLUTION
(a) For equivalent single force at distance d from A Have
Fy :  300 lb = R
or R = 300 lb and
M C :
( 300 lb )( d )  450 lb ft
=0
or d = 1.500 ft (b) Have
Fy :  200 lb + 400 lb = R
or R = 200 lb and
M C :
( 200 lb )( d ) + ( 400 lb )( 4  d )  1150 lb ft
=0
or d = 2.25 ft (c) Have
Fy :  200 lb  100 lb = R
or R = 300 lb and
M C : 100 lb ft + ( 200 lb )( d )  (100 lb )( 4  d ) + 200 lb ft = 0
or d = 0.333 ft
PROBLEM 3.101
Five separate forcecouple 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 F = (10 N ) j and a couple of moment M = ( 6 N m ) i + ( 4 N m ) k located at point A.
SOLUTION
The equivalent forcecouple system at A for each of the five forcecouple systems will be determined. Each will then be compared to the given forcecouple system to determine if they are equivalent.
Forcecouple system at B
Have or and
F :
(10 N ) j = F
F = (10 N ) j
M A : M B + rB/ A F = M
(
)
( 4 N m ) i + ( 2 N m ) k + ( 0.2 m ) i (10 N ) j = M
M = ( 4 Nm) i + ( 4 Nm)k
Comparing to given forcecouple system at A, Is Not Equivalent
Forcecouple system at C
Have or and
F :
(10 N ) j = F
F = (10 N ) j
M A : M C + rC/ A F = M
(
)
(8.5 N m ) i + ( 0.2 m ) i + ( 0.25 m ) k (10 N ) j = M
M = ( 6 N m ) i + ( 2.0 N m ) k
Comparing to given forcecouple system at A, Is Not Equivalent
PROBLEM 3.101 CONTINUED
Forcecouple system at E
Have or and
F :
(10 N ) j = F
F = (10 N ) j
M A : M E + rE/ A F = M
(
)
( 6 N m ) i + ( 0.4 m ) i  ( 0.08 m ) j (10 N ) j = M
M = (6 Nm) i + ( 4 Nm)k
Comparing to given forcecouple system at A, Is Equivalent
Forcecouple system at G
Have or
F :
(10 N ) i + (10 N ) j = F
F = (10 N ) i + (10 N ) j
F has two force components
forcecouple system at G Is Not Equivalent Forcecouple system at I
Have or and
F :
(10 N ) j = F
F = (10 N ) j
M A : M I + rI / A F = M
(
)
(10 N m ) i  ( 2 N m ) k
+ ( 0.4 m ) i  ( 0.2 m ) j + ( 0.4 m ) k (10 N ) j = M
or
M = (6 N m) i + (2 N m)k
Comparing to given forcecouple 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
WA = mA g = ( 38 kg ) g WB = mB g = ( 29 kg ) g
(a)
WC = mC g = ( 27 kg ) g
For resultant weight to act at C,
M C = 0
Then ( 38 kg ) g ( 2 m )  ( 27 kg ) g ( d )  ( 29 kg ) g ( 2 m ) = 0
d =
76  58 = 0.66667 m 27 or d = 0.667 m
(b)
WC = mC g = ( 24 kg ) g
For resultant weight to act at C,
M C = 0
Then ( 38 kg ) g ( 2 m )  ( 24 kg ) g ( d )  ( 29 kg ) g ( 2 m ) = 0
d =
76  58 = 0.75 m 24 or d = 0.750 m
PROBLEM 3.103
Three stage lights are mounted on a pipe as shown. The mass of each light is mA = mB = 1.8 kg and mC = 1.6 kg . (a) If d = 0.75 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
WA = WB = m A g = (1.8 kg ) g WC = mC g = (1.6 kg ) g
(a) Have
d = 0.75 m R = WA + WB + WC
R = (1.8 + 1.8 + 1.6 ) kg g
or Have
R = ( 5.2 g ) N
M D : 1.8g ( 0.3 m )  1.8g (1.3 m )  1.6 g ( 2.05 m ) = 5.2 g ( D ) D = 1.18462 m
or D = 1.185 m (b) Have
M D :  (1.8 g )( 0.3 m )  (1.8g )(1.3 m )  (1.6 g )(1.3 m + d ) =  ( 5.2 g )(1.25 m ) d = 0.9625 m D= L = 1.25 m 2
or d = 0.963 m
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 a = 1.1 m, (b) the value of a so that the loads on the bridge supports at A and B are equal.
SOLUTION
(a) Have
a = 1.1 m F : WC  WD  WE = R R = 800 N  700 N  540 N R = 2040 N
(a)
or Have
R = 2040 N
M A :  (800 N )(1.5 m )  (700 N )(2.6 m )  (540 N )(4.25 m ) = R ( d )  5315 N m =  ( 2040 N ) d d = 2.6054 m
and
or d = 2.61 m to the right of A (b) For equal reaction forces at A and B, the resultant, R, must act at the center of the span. (b) From
L M A =  R 2
 ( 800 N )(1.5 m )  ( 700 N )(1.5 m + a )  ( 540 N )(1.5 m + 2.5a ) =  ( 2040 N )( 3 m )
3060 + 2050a = 6120 and
a = 1.49268 m
or a = 1.493 m
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
Fx :  ( 90 N ) sin 30 + (125 N ) cos 40 = Rx
or Rx = 50.756 N
Fy :  ( 90 N ) cos30  200 N  (125 N ) sin 40 = Ry
or Ry = 358.29 N Then and tan =
R= Ry Rx =
( 50.756 )2 + ( 358.29 )2
358.29 = 7.0591 50.756
= 361.87 N = 81.937
or R = 362 N Also
81.9
M A : M  ( 90 N ) sin 35 ( 0.6 m )  ( 200 N ) cos 25 ( 0.85 m )  (125 N ) sin 65 (1.25 m ) = 0
M = 326.66 N m
or M = 327 N m
PROBLEM 3.106
To test the strength of a 25 20in. suitcase, forces are applied as shown. If P = 18 lb, (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 = 18 lb Have
F :  ( 20 lb ) i + 42 lb ( 3i + 2 j) + (18 lb ) j + ( 36 lb ) i = Rxi + Ry j 13
 (18.9461 lb ) i + ( 41.297 lb ) j = Rxi + Ry j or
R=
2 2 Rx + Ry =
R =  (18.95 lb ) i + ( 41.3 lb ) j
(18.9461)2 + ( 41.297 )2
= 45.436 lb
x = tan 1
Ry 1 41.297 = tan = 65.355 18.9461 Rx
or R = 45.4 lb (b) Have
65.4
M B = M B
42 lb M B = ( 4 in.) j ( 20 lb ) i + ( 21 in.) i ( 3i + 2 j) + (12 in.) j ( 36 lb ) i + ( 3 in.) i (18 lb ) j 13
M B = (191.246 lb in.) k
PROBLEM 3.106 CONTINUED
Since
M B = rB R
i j k (191.246 lb in.) k = x y 0 = ( 41.297 x + 18.9461y ) k 18.9461 41.297 0
For
y = 0,
x=
191.246 = 4.6310 in. 41.297 191.246 = 10.0942 in. 18.9461
or x = 4.63 in. or y = 10.09 in.
For
x = 0,
y =
PROBLEM 3.107
Solve Problem 3.106 assuming that P = 28 lb.
Problem 3.106: To test the strength of a 25 20in. suitcase, forces are applied as shown. If P = 18 lb, (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
F :  ( 20 lb ) i + 42 ( 3i + 2 j) + ( 28 lb ) j + ( 36 lb ) i = Rxi + Ry j 13
 (18.9461 lb ) i + ( 51.297 lb ) j = Rxi + Ry j or
R=
R =  (18.95 lb ) i + ( 51.3 lb ) j
2 2 Rx + Ry =
(18.9461)2 + ( 51.297 )2
= 54.684 lb
x = tan 1
Ry 1 51.297 = tan = 69.729 18.9461 Rx
or R = 54.7 lb (b) Have M B = M B
69.7
42 lb M B = ( 4 in.) j ( 20 lb ) i + ( 21 in.) i ( 3i + 2j) + (12 in.) j ( 36 lb ) i + ( 3 in.) i ( 28 lb ) j 13
M B = ( 221.246 lb in.) k
PROBLEM 3.107 CONTINUED
Since
M B = rB R i j k ( 221.246 lb in.) k = x y 0 = ( 51.297 x + 18.9461y ) k 18.9461 51.297 0
For For
y = 0, x = 0,
x= y =
221.246 = 4.3130 in. 51.297 221.246 = 11.6776 in. 18.9461
or x = 4.31 in. or y = 11.68 in.
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 = 45 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 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 F = R
R =  2.6 j  5.25 j  10.5 ( cos 45i + sin 45 j)  3.2i kN
R =  (10.6246 kN ) i  (15.2746 kN ) j
R=
2 2 Rx + Ry =
(10.6246 )2 + (15.2746 )2
= 18.6064 kN
= tan 1
Ry 1 15.2746 = tan = 55.179 10.6246 Rx
or R = 18.61 kN Have where
55.2
M EF = M EF
MEF = (2.6 kN )(90 mm ) + (5.25 kN )(40 mm )
 (10.5 kN )(20 mm )  ( 3.2 kN ) ( 40 mm ) sin 45 + 40 mm
MEF = 15.4903 N m
To obtain distance d left of EF, Have
d =
M EF = dRy = d ( 15.2746 kN ) 15.4903 N m = 1.01412 103 m 3 15.2746 10 N or d = 1.014 mm left of EF
PROBLEM 3.108 CONTINUED
(b) Have M EF = 0 M EF = 0 = ( 2.6 kN )( 90 mm ) + ( 5.25 kN )( 40 mm )
 (10.5 kN )( 20 mm )
 ( 3.2 kN ) ( 40 mm ) sin + 40 mm
(128 N m ) sin = 106 N m
sin = 0.828125
= 55.907
or = 55.9
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 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,
=
tan = tan =
Ry Rx
where Rx = 3.2 kN  (10.5 kN ) sin Ry = 2.6 kN  5.25 kN  (10.5 kN ) cos
tan =
3.2 + 10.5 sin 7.85 + 10.5cos 3.2 = 0.40764 7.85 or = 22.2
and
tan =
= 22.178
(b) From
= 22.178
Rx = 3.2 kN  (10.5 kN ) sin 22.178 = 7.1636 kN Ry = 7.85 kN  (10.5 kN ) cos 22.178 = 17.5732 kN
R=
2 2 Rx + Ry =
( 7.1636 )2 + (17.5732 )2
= 18.9770 kN
or Then M EF = M EF where
R = 18.98 kN
67.8
M EF = ( 2.6 kN )( 90 mm ) + ( 5.25 kN )( 40 mm )  (10.5 kN )( 20 mm )
 ( 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
d =
M EF = dRy = d ( 17.5732 ) 57.682 N m = 3.2824 103 m 17.5732 103 N or d = 3.28 mm left of 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
R = F R = ( 240 N )( cos 70i  sin 70 j)  (160 N ) j
+ ( 300 N )(  cos 40i  sin 40 j)  (180 N ) j R =  (147.728 N ) i  ( 758.36 N ) j
R=
2 2 Rx + Ry =
(147.728)2 + ( 758.36 )2
= 772.62 N
= tan 1
Ry 1 758.36 = tan = 78.977 147.728 Rx
or R = 773 N Have where
M A =  [ 240 N cos 70] ( 6 m )  [ 240 N sin 70] ( 4 m )  (160 N )(12 m ) + M A = dRy
79.0
[300 N cos 40] ( 6 m )
 [300 N sin 40] ( 20 m )  (180 N )( 8 m ) = 7232.5 N m d = 7232.5 N m = 9.5370 m 758.36 N
or d = 9.54 m to the right of A
PROBLEM 3.111
Three forces and a couple act on crank ABC. For P = 5 lb and = 40, (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) Have
P = 5 lb,
= 40
R = F = ( 5 lb )( cos 40i + sin 40 j)  ( 3 lb ) i  ( 2 lb ) j R = ( 0.83022 lb ) i + (1.21394 lb ) j
R=
2 2 Rx + Ry =
( 0.83022 )2 + (1.21394 )2
= 1.47069 lb
= tan 1
Ry 1 1.21394 = tan = 55.632 0.83022 Rx
or R = 1.471 lb (b) From where MB = MB = dRy
55.6
MB =  (5 lb )cos 40 (15 in. )sin 50  (5 lb )sin 40 (15 in. )sin 50 + ( 3 lb ) ( 6 in.) sin 50
 ( 2 lb )( 6 in.) + 50 lb in.
M B = 23.211 lb in. and
d = MB 23.211 lb in. = = 19.1205 in. Ry 1.21394 lb
or d = 19.12 in. to the left of B
PROBLEM 3.111 CONTINUED
(c) From
M B = rD/B R
 ( 23.211 lb in.) k = ( d1 cos 50i + d1 sin 50 j)
( 0.83022 lb ) i + (1.21394 lb ) j
 ( 23.211 lb in.) k = ( 0.78028d1  0.63599d1 ) k
d1 = or 23.211 = 16.3889 in. 1.41627
d1 = 16.39 in. from B along line 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 Fx = 0 P cos  3 lb = 0 P cos = 3 lb and Fy = 0 P sin  2 lb = 0 P sin = 2 lb Dividing Equation (2) by Equation (1), tan = 2 3 (2) (1)
= 33.690 Substituting into Equation (1), P= or (a) Based on 3 lb = 3.6056 lb cos 33.690
P = 3.61 lb
33.7
M B = 0
 ( 3.6056 lb ) cos 33.690 ( d + 6 in.) sin 50  ( 3.6056 lb ) sin 33.690 ( d + 6 in.) cos 50 + ( 3 lb ) ( 6 in.) sin 50  ( 2 lb )( 6 in.) + 50 lb in. = 0
3.5838d = 30.286 d = 8.4509 in. or d = 8.45 in.
PROBLEM 3.112 CONTINUED
(b) Based on M D = 0
 ( 3.6056 lb ) cos 33.690 ( d + 6 in.) sin 50  ( 3.6056 lb ) sin 33.690 ( d + 6 in.) cos 50 + 6 in. + ( 3 lb ) ( 6 in.) sin 50 + 50 lb in. = 0
3.5838d = 30.286 d = 8.4509 in. or d = 8.45 in. This result is expected, since R = 0 and M R = 0 for B d = 8.45 in. implies that R = 0 and M = 0 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 forcecouple at A due to belts on pulley A Have F : 120 N  160 N = RA R A = 280 N Have M A :  40 N ( 0.02 m ) = M A M A = 0.8 N m Equivalent forcecouple at B due to belts on pulley B Have F :
( 210 N + 150 N )
25
25 = R B
R B = 360 N Have
M B :  60 N ( 0.015 m ) = M B M B = 0.9 N m
Equivalent forcecouple at F Have F : R F = ( 280 N ) j + ( 360 N )( cos 25i + sin 25 j) = ( 326.27 N ) i  (127.857 N ) j
R = RF =
2 2 RFx + RFy =
( 326.27 )2 + (127.857 )2
= 350.43 N
= tan 1
RFy 1 127.857 = tan = 21.399 326.27 RFx
or R F = R = 350 N
21.4
PROBLEM 3.113 CONTINUED
Have M F : M F =  ( 280 N )( 0.06 m )  0.80 N m
 ( 360 N ) cos 25 ( 0.010 m ) + ( 360 N ) sin 25 ( 0.120 m )  0.90 N m
M F =  ( 3.5056 N m ) k
To determine where a single resultant force will intersect line FE, M F = dR y
d = MF 3.5056 N m = = 0.027418 m = 27.418 mm Ry 127.857 N
or d = 27.4 mm
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 forcecouple system at the point D obtained by drawing the perpendicular from the point of contact to the x axis (b) For a = 1 m and b = 2 m, determine the value of x for which the moment of the equivalent forcecouple system at D is maximum.
SOLUTION
(a) The slope of any tangent to the surface of member C is
dy d x 2 2b = b 1  2 = 2 x dx dx a a Since the force F is perpendicular to the surface,
dy tan =  dx For equivalence
1
=
a2 1 2b x
F : F = R M D : where
cos = 2bx , x2 y A = b 1  2 a
( F cos )( y A ) = M D
(a )
2
2
+ ( 2bx )
2
MD =
x3 2 Fb 2 x  2 a a 4 + 4b 2 x 2
Therefore, the equivalent forcecouple system at D is
R = F
a2 tan 1 2bx
x3 2Fb 2 x  2 a M = 4 2 2 a + 4b x
PROBLEM 3.114 CONTINUED
(b) To maximize M, the value of x must satisfy where, for a = 1 m, b = 2 m
M = 8F x  x 3 1 + 16 x 2
dM =0 dx
(
)
dM = 8F dx
1 1 + 16 x 2 1  3x 2  x  x3 ( 32 x ) 1 + 16 x 2 2
) ( ) ( (1 + 16x ) (1 + 16x )(1  3x )  16x ( x  x ) = 0
2
(
)

1 2
=0
2
2
3
or x2 = 3 9  4 ( 32 )( 1) 2 ( 32 )
32 x 4 + 3x 2  1 = 0 = 0.136011 m 2 and  0.22976 m 2
Using the positive value of x 2 , x = 0.36880 m or x = 369 mm
PROBLEM 3.115
As plastic bushings are inserted into a 3in.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 forcecouple system at C.
SOLUTION
For equivalence F: FA + FB + FC + FD = R C
R C =  ( 5 lb ) j  ( 3 lb ) j  ( 4 lb ) k  ( 7 lb ) i
R C = ( 7 lb ) i  ( 8 lb ) j  ( 4 lb ) k Also for equivalence M C : rA/C FA + rB/C FB + rD/C FD = M C or
i j k i j k i j k = 0 0 1.5 in. + 1 in. 0 1.5 in. + 0 1.5 in. 1.5 in. 0 5 lb 0 0 3 lb 0 7 lb 0 0
= ( 7.50 lb in.  0 ) i + ( 0  4.50 lb in.) i + ( 3.0 lb in.  0 ) k + (10.5 lb in.  0 ) j + ( 0 + 10.5 lb in.) k
MC
or M C =  (12.0 lb in.) i + (10.5 lb in.) j + ( 7.5 lb in.) k
PROBLEM 3.116
Two 300mmdiameter 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 forcecouple system at A.
SOLUTION
Equivalent forcecouple at each pulley Pulley B R B = ( 290 N )(  cos 20 j + sin 20k )  430 Nj =  ( 702.51 N ) j + ( 99.186 N ) k M B =  ( 430 N  290 N )( 0.15 m ) i =  ( 21 N m ) i Pulley C R C = ( 310 N + 480 N )(  sin10 j  cos10k ) =  (137.182 N ) j  ( 778.00 N ) k M C = ( 480 N  310 N )( 0.15 m ) i = ( 25.5 N m ) i Then R = R B + R C =  ( 839.69 N ) j  ( 678.81 N ) k or R =  ( 840 N ) j  ( 679 N ) k M A = M B + M C + rB/ A R B + rC/ A R C
i j k =  ( 21 N m ) i + ( 25.5 N m ) i + 0.45 0 0 Nm 0 702.51 99.186 i j k + 0.90 0 0 Nm 0 137.182 778.00 = ( 4.5 N m ) i + ( 655.57 N m ) j  ( 439.59 N m ) k or M A = ( 4.50 N m ) i + ( 656 N m ) j  ( 440 N m ) 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 forcecouple system at C consisting of the force C =  ( 40 N ) i + ( 20 N ) k and the couple M C = ( 40 N m ) i , determine the forces applied at A and B when
Az = 10 N.
SOLUTION
Have or F : A + B = C Fx : Ax + Bx = 40 N Bx =  ( Ax + 40 N ) Fy : Ay + By = 0 or Ay =  By Fz : 10 N + Bz = 20 N or Have Bz = 10 N M C : rB/C B + rA/C A = M C
i j k i j k 0.2 0 0.05 + 0.2 0 0.2 N m = ( 40 N m ) i Bx By 10 Ax Ay 10
(1)
(2)
(3)
or
( 0.05By  0.2 Ax ) i + ( 0.05Bx  2 + 0.2 Ax  2) j
+ 0.2By + 0.2 Ay k = ( 40 N m ) i
(
)
From
i  coefficient
0.05By  0.2 Ay = 40 N m  0.05Bx + 0.2 Ax = 4 N m 0.2 By + 0.2 Ay = 0
(4) (5) (6)
j  coefficient k  coefficient
PROBLEM 3.117 CONTINUED
From Equations (2) and (4): 0.05By  0.2  By = 40
(
)
By = 160 N, Ay = 160 N From Equations (1) and (5): 0.05 (  Ax  40 ) + 0.2 Ax = 4 Ax = 8 N From Equation (1): Bx =  ( 8 + 40 ) = 48 N A = ( 8 N ) i  (160 N ) j + (10 N ) k B =  ( 48 N ) i + (160 N ) j + (10 N ) 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 forcecouple system at A consisting of the force R = ( 3.9 lb ) i + Ry j  (1.1 lb ) k and the couple
M R = M xi + (1.5 lb ft ) j  (1.1 lb ft ) k. . (b) Find the corresponding A values of Ry and M x .
SOLUTION
Have F : B + C = R Fx : Bx + Cx = 3.9 lb Fy : C y = Ry Fz : C z = 1.1 lb Have
M A : rB/ A B + rC/ A C + M B = M R A i j k i j k 1 1 4 0 2.0 + ( 2 lb ft ) i = M xi + (1.5 lb ft ) j  (1.1 lb ft ) k x 0 4.5 + 12 12 Bx 0 0 C x C y 1.1
or
Bx = 3.9 lb  Cx
(1) (2) (3)
( 2  0.166667C y ) i + ( 0.375Bx + 0.166667Cx + 0.36667 ) j + ( 0.33333C y ) k
= M xi + (1.5 ) j  (1.1) k From i  coefficient j  coefficient k  coefficient (a) From Equations (1) and (5): 0.375 ( 3.9  Cx ) + 0.166667Cx = 1.13333
Cx = From Equation (1): 0.32917 = 1.58000 lb 0.20833 Bx = 3.9  1.58000 = 2.32 lb B = ( 2.32 lb ) i
2  0.166667C y = M x 0.375Bx + 0.166667Cx + 0.36667 = 1.5 0.33333C y = 1.1 or C y = 3.3 lb
(4) (5) (6)
C = (1.580 lb ) i  ( 3.30 lb ) j  (1.1 lb ) k
(b) From Equation (2): From Equation (4): Ry = C y = 3.30 lb or R y =  ( 3.30 lb ) j
M x = 0.166667 ( 3.30 ) + 2.0 = 2.5500 lb ft or M x = ( 2.55 lb ft ) 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 10lb force at F lies in a plane parallel to the yz plane, determine (a) the angle the force at F should form with the horizontal if duct AB is not to tend to rotate about the vertical, (b) the forcecouple 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 yaxis if:
R M By = j M R = j rF /B FF + rE/B FE = 0 B
(
)
where
rF /B = ( 45 in.) i  ( 23 in.) j + ( 28 in.) k rE/B = ( 54 in.) i  ( 34 in.) j + ( 28 in.) k
FF = 10 lb ( sin ) j + ( cos ) k
FE =  ( 5 lb ) k i j k i j k = (10 lb ) 45 in. 23 in. 28 in. + ( 5 lb )( 2 in.) 27 17 14 0 sin cos 0 0 1
= ( 230 cos  280sin + 170 ) i  ( 450 cos  270 ) j + ( 450sin ) k lb in.
M R B
Thus,
R M By = 450 cos + 270 = 0
cos = 0.60
= 53.130
or = 53.1
PROBLEM 3.119 CONTINUED
(b) R = FE + FF where
FE =  ( 5 lb ) k FF = (10 lb )( sin 53.130 j + cos 53.130k ) = ( 8 lb ) j + ( 6 lb ) k
R = ( 8 lb ) j + (1 lb ) k and
M = M R =  230 ( 0.6 ) + 280 ( 0.8 )  170 i  450 ( 0.6 )  270 j + 450 ( 0.8 ) k B
=  (192 lb in.) i  ( 0 ) j + ( 360 lb in.) k or M =  (192 lb in.) i + ( 360 lb in.) 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 10lb force at F lies in a plane parallel to the yz plane and that = 60, (a) replace the given force system with an equivalent forcecouple 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 where
R = F = FF + FE
FF = 10 lb ( sin 60 ) j + ( cos 60 ) k = ( 8.6603 lb ) j + ( 5.0 lb ) k
FE =  ( 5 lb ) k
R = ( 8.6603 lb ) j Have where
R M C = ( r F ) = rF /C FF + rE/C FE
or R = ( 8.66 lb ) j
rF /C = ( 9 in.) i  ( 2 in.) j rE/C = (18 in.) i  (13 in.) j
R MC
i j k i j k = 9 2 0 lb in. + 18 13 0 lb in. 0 8.6603 5.0 0 0 5
= ( 55 lb in.) i + ( 45 lb in.) j + ( 77.942 lb in.) k
R or M C = ( 55.0 lb in.) i + ( 45.0 lb in.) j + ( 77.9 lb in.) k
(b) To determine which direction duct section CD has a tendency to turn, have
R R M CD = DC M C
where
DC = Then  (18 in.) i + ( 4 in.) j 2 85 in. = 1 ( 9i + 2 j) 85
R M CD =
1 ( 9i + 2 j) ( 55i + 45j + 77.942k ) lb in. 85
= ( 53.690 + 9.7619 ) lb in. = 43.928 lb in.
R Since DC M C < 0, duct DC tends to rotate clockwise relative to elbow C as viewed from D to C.
PROBLEM 3.121
The headandmotor 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 25o about the y axis and 20o 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 forcecouple system at the center O of the base of the vertical column.
SOLUTION
Have
R =F
= ( 44 N ) ( sin 20 cos 25 ) i  ( cos 20 ) j  ( sin 20 sin 25 ) k
= (13.6389 N ) i  ( 41.346 N ) j  ( 6.3599 N ) k
or R = (13.64 N ) i  ( 41.3 N ) j  ( 6.36 N ) k Have where
rB/O = ( 0.280 m ) sin 25 i + ( 0.300 m ) j + ( 0.280 m ) cos 25 k
M O = rB/O F + M C
= ( 0.118333 m ) i + ( 0.300 m ) j + ( 0.25377 m ) k
M C = ( 7.2 N m ) ( sin 20 cos 25 ) i  ( cos 20 ) j  ( sin 20 sin 25 ) k
= ( 2.2318 N m ) i  ( 6.7658 N m ) j  (1.04072 N m ) k
MO
i j k = 0.118333 0.300 0.25377 N m 13.6389 41.346 6.3599 + ( 2.2318i  6.7658 j  1.04072k ) N m = (10.8162 N m ) i  ( 2.5521 N m ) j  (10.0250 N m ) k or M O = (10.82 N m ) i  ( 2.55 N m ) j  (10.03 N m ) 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 forcecouple system shown, where R = 300 N and M = 37.5 N m. Replace this forcecouple system with an equivalent forcecouple system at C.
SOLUTION
From
 ( 0.2 m ) i + (1.4 m ) j  ( 0.5 m ) k R C = R = ( 300 N ) AB = 300 N 1.50 m
R C =  ( 40.0 N ) i + ( 280 N ) j  (100 N ) k
From where
M C = rA/C R + M
rA/C = ( 2.6 m ) i + ( 0.5 m ) k
( 0.2 m ) i  (1.4 m ) j + ( 0.5 m ) k M = ( 37.5 N m ) BA = ( 37.5 N m ) 1.50 m
= ( 5.0 N m ) i  ( 35.0 N m ) j + (12.5 N m ) k i j k = (10 N m ) 2.6 0 0.5 + ( 5.0 N m ) i  ( 35.0 N m ) j + (12.5 N m ) k 4 28 10
= ( 140 + 5 ) N m i + ( 20 + 260  35 ) N m j + ( 728 + 12.5 ) N m k
MC
or M C =  (135.0 N m ) i + ( 205 N m ) j + ( 741 N m ) k
PROBLEM 3.123
Three children are standing on a 15 15ft 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
F : FA + FB + FC = R  ( 85 lb ) j  ( 60 lb ) j  ( 90 lb ) j = R  ( 235 lb ) j = R
or R = 235 lb
Have
M x : FA ( z A ) + FB ( z B ) + FC ( zC ) = R ( z D )
(85 lb)( 9 ft ) + ( 60 lb )(1.5 ft ) + ( 90 lb )(14.25 ft ) = ( 235 lb )( zD )
z D = 9.0957 ft
Have or z D = 9.10 ft
M z : FA ( x A ) + FB ( xB ) + FC ( xC ) = R ( xD )
(85 lb )( 3 ft ) + ( 60 lb )( 4.5 ft ) + ( 90 lb )(14.25 ft ) = ( 235 lb )( xD )
xD = 7.6915 ft
or xD = 7.69 ft
PROBLEM 3.124
Three children are standing on a 15 15ft 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
F : FA + FB + FC + FD = R  ( 85 lb ) j  ( 60 lb ) j  ( 90 lb ) j  ( 95 lb ) j = R R =  ( 330 lb ) j
Have
M x : FA ( z A ) + FB ( z B ) + FC ( zC ) + FD ( z D ) = R ( z H )
(85 lb )( 9 ft ) + ( 60 lb )(1.5 ft ) + ( 90 lb )(14.25 ft ) + ( 95 lb )( zD ) = ( 330 lb )( 7.5 ft )
z D = 3.5523 ft
Have or z D = 3.55 ft
M z : FA ( x A ) + FB ( xB ) + FC ( xC ) + FD ( xD ) = R ( xH )
(85 lb )( 3 ft ) + ( 60 lb )( 4.5 ft ) + ( 90 lb )(14.25 ft ) + ( 95 lb )( xD ) = ( 330 lb )( 7.5 ft )
xD = 7.0263 ft
or xD = 7.03 ft
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
F : FA + FB + FC + FD = R  ( 580 kN ) j  ( 2350 kN ) j  ( 330 kN ) j  (140 kN ) j = R R =  ( 3400 kN ) j R = 3400 kN
Have
M x : FA ( z A ) + FB ( z B ) + FC ( zC ) + FD ( z D ) = R ( z E )
( 580 kN )(8 m ) + ( 2350 kN )(16 m ) + ( 330 kN )( 6 m ) + (140 kN )( 33.5 m ) = ( 3400 kN )( zE )
z E = 14.3853 m
Have or z E = 14.39 m
M z : FA ( x A ) + FB ( xB ) + FC ( xC ) + FD ( xD ) = R ( xE )
( 580 kN )(10 m ) + ( 2350 kN )( 32 m ) + ( 330 kN )( 54 m ) + (140 kN )( 32 m ) = ( 3400 kN )( xE )
xE = 30.382 m
or xE = 30.4 m
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 580kN 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
F : FB + FC + FD + FE = R  ( 2350 kN ) j  ( 330 kN ) j  (140 kN ) j  ( 580 kN ) j = R R =  ( 3400 kN ) j
Have
M x : FB ( z B ) + FC ( zC ) + FD ( z D ) + FE ( z E ) = R ( z B )
( 2350 kN )(16 m ) + ( 330 kN )( 6 m ) + (140 kN )( 33.5 m ) + ( 580 kN )( b ) = ( 3400 kN )(16 m )
b = 17.4655 m
Have or b = 17.47 m
M z : FB ( xB ) + FC ( xC ) + FD ( xD ) + FE ( xE ) = R ( xB )
( 2350 kN )( 32 m ) + ( 330 kN )( 54 m ) + (140 kN )( 32 m ) + ( 580 kN )( a ) = ( 3400 kN )( 32 m )
a = 19.4828 m
or a = 19.48 m
PROBLEM 3.127
A group of students loads a 2 4m flatbed trailer with two 0.6 0.6 0.6m boxes and one 0.6 0.6 1.2m 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 1.2m 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 1.2m box should be placed adjacent to one of the edges of the trailer with the 0.6 0.6m side on the bottom. The edges to be considered are based on the location of the resultant for the three given weights. Have
F :  ( 200 N ) j  ( 400 N ) j  (180 N ) j = R R =  ( 780 N ) j
Have
M z :
( 200 N )( 0.3 m ) + ( 400 N )(1.7 m ) + (180 N )(1.7 m ) = ( 780 N )( x )
x = 1.34103 m
Have
M x :
( 200 N )( 0.3 m ) + ( 400 N )( 0.6 m ) + (180 N )( 2.4 m ) = ( 780 N )( z )
z = 0.93846 m
From the statement of the problem, it is known that the resultant of R from the original loading and the lightest load W passes through G, the point of intersection of the two center lines. Thus, M G = 0. Further, since the lightest load W is to be as small as possible, the fourth box should be placed as far from G as possible without the box overhanging the trailer. These two requirements imply
( 0.3 m
x 1 m ) (1.8 m z 3.7 m )
PROBLEM 3.127 CONTINUED
Let x = 0.3 m,
M Gz :
( 200 N )( 0.7 m )  ( 400 N )( 0.7 m )  (180 N )( 0.7 m ) + W ( 0.7 m ) = 0
W = 380 N
M Gx :  ( 200 N )(1.5 m )  ( 400 N )(1.2 m ) + (180 N )( 0.6 m ) + ( 380 N )( z  1.8 m ) = 0 z = 3.5684 m < 3.7 m
Let z = 3.7 m,
acceptable
M Gx :  ( 200 N )(1.5 m )  ( 400 N )(1.2 m ) + (180 N )( 0.6 m ) + W (1.7 m ) = 0 W = 395.29 N > 380 N
Since the weight W found for x = 0.3 m is less than W found for z = 3.7 m, x = 0.3 m results in the smallest weight W. or W = 380 N at
( 0.3 m, 0, 3.57 m )
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 4m flatbed trailer with two 0.6 0.6 0.6m boxes and one 0.6 0.6 1.2m 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 1.2m 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 x = 0.6 m, with a small side of the box touching the zaxis, satisfies this condition. Let x = 0.6 m,
M Gz :
( 200 N )( 0.7 m )  ( 400 N )( 0.7 m )  (180 N )( 0.7 m ) + W ( 0.4 m ) = 0
W = 665 N
and
M GX :  ( 200 N )(1.5 m )  ( 400 N )(1.2 m ) + (180 N )( 0.6 m ) + ( 665 N )( z  1.8 m ) = 0 z = 2.8105 m
(2 m <
z < 4 m)
acceptable
or W = 665 N at
( 0.6 m, 0, 2.81 m )
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 forcecouple at the origin. Have
F : Pi  Pi  Pk = R R =  Pk
Have
R M O :  P ( 3a ) k  P ( 3a ) j + P ( ai + 3aj) = M O R M O = Pa ( i  3k )
Then let vectors ( R, M1 ) represent the components of the wrench, where their directions are the same. (a)
R =  Pk
or Magnitude of R = P Direction of R : x = 90, y = 90, z = 180
(b) Have
R M1 = R M O
= k Pa ( i  3k )
= 3Pa
and pitch
p= M1 3Pa = = 3a R P or p = 3a
PROBLEM 3.129 CONTINUED
(c) Have
R M O = M1 + M 2 R M 2 = M O  M1 = Pa ( i  3k )  ( 3Pak ) =  Pai
Require
M 2 = rQ/O R
 Pai = ( xi + yj) (  P ) k = Pxj  Pyi
From
i :  Pa =  Py j: x = 0
or
y =a
The axis of the wrench is parallel to the zaxis and intersects the xy plane at x = 0, 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 forcecouple system at the origin. Have
F :
( 2P ) i  ( P ) j + ( P ) j = R
R = ( 2P ) i
Have
R M O : ( rO F ) = M O
R MO
i j k i j k = Pa 2 2 2.5 + 0 0 4 = Pa ( 1.5i + 5j  6k ) 2 1 0 0 1 0 R = 2 Pi
or Magnitude of R = 2 P Direction of R : x = 0, y = 90, z = 90
(a)
(b) Have
R M1 = R MO
R =
R R
= i ( 1.5Pai + 5Paj  6 Pak ) = 1.5Pa
and pitch
p= M1 1.5Pa = = 0.75a R 2P
or p = 0.75a
PROBLEM 3.130 CONTINUED
(c) Have
R M O = M1 + M 2 R M 2 = M O  M1 = ( 5Pa ) j  ( 6Pa ) k
Require
M 2 = rQ/O R
( 5Pa ) j  ( 6Pa ) k = ( yj + zk ) ( 2Pi ) =  ( 2Py ) k + ( 2Pz ) j
From i : 5Pa = 2 Pz
z = 2.5a
From k :  6 Pa = 2 Py
y = 3a The axis of the wrench is parallel to the xaxis and intersects the yzplane at y = 3a, z = 2.5a
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 forcecouple at the origin. Have F :  (10 N ) j  (11 N ) j = R R =  ( 21 N ) j Have
R M O : ( rO F ) + M C = M O
R MO
i j k i j k = 0 0 0.5 N m + 0 0 0.375 N m  (12 N m ) j 0 10 0 0 11 0 = ( 0.875 N m ) i  (12 N m ) j
(a) (b) Have
R =  ( 21 N ) j
R M1 = R M O
or R =  ( 21 N ) j
R =
R R
= (  j) ( 0.875 N m ) i  (12 N m ) j
= 12 N m and pitch p=
and
M1 =  (12 N m ) j or p = 0.571 m
M1 12 N m = = 0.57143 m R 21 N
PROBLEM 3.131 CONTINUED
(c) Have
R M O = M1 + M 2 R M 2 = M O  M1 = ( 0.875 N m ) i
Require
M 2 = rQ/O R
( 0.875 N m ) i = ( xi + zk )  ( 21 N ) j
0.875i =  ( 21x ) k + ( 21z ) i
From i:
0.875 = 21z z = 0.041667 m
From k:
0 = 21x z =0
The axis of the wrench is parallel to the yaxis and intersects the xzplane at x = 0, z = 41.7 mm
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 forcecouple system. Have Have
F :  ( 6 lb ) i  ( 4.5 lb ) j = R
R = 7.5 lb
R M O : ( rO F ) + M C = M O R M O = 6 lb ( 8 in.) j  (160 lb in.) i  ( 72 lb in.) j
=  (160 lb in.) i  (120 lb in.) j
R M O = 200 lb in.
(a) (b) Have
R M1 = R M O
R =  ( 6 lb ) i  ( 4.5 lb ) j
=
R R
= ( 0.8i  0.6 j)  (160 lb in.) i  (120 lb in.) j
= 200 lb in.
and Pitch
M1 = 200 lb in. ( 0.8i  0.6j)
p=
M1 200 lb in. = = 26.667 in. R 7.50 lb or p = 26.7 in.
(c) From above note that
R M1 = M O
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 dy 3 = dx 4
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 forcecouple at the origin. Have and
F :  ( 20 lb ) k  ( 21 lb ) j =  ( 21 lb ) j  ( 20 lb ) k = R
R M O : ( rO F ) + M C = M O
R = 29 lb
i j k i j k R 20 lb ( 4 in.) 4 3 0 + 21 lb ( 4 in.) 6 0 1 + ( 300 j  320k ) lb in. = M O 0 0 1 0 1 0
R M O =  (156 lb in.) i + ( 20 lb in.) j  ( 824 lb in.) k
(a) (b) Have
R M1 = R M O
R =  ( 21 lb ) j  ( 20 lb ) k
R =
R R
=
21j  20k  (156 lb in.) i + ( 20 lb in.) j  ( 824 lb in.) k 29
= 553.80 lb in.
PROBLEM 3.133 CONTINUED
and Then pitch (c) Have
M1 = M1 R =  ( 401.03 lb in.) j  ( 381.93 lb in.) k
p=
M1 553.80 lb in. = = 19.0964 in. R 29 lb
R M O = M1 + M 2
or p = 19.10 in.
R M 2 = M O  M1 = ( 156i + 20 j  824k )  ( 401.03j  381.93k ) lb in.
=  (156.0 lb in.) i + ( 421.03 lb in.) j  ( 442.07 lb in.) k
Require
M 2 = rQ/O R
( 156i + 421.03j  442.07k ) = ( xi + zk ) ( 21j  20k )
= ( 21z ) i + ( 20 x ) j  ( 21x ) k
From i:
156 = 21z z = 7.4286 in.
or From k:
z = 7.43 in.
442.07 = 21x x = 21.051 in.
or
x = 21.1 in.
The axis of the wrench intersects the xzplane at
x = 21.1 in., z = 7.43 in.
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 forcecouple at the origin at B. (a) Have
15 8 F :  ( 79.2 lb ) k  ( 51 lb ) i + j = R 17 17 R =  ( 24.0 lb ) i  ( 45.0 lb ) j  ( 79.2 lb ) k and Have
R = 94.2 lb
M B : rA/B FA + M A + M B = M R B
MR B
i j k 15 8 0  660k  714 i + = 0 20 j = 1584i  660k  42 ( 8i + 15 j) 17 17 0 0 79.2 M R = (1248 lb in.) i  ( 630 lb in.) j  ( 660 lb in.) k B
(b) Have
R M1 = R M O
R =
R R
=
24.0i  45.0 j  79.2k (1248 lb in.) i  ( 630 lb in.) j  ( 660 lb in.) k 94.2
= 537.89 lb in.
PROBLEM 3.134 CONTINUED
and
M1 = M1 R =  (137.044 lb in.) i  ( 256.96 lb in.) j  ( 452.24 lb in.) k
Then pitch (c) Have
p=
M1 537.89 lb in. = = 5.7101 in. R 94.2 lb M R = M1 + M 2 B
or p = 5.71 in.
M 2 = M R  M1 = (1248i  630 j  660k )  ( 137.044i  256.96 j  452.24k ) B = (1385.04 lb in.) i  ( 373.04 lb in.) j  ( 207.76 lb in.) k
Require
M 2 = rQ/B R i j k 1385.04i  373.04 j  207.76k = x 0 z 24 45 79.2 = ( 45 z ) i  ( 24 z ) j + ( 79.2 x ) j  ( 45 x ) k
From i: From k:
1385.04 = 45 z
z = 30.779 in. x = 4.6169 in. The axis of the wrench intersects the xzplane at
x = 4.62 in., z = 30.8 in.
207.76 = 45x
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 forcecouple at the origin. Have
F : P BA + P DC + P DE = R 4 3 3 4 9 4 12 R = P j  k + i  j + i  j + k 5 5 5 5 25 5 25 R = 3P 25 3P ( 2i  20 j  k ) 25
R=
( 2 )2 + ( 20 )2 + (1)2
=
27 5 P 25
Have
R M : ( rO P ) = M O
( 24a ) j
4 P 3P 4P 4P 12 P 3P 9 P R j k + ( 20a ) j i j + ( 20a ) j i j+ k = MO 5 5 5 25 5 5 25
R MO =
24 Pa ( i  k ) 5
R M1 = R MO
(b) Have where
R =
3P 25 1 R = = ( 2i  20 j  k ) ( 2i  20 j  k ) R 25 27 5 P 9 5
PROBLEM 3.135 CONTINUED
Then
M1 = 1 9 5
( 2i  20 j  k )
24 Pa 8Pa ( i  k ) = 5 15 5
and pitch
p=
M1 8Pa 25 8a = = R 81 15 5 27 5 P
or p = 0.0988a
(c)
M1 = M 1 R =
8Pa 1 8Pa ( 2i + 20 j + k ) ( 2i  20 j  k ) = 675 15 5 9 5
Then Require
R M 2 = M O  M1 =
24Pa 8Pa 8Pa ( i  k )  ( 2i + 20 j + k ) = ( 403i  20 j  406k ) 5 675 675 M 2 = rQ/O R
8Pa 3P ( 403i  20 j  406k ) = ( xi + zk ) ( 2i  20 j  k ) 675 25 3P = 20 zi + ( x + 2 z ) j  20 xk 25 From i: 8 ( 403) 8 ( 406 )
Pa 3P = 20 z 675 25 Pa 3P = 20 x 675 25
z = 1.99012a
From k:
x = 2.0049a
The axis of the wrench intersects the xzplane at
x = 2.00a, z = 1.990a
PROBLEM 3.136
Determine whether the forceandcouple 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 forcecouple at D. Have
F : FDA + FED = FDA DA + FED ED = R
 ( 0.300 m ) i + ( 0.225 m ) j + ( 0.200 m ) k FDA = 136 N 0.425 m
where
=  ( 96 N ) i + ( 72 N ) j + ( 64 N ) k
 ( 0.150 m ) i  ( 0.200 m ) k FED = 120 N =  ( 72 N ) i  ( 96 N ) k 0.250 m
R =  (168 N ) i + ( 72 N ) j  ( 32 N ) k
Have or
R M D : M A = M D
 ( 0.150 m ) i  ( 0.150 m ) j + ( 0.450 m ) k 16 N m M R = (16 N m ) ( i  j + 3k ) = D 0.150 11 m 11
PROBLEM 3.136 CONTINUED
The forcecouple at D can be replaced by a single force if R is perpendicular to M R . To be perpendicular, D R M R = 0. D Have
R M R = ( 168i + 72 j  32k ) D = 128 ( 21  9  12 ) 11 16 ( i  j + 3k ) 11
=0
Forcecouple can be reduced to a single equivalent force.
To determine the coordinates where the equivalent single force intersects the yzplane, M R = rQ/D R D where
rQ/D = ( 0  0.300 ) m i + ( y  0.075 ) m j + ( z  0 ) m k
i 16 N m ( i  j + 3k ) = (8 N ) 0.3 11 21 or
j k ( y  0.075) z m 9 4
16 N m ( i  j + 3k ) = (8 N ) 4 ( y  0.075)  9 z i + ( 21z  1.2 ) j + 2.7 + 21 ( y  0.075) k m 11
{
}
From j:
16 = 8 ( 21z  1.2 ) 11 48 = 8 2.7 + 21( y  0.075 ) 11
z = 0.028427 m = 28.4 mm
y = 0.28972 m = 290 mm
From k:
line of action of R intersects the yzplane at
y = 290 mm, z = 28.4 mm
PROBLEM 3.137
Determine whether the forceandcouple 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 forcecouple at the origin. Have F : FA + FG = R
( 4 in.) i + ( 6 in.) j  (12 in.) k R = (10 lb ) k + 14 lb = ( 4 lb ) i + ( 6 lb ) j  ( 2 lb ) k 14 in.
and Have
R=
56 lb
R M O : ( rO F ) + M C = M O
R M O = (12 in.) j (10 lb ) k + (16 in.) i ( 4 lb ) i + ( 6 lb ) j  (12 lb ) k
{
}
(16 in.) i  (12 in.) j ( 4 in.) i  (12 in.) j + ( 6 in.) k + ( 84 lb in.) + ( 120 lb in.) 20 in. 14 in.
R M 0 = ( 221.49 lb in.) i + ( 38.743 lb in.) j + (147.429 lb in.) k
= (18.4572 lb ft ) i + ( 3.2286 lb ft ) j + (12.2858 lb ft ) k
PROBLEM 3.137 CONTINUED
R The forcecouple at O can be replaced by a single force if the direction of R is perpendicular to M O . R To be perpendicular R M O = 0
Have
R R M O = ( 4i + 6 j  2k ) (18.4572i + 3.2286 j + 12.2858k ) = 0?
= 73.829 + 19.3716  24.572 0 System cannot be reduced to a single equivalent force. To reduce to an equivalent wrench, the moment component along the line of action of P is found.
R M1 = R M O
R =
R R
( 4i + 6 j  2k ) = (18.4572i + 3.2286 j + 12.2858k ) 56
= 9.1709 lb ft and And pitch M1 = M1 R = ( 9.1709 lb ft )( 0.53452i + 0.80178 j  0.26726k )
p= M1 9.1709 lb ft = = 1.22551 ft R 56 lb
or p = 1.226 ft Have
R M 2 = M O  M1 = (18.4572i + 3.2286 j + 12.2858k )  ( 9.1709 )( 0.53452i + 0.80178 j  0.26726k )
= (13.5552 lb ft ) i  ( 4.1244 lb ft ) j + (14.7368 lb ft ) k Require M 2 = rQ/O R
(13.5552i  4.1244 j + 14.7368k ) = ( yj + zk ) ( 4i + 6 j  2k )
=  ( 2 y + 6z ) i + ( 4z ) j  ( 4 y ) k From j: From k: 4.1244 = 4z 14.7368 = 4 y or or z = 1.0311 ft y = 3.6842 ft
line of action of the wrench intersects the yz plane at y = 3.68 ft, z = 1.031 ft
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 A = Axi + Az k B = Bxi + Bzk Then, for equivalence to the given force system Fx : Ax + Bx = 0 Fz : Az + Bz = R M x : Az ( a ) + Bz ( a + b ) = 0 M z :  Ax ( a )  Bx ( a + b ) = M From Equation (1), Substitute into Equation (4)  Ax ( a ) + Ax ( a + b ) = M Ax = From Equation (2), and Equation (3), M b and Bx =  M b Bx =  Ax (1) (2) (3) (4)
Bz = R  Az Az a + ( R  Az )( a + b ) = 0
a Az = R 1 + b
PROBLEM 3.138 CONTINUED
and a Bz = R  R 1 + b a Bz =  R b Then M A= b a i + R 1 + k b a i  Rk b
M B =  b
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 xaxis intercept of B. The known components of the wrench can be expressed as R = Rxi + Ry j + Rzk and M = M xi + M y j + M zk
while the unknown forces A and B can be expressed as A = Axi + Ay j + Azk and B = Bxi + Bzk
Since the position vector of point P is given, it follows that the scalar components (x, y, z) of the position vector rP are also known. Then, for equivalence of the two systems Fx : Rx = Ax + Bx Fy : Ry = Ay Fz : Rz = Az + Bz M x : M x = yAz  zAy M y : M y = zAx  xAz  bBz M z : M z = xAy  yAx (1) (2) (3) (4) (5) (6)
PROBLEM 3.139 CONTINUED
Based on the above six independent equations for the six unknowns Ax , Ay , Az , Bx , Bz , b , there exists a unique solution for A and B. From Equation (2) Equation (6) Ay = Ry 1 Ax = xRy  M z y
(
)
(
) ) ) )
Equation (1)
1 Bx = Rx  xRy  M z y
(
Equation (4)
1 Az = M x + zRy y
(
Equation (3)
1 Bz = Rz  M x + zRy y
(
Equation (5)
b=
( xM x + yM y + zM z ) ( M x  yRz + zRy )
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 R = Rj and M = Mj and are known.
The unknown forces A and B can be expressed as A = Axi + Ay j + Azk and B = Bxi + By j + Bzk
The distance a is known. It is assumed that force B intersects the xz plane at (x, 0, z). Then for equivalence Fx : 0 = Ax + Bx Fy : R = Ay + By Fz : 0 = Az + Bz M x : 0 =  zBy M y : M = aAz  xBz + zBx M z : 0 = aAy + xBy Since A and B are made perpendicular, AB = 0 There are eight unknowns: or Ax Bx + Ay By + Az Bz = 0 Ax , Ay , Az , Bx , By , Bz , x, z (7) (1) (2) (3) (4) (5) (6)
But only seven independent equations. Therefore, there exists an infinite number of solutions.
PROBLEM 3.140 CONTINUED
Next consider Equation (4): If By = 0, Equation (7) becomes Using Equations (1) and (3) this equation becomes 0 =  zBy Ax Bx + Az Bz = 0
2 2 Ax + Az = 0
Since the components of A must be real, a nontrivial solution is not possible. Thus, it is required that By 0, so that from Equation (4), z = 0. To obtain one possible solution, arbitrarily let Ax = 0. (Note: Setting Ay , Az , or Bz equal to zero results in unacceptable solutions.) The defining equations then become. 0 = Bx R = Ay + By 0 = Az + Bz M = aAz  xBz 0 = aAy + xBy Ay By + Az Bz = 0 Then Equation (2) can be written Equation (3) can be written Equation (6) can be written Substituting into Equation (5),
R  By M = aAz  a (  Az ) By
(1) (2) (3) (5) (6) (7) Ay = R  By Bz =  Az
x= aAy By
or Substituting into Equation (7),
Az = 
M By aR
(8)
M M ( R  By ) By +  aR By aR By = 0
PROBLEM 3.140 CONTINUED
or Then from Equations (2), (8), and (3)
Ay = R 
Az = 
By =
a 2 R3 a R2 + M 2
2
a 2 R3 RM 2 = 2 2 a2R2 + M 2 a R + M2
M a 2 R3 aR 2 M 2 2 = 2 2 aR a R + M 2 a R + M2
Bz =
aR 2 M a2R2 + M 2
In summary A=
RM ( Mj  aRk ) a R2 + M 2
2
B=
aR 2 ( aRj + Mk ) a2R2 + M 2
Which shows that it is possible to replace a wrench with two perpendicular forces, one of which is applied at a given point. Lastly, if R > 0 and M > 0, it follows from the equations found for A and B that Ay > 0 and By > 0. From Equation (6), x < 0 (assuming a > 0). Then, as a consequence of letting Ax = 0, force A lies in a plane parallel to the yz plane and to the right of the origin, while force B lies in a plane parallel to the yz plane but to the left of the origin, as shown in the figure below.
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 P ( x, 0, z ) . Denoting the known direction of line AA by A = xi + y j + zk it follows that force A can be expressed as A = A A = A xi + y j + z k Force B can be expressed as B = Bxi + By j + Bzk Next, observe that since the axis of the wrench and the prescribed line of action AA are known, it follows that the distance a can be determined. In the following solution, it is assumed that a is known. Then, for equivalence
(
)
Fx : 0 = Ax + Bx Fy : R = A y + By Fz : 0 = Az + Bz M x : 0 =  zBy M y : M = aAz + zBx  xBz M z : 0 = aA y + xBy Since there are six unknowns A, Bx , By , Bz , x, z and six independent equations, it will be possible to obtain a solution.
(1) (2) (3) (4) (5) (6)
(
)
PROBLEM 3.141 CONTINUED
Case 1: Let z = 0 to satisfy Equation (4)
Now Equation (2) Equation (3) Equation (6) Substitution into Equation (5)
x=
A y = R  By Bz =  Az
aA y By a =  R  By By
(
)
a M = aAz   R  By (  Az ) By
(
)
A= Substitution into Equation (2) R=
z aR
1 M
By
1 M B + By z aR y y
By = Then
A=
z aR 2 z aR  y M
MR R = aR z aR  y M y  z M
Bx =  Ax = Bz =  Az =
x MR z aR  y M z MR z aR  y M
P A aR y  z M
In summary
A=
B=
R ( Mi + z aRj + z Mk ) z aR  y M x
and
z aR  y M R x = a 1  = a 1  R By z aR 2
or x = Note that for this case, the lines of action of both A and B intersect the x axis.
y M z R
PROBLEM 3.141 CONTINUED
Case 2: Let By = 0 to satisfy Equation (4)
Now Equation (2)
A=
R
y
Equation (1)
Bx =  R x y Bz =  R z y aA y = 0
Equation (3) Equation (6) Substitution into Equation (5)
M = z R x y  x R z y
which requires a = 0
or
z x  x z = y R
M
This last expression is the equation for the line of action of force B. In summary
R A= y R B= y A
(  x i  z k )
Assuming that x , y , z > 0, 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 360N force to a steel bar as shown. (a) Replace that force with an equivalent forcecouple 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
F : 360 N (  sin 40i  cos 40 j) =  ( 231.40 N ) i  ( 275.78 N ) j = F
or F = 360 N Have where
M D : rB/D R = M
rB/D =  ( 0.65 m ) cos 30 i + ( 0.65 m ) sin 30 j
50
=  ( 0.56292 m ) i + ( 0.32500 m ) j
i j k M = 0.56292 0.32500 0 N m = (155.240 + 75.206 ) N m k 231.40 275.78 0 = ( 230.45 N m ) k (b) Have where M D : M = rA/D FA
rA/D =  (1.05 m ) cos 30 i + (1.05 m ) sin 30 j
or M = 230 N m
=  ( 0.90933 m ) i + ( 0.52500 m ) j
PROBLEM 3.142 CONTINUED
i j k FA 0.90933 0.52500 0 N m = [ 230.45 N m ] k 0 1 0 or
( 0.90933FA ) k
FA = 253.42 N
= 230.45k or FA = 253 N
Have
F : F = FA + FD  ( 231.40 N ) i  ( 275.78 N ) j =  ( 253.42 N ) j + FD (  cos i  sin j)
From
i : 231.40 N = FD cos j: 22.36 N = FD sin
(1) (2)
Equation (2) divided by Equation (1) tan = 0.096629 = 5.5193 Substitution into Equation (1) or
= 5.52
FD =
231.40 = 232.48 N cos5.5193 or FD = 232 N 5.52
PROBLEM 3.143
A worker tries to move a rock by applying a 360N 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 360N force shown in the figure.
SOLUTION
Have
F : R = FA + FC
 ( 360 N ) sin 40 i  ( 360 N ) cos 40 j =  ( FA + FC ) sin i  ( FA + FC ) cos j
From
i: j:
( 360 N ) sin 40 = ( FA + FC ) sin ( 360 N ) cos 40 = ( FA + FC ) cos
tan 40 = tan = 40
(1) (2)
Dividing Equation (1) by Equation (2),
Substituting = 40 into Equation (1), FA + FC = 360 N Have where M C : rB/C R = rA/C FA (3)
rB/C = ( 0.35 m )( cos30i + sin 30 j) =  ( 0.30311 m ) i + ( 0.175 m ) j
PROBLEM 3.143 CONTINUED
R = ( 360 N )( sin40i  cos 40 j) =  ( 231.40 N ) i  ( 275.78 N ) j rA/C = ( 0.75 m )( cos30i + sin 30 j) =  ( 0.64952 m ) i + ( 0.375 m ) j FA = FA (  sin 40i  cos 40 j) = FA ( 0.64279i  0.76604 j) i j k i j k 0.30311 0.175 0 N m = FA 0.64952 0.375 0 N m 231.40 275.78 0 0.64279 0.76604 0 83.592 + 40.495 = ( 0.49756 + 0.24105 ) FA FA = 168.002 N Substituting into Equation (3),
FC = 360  168.002 = 191.998 N
or
FA = 168.0 N FC = 192.0 N
or
or FA = 168.0 N FC = 192.0 N
50 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 360N forces are reversed.
SOLUTION
(a) (a) Have F : F = ( 360 N ) j  ( 360 N ) j  ( 600 N ) k or F =  ( 600 N ) k and M D :
( 360 N )( 0.15 m ) = ( 600 N )( d )
or d = 90.0 mm below ED
d = 0.09 m
(b) Have from part a (b) and
F =  ( 600 N ) k M D :  ( 360 N )( 0.15 m ) =  ( 600 N )( d ) d = 0.09 m or d = 90.0 mm above 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
Have W = mg = 80 kg 9.81 m/s 2 = 784.8 N M E : M E = ( 784.8 N )( 0.25 m ) M E = 196.2 N m (b) For the force at B to be the smallest, resulting in a moment ( M E ) about E, the line of action of force FB must be perpendicular to the line connecting E to B. The sense of FB must be such that the force produces a counterclockwise moment about E. Note: Have d =
(
)
( 0.85 m )2 + ( 0.5 m )2
= 0.98615 m
M E : 196.2 N m = FB ( 0.98615 m ) FB = 198.954 N
and
= tan 1 = 59.534 0.5 m
or FB = 199.0 N 59.5
0.85 m
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 Have W = mg = 80 kg 9.81 m/s 2 = 784.8 N M E : M E = ( 784.8 N )( 0.25 m ) M E = 196.2 N m (b) For the force at A to be the smallest, resulting in a moment about E, the line of action of force FA must be perpendicular to the line connecting E to A. The sense of FA must be such that the force produces a counterclockwise moment about E. Note: Have d =
(
)
( 0.35 m )2 + ( 0.5 m )2
= 0.61033 m
M E : 196.2 N m = FA ( 0.61033 m ) FA = 321.47 N
and
= tan 1 = 34.992 0.5 m
or FA = 321 N 35.0
0.35 m
(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 M E : M E = rD/E FD
(196.2 N m ) k = ( 0.85 m ) i ( FD ) j (196.2 N m ) k = ( 0.85FD ) k
FD = 230.82 N or FD = 231 N
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 TDE when TAB = 255 lb.
SOLUTION
The moment about the xaxis due to the two cable forces can be found using the zcomponents of each force acting at their intersection with the xyplane (A and D). The xcomponents of the forces are parallel to the xaxis, and the ycomponents of the forces intersect the xaxis. Therefore, neither the x or y components produce a moment about the xaxis. Have
where M x :
(TAB ) z ( y A ) + (TDE ) z ( yD ) = M x
(TAB ) z
= k TAB = k (TAB AB ) i  12 j + 12k = k 255 lb = 180 lb 17
(TDE ) z
= k TDE = k (TDE DE ) 1.5i  14 j + 12k = k TDE = 0.64865TDE 18.5
y A = 12 ft yD = 14 ft M x = 4728 lb ft
and
(180 lb )(12 ft ) + ( 0.64865TDE )(14 ft ) = 4728 lb ft
TDE = 282.79 lb
or TDE = 283 lb
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 TDE when TAB = 255 lb.
SOLUTION
The moment about the xaxis 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 where M x :
(TAB ) z ( y A ) + (TDE ) z ( yD ) = M x
(TAB ) z
= k TAB = k (TAB AB ) i  12 j + 12k = k 306 lb = 216 lb 17
(TDE ) z
= k TDE = k (TDE DE ) 1.5i  14 j + 12k = k TDE = 0.64865TDE 18.5
y A = 12 ft yD = 14 ft M x = 4728 lb ft
and
( 216 lb )(12 ft ) + ( 0.64865TDE )(14 ft ) = 4728 lb ft
TDE = 235.21 lb
or TDE = 235 lb
PROBLEM 3.149
As an adjustable brace BC is used to bring a wall into plumb, the forcecouple system shown is exerted on the wall. Replace this forcecouple system with an equivalent forcecouple system at A knowing that R = 21.2 lb and M = 13.25 lb ft.
SOLUTION
Have where BC =
RA =
F : R = R A = R BC
( 42 in.) i  ( 96 in.) j  (16 in.) k
106 in.
21.2 lb ( 42i  96 j  16k ) 106
or R A = ( 8.40 lb ) i  (19.20 lb ) j  ( 3.20 lb ) k Have where M A : rC/ A R + M = M A rC/ A = ( 42 in.) i + ( 48 in.) k = = ( 3.5 ft ) i + ( 4.0 ft ) k R = ( 8.40 lb ) i  (19.20 lb ) j  ( 3.20 lb ) k M =  BC M = 42i + 96 j + 16k (13.25 lb ft ) 106 1 ( 42i + 48k ) ft 12
=  ( 5.25 lb ft ) i + (12 lb ft ) j + ( 2 lb ft ) k
PROBLEM 3.149 CONTINUED
Then
i j k 3.5 0 4.0 lb ft + ( 5.25i + 12 j + 2k ) lb ft = M A 8.40 19.20 3.20 M A = ( 71.55 lb ft ) i + ( 56.80 lb ft ) j  ( 65.20 lb ft ) k or M A = ( 71.6 lb ft ) i + ( 56.8 lb ft ) j  ( 65.2 lb ft ) k
PROBLEM 3.150
Two parallel 60N 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 where
M B :  d1Cx + d 2C y = M
d1 = ( 0.360 m ) sin 55 = 0.29489 m d 2 = ( 0.360 m ) cos 55 = 0.20649 m Cx = ( 60 N ) cos 20 = 56.382 N C y = ( 60 N ) sin 20 = 20.521 N
M =  ( 0.29489 m )( 56.382 N ) k + ( 0.20649 m )( 20.521 N ) k =  (12.3893 N m ) k or M = 12.39 N m
(b) Have
M = Fd ( k ) = 60 N ( 0.360 m ) sin ( 55  20 ) ( k )
=  (12.3893 N m ) k or M = 12.39 N m
PROBLEM 3.150 CONTINUED
(c) Have M A : ( rA F ) = rB/ A FB + rC/ A FC = M
i j k i j k M = ( 0.520 m )( 60 N ) cos 55 sin 55 0 + ( 0.880 m )( 60 N ) cos 55 sin 55 0  cos 20  sin 20 0 cos 20 sin 20 0 = (17.8956 N m  30.285 N m ) k =  (12.3892 N m ) k or M = 12.39 N m
PROBLEM 3.151
A 32lb 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
F :
( 60 lb ) i  ( 32 lb ) j + (140 lb )( cos 30i + sin 30j) = R
R = (181.244 lb ) i + ( 38.0 lb ) j or R = 185.2 lb 11.84
Have
M O : M O = xRy
 (140 lb ) cos 30 ( 4 + 2 cos 30 ) in.  (140 lb ) sin 30 ( 2 in.) sin 30
 ( 60 lb )( 2 in.) = x ( 38.0 lb )
x=
and
1 ( 694.97  70.0  120 ) in. 38.0
x = 23.289 in.
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, M x = 61 lb ft, and
M z = 43 lb ft, determine and d.
SOLUTION
Have where M O : rA/O F = M O
rA/O =  ( 4 in.) i + (11 in.) j  ( d ) k F = F ( cos cos i  sin j + cos sin k )
For
F = 70 lb, = 25
F = ( 70 lb ) ( 0.90631cos ) i  0.42262 j + ( 0.90631sin ) k
MO
i j k = ( 70 lb ) 4 11 d in. 0.90631cos 0.42262 0.90631sin
= ( 70 lb ) ( 9.9694sin  0.42262d ) i + ( 0.90631d cos + 3.6252sin ) j + (1.69048  9.9694 cos ) k in.
and
M x = ( 70 lb )( 9.9694sin  0.42262d ) in. =  ( 61 lb ft )(12 in./ft ) M y = ( 70 lb )( 0.90631d cos + 3.6252sin ) in. M z = ( 70 lb )(1.69048  9.9694cos ) in. = 43 lb ft (12 in./ft )
(1) (2) (3)
PROBLEM 3.152 CONTINUED
From Equation (3)
= cos 1 = 24.636 697.86
or = 24.6 From Equation (1) 1022.90 d = = 34.577 in. 29.583 or d = 34.6 in.
634.33
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, M x = 77 lb ft and M z = 81 lb ft. For d = 27 in., determine the moment M y of F about the y axis.
SOLUTION
Have where M O : rA/O F = M O
rA/O =  ( 4 in.) i + (11 in.) j  ( 27 in.) k F = F ( cos cos i  sin j + cos sin k )
MO
i j k = F 4 11 27 lb in. cos cos  sin cos sin
= F (11cos sin  27sin ) i + ( 27 cos cos + 4cos sin ) j + ( 4sin  11cos cos ) k ( lb in.)
and
M x = F (11cos sin  27sin )( lb in.) M y = F ( 27 cos cos + 4cos sin )( lb in.) M z = F ( 4sin  11cos cos )( lb in.)
(1) (2) (3) (4) (5)
Now, Equation (1) and Equation (3)
cos sin = cos cos =
1 Mx + 27sin 11 F 1 Mz 4sin  F 11
Substituting Equations (4) and (5) into Equation (2),
1 1 M M M y = F 27 4sin  z + 4 x + 27sin F 11 11 F
or
My =
1 ( 27M z + 4M x ) 11
PROBLEM 3.153 CONTINUED
Noting that the ratios 27 4 and are the ratios of lengths, have 11 11 My = 27 4 ( 81 lb ft ) + ( 77 lb ft ) = 226.82 lb ft 11 11 or M y = 227 lb ft
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