PROBLEM 3.1
A 13.2-N force P is applied to the lever which controls the auger of a
snowblower. Determine the moment of P about A when α is equal to 30°.
SOLUTION
First note
Px = P sin α = (13.2 N ) sin 30° = 6.60 N
Py = P cos α = (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
W
PROBLEM 3.2
The force P is applied to the lever which controls the auger of a
snowblower. Determine the magnitude and the direction of the smallest
force P which has a 2.20- N ⋅ m counterclockwise moment about A.
SOLUTION
For P to be a minimum, it must be perpendicular to the line joining points
A and B.
rAB =
(86 mm )2 + (122 mm )2
 y
= 149.265 mm
 122 mm 
α = θ = tan −1   = tan −1 
 = 54.819°
x
 86 mm 
M A = rAB Pmin
Then
or
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° W
PROBLEM 3.3
A 13.1-N force P is applied to the lever which controls the auger of a
snowblower. Determine the value of α knowing that the moment of P
about A is counterclockwise and has a magnitude of 1.95 N ⋅ m.
SOLUTION
M A = rB/ A P sin θ
By definition
θ = φ + ( 90° − α )
where
 122 mm 
φ = tan −1 
 = 54.819°
 86 mm 
and
Also
Then
rB/ A =
(86 mm )2 + (122 mm )2
= 149.265 mm
1.95 N ⋅ m = ( 0.149265 m )(13.1 N ) sin ( 54.819° + 90° − α )
or
sin (144.819° − α ) = 0.99725
or
144.819° − α = 85.752°
and
144.819° − α = 94.248°
∴ α = 50.6°, 59.1° W
PROBLEM 3.4
A foot valve for a pneumatic system is hinged at B. Knowing that
α = 28°, determine the moment of the 4-lb force about point B by
resolving the force into horizontal and vertical components.
SOLUTION
Note that
θ = α − 20° = 28° − 20° = 8°
and
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. W
PROBLEM 3.5
A foot valve for a pneumatic system is hinged at B. Knowing that
α = 28°, determine the moment of the 4-lb force about point B by
resolving the force into components along ABC and in a direction
perpendicular to ABC.
SOLUTION
First resolve the 4-lb force into components P and Q, where
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. W
PROBLEM 3.6
It is known that a vertical force of 800 N is required to remove the nail at
C from the board. As the nail first starts moving, determine (a) the
moment about B of the force exerted on the nail, (b) the magnitude of the
force P which creates the same moment about B if α = 10°, (c) the
smallest force P which creates the same moment about B.
SOLUTION
(a) Have
M B = rC/B FN
= ( 0.1 m )( 800 N )
= 80.0 N ⋅ m
or M B = 80.0 N ⋅ m
W
(b) By definition
M B = rA/B P sin θ
where
θ = 90° − ( 90° − 70° ) − α
= 90° − 20° − 10°
= 60°
∴ 80.0 N ⋅ m = ( 0.45 m ) P sin 60°
P = 205.28 N
or P = 205 N W
(c) For P to be minimum, it must be perpendicular to the line joining
points A and B. Thus, P must be directed as shown.
Thus
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° W
PROBLEM 3.7
A sign is suspended from two chains AE and BF. Knowing that the
tension in BF is 45 lb, determine (a) the moment about A of the force
exert by the chain at B, (b) the smallest force applied at C which creates
the same moment about A.
SOLUTION
M A = rB/ A × TBF
(a) Have
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
W
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
 4.4 ft 
φ = tan −1 
 = 34.095°
 6.5 ft 
θ = 90° − φ = 90° − 34.095° = 55.905°
or
( FC )min
= 36.0 lb
55.9° W
PROBLEM 3.8
A sign is suspended from two chains AE and BF. Knowing that the
tension in BF is 45 lb, determine (a) the moment about A of the force
exerted by the chain at B, (b) the magnitude and sense of the vertical
force applied at C which creates the same moment about A, (c) the
smallest force applied at B which creates the same moment about A.
SOLUTION
M A = rB/ A × TBF
(a) Have
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
W
M A = rC/ A × FC
(b) Have
M A = xFC
or
∴ FC =
MA
282.56 lb ⋅ ft
=
= 43.471 lb
x
6.5 ft
or FC = 43.5 lb W
M A = rB/ A × ( FB )min
(c) Have
For FB to be minimum, it must be perpendicular to the line joining
points A and B.
∴ M A = d ( FB )min
d =
where
∴
and
( 6.5 ft )2 + ( 4.4 ft
( FB )min
=
− 3.1 ft ) = 6.6287 ft
2
MA
282.56 lb ⋅ ft
=
= 42.627 lb
d
6.6287 ft

6.5 ft

θ = tan −1 
 = 78.690°
 4.4 ft − 3.1 ft 
or
( FB )min
= 42.6 lb
78.7° W
PROBLEM 3.9
The tailgate of a car is supported by the hydraulic lift BC. If the lift exerts
a 125-N force directed along its center line on the ball and socket at B,
determine the moment of the force about A.
SOLUTION
First note
dCB =
( 240 mm )2 + ( 46.6 mm )2
= 244.48 mm
Then
and
where
240 mm
244.48 mm
sin θ =
46.6 mm
244.48 mm
FCB = FCB cosθ i − FCB sin θ j
=
Now
cosθ =
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
Then
125 N
M A = ( 306 mm ) i − ( 286.6 mm ) j ×
( 240i − 46.6 j)
244.48
= ( 27878 N ⋅ mm ) k = ( 27.878 N ⋅ m ) k
or M A = 27.9 N ⋅ m
W
PROBLEM 3.10
The tailgate of a car is supported by the hydraulic lift BC. If the lift exerts
a 125-N force directed along its center line on the ball and socket at B,
determine the moment of the force about A.
SOLUTION
First note
Then
and
dCB =
( 344 mm )2 + (152.4 mm )2
cosθ =
344 mm
376.25 mm
152.4 mm
376.25 mm
FCB = ( FCB cosθ ) i − ( FCB sin θ ) j
=
Now
sin θ =
= 376.25 mm
125 N
( 344 mm ) i + (152.4 mm ) j
376.25 mm 
M A = rB/ A × FCB
where
rB/ A = ( 410 mm ) i − ( 87.6 mm ) j
Then
125 N
M A = ( 410 mm ) i − ( 87.6 mm ) j ×
( 344i − 152.4 j)
376.25
= ( 30770 N ⋅ mm ) k
= ( 30.770 N ⋅ m ) k
or M A = 30.8 N ⋅ m
W
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
Slope of line EC =
(a)
Then
and
Then
TABx =
35 in.
5
=
76 in. + 8 in. 12
12
(TAB )
13
=
12
( 260 lb ) = 240 lb
13
TABy =
5
( 260 lb ) = 100 lb
13
M D = TABx ( 35 in.) − TABy ( 8 in.)
= ( 240 lb )( 35 in.) − (100 lb )( 8 in.)
= 7600 lb ⋅ in.
or M D = 7600 lb ⋅ in.
(b) Have
M D = TABx ( y ) + TABy ( x )
= ( 240 lb )( 0 ) + (100 lb )( 76 in.)
= 7600 lb ⋅ in.
or M D = 7600 lb ⋅ in.
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
Then
TABx =
24
TAB
25
and
TABy =
7
TAB
25
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 )
M D = 1152 N ⋅ m
where
(TAB max ) y
= TAB max sin θ = ( 2880 N ) sin θ
1.05 m
sin θ =
Now
(d

∴ 1152 N ⋅ m = 2880 N 


or
or
or
+ 0.24 ) + (1.05 ) m
2
1.05
( d + 0.24 )2 + (1.05)2
( d + 0.24 )2 + (1.05)2
(d
2

 (d )


= 2.625d
+ 0.24 ) + (1.05 ) = 6.8906d 2
2
2
5.8906d 2 − 0.48d − 1.1601 = 0
Using the quadratic equation, the minimum values of d are 0.48639 m and −0.40490 m.
Since only the positive value applies here, d = 0.48639 m
or d = 486 mm
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
M C = rB/C × FB
Have
Noting the direction of the moment of each force component about C is
clockwise,
M C = xFBy + yFBx
where
x = 144 mm − 78 mm = 66 mm
y = 86 mm + 108 mm = 194 mm
and
FBx =
FBy =
78
( 78)
2
+ ( 86 )
2
86
( 78) + (86 )
2
2
( 580 N ) = 389.65 N
( 580 N ) = 429.62 N
∴ M C = ( 66 mm )( 429.62 N ) + (194 mm )( 389.65 N )
= 103947 N ⋅ mm
= 103.947 N ⋅ m
or M C = 103.9 N ⋅ m
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 α cos β =
SOLUTION
1
sin
2
(α + β ) +
1
sin
2
(α − β ) .
B = B ( cos β i + sin β j)
First note
B′ = B ( cos β i − sin β j)
C = C ( cos α i + sin α j)
By definition
Now
B × C = BC sin (α − β )
(1)
B′ × C = BC sin (α + β )
(2)
B × C = B ( cos β i + sin β j) × C ( cos α i + sin α j)
= BC ( cos β sin α − sin β cos α ) k
(3)
B × C = B ( cos β i − sin β j) × C ( cos α i + sin α j)
= BC ( cos β sin α + sin β cos α ) k
Equating magnitudes of B × C from Equations (1) and (3),
(4)
(5)
sin (α − β ) = cos β sin α − sin β cos α
Similarly, equating magnitudes of B′ × C from Equations (2) and (4),
sin (α + β ) = cos β sin α + sin β cos α
(6)
Adding Equations (5) and (6)
sin (α − β ) + sin (α + β ) = 2cos β sin α
∴ sin α cos β =
1
1
sin (α + β ) + sin (α − β )
2
2
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
d = λ AB × rO/ A
Have
λ AB =
where
rB/ A
rB/ A
rB/ A = ( −140 mm − 420 mm ) i + 180 mm − ( −150 mm )  j
and
= − ( 560 mm ) i + ( 330 mm ) j
rB/ A =
∴ λ AB =
( −560 )2 + ( 330 )2 mm = 650 mm
− ( 560 mm ) i + ( 330 mm ) j
1
=
( −56i + 33j)
650 mm
65
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
λ =
(a) Have
A×B
A×B
A = 4i − 2 j + 3k
where
B = −2i + 6 j − 5k
Then
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
A×B = 2
and
∴ λ =
( −4 )2 + ( 7 )2 + (10 )2
2 ( −4i + 7 j + 10k )
or λ =
2 165
λ =
(b) Have
= 2 165
1
( −4i + 7 j + 10k )
165
A×B
A×B
A = 7i + j − 4k
where
B = −6i − 3j + 2k
Then
and
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
A×B = 5
∴ λ =
( −2 )2 + ( 2 )2 + ( −3)2
5 ( −2i + 2 j − 3k )
5 17
= 5 17
or λ =
1
( −2i + 2 j − 3k )
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 = P×Q
(a) Have
P = ( 8 in.) i + ( 2 in.) j − (1 in.) k
where
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
∴ Α=
(8)2 + ( −13)2 + ( 38)2 in 2
= 40.951 in 2
or A = 41.0 in 2
A = P×Q
(b) Have
P = − ( 3 in.) i + ( 6 in.) j + ( 4 in.) k
where
Q = ( 2 in.) i + ( 5 in.) j − ( 3 in.) k
Then
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
∴ Α=
( −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
MO = r × F
(a) Have
F = − (5 N ) i − ( 2 N ) j + (3 N ) k
where
r = ( 4 m ) i − ( 2 m ) j − (1 m ) k
∴ MO
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
MO = r × F
(b) Have
F = − (5 N ) i − ( 2 N ) j + (3 N ) k
where
r = − (8 m ) i + ( 3 m ) j − ( 4 m ) k
∴ MO
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
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
MO = r × F
(a) Have
F = − (1.5 lb ) i + ( 3 lb ) j + ( 2 lb ) k
where
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
MO = r × F
(b) Have
F = − (1.5 lb ) i + ( 3 lb ) j − ( 2 lb ) k
where
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.
MO = r × F
(c) Have
F = − (1.5 lb ) i − ( 3 lb ) j − ( 2 lb ) k
where
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
M O = rB/O × FB
Have
rB/O = ( 8.4 m ) j
where
FB = TAB + TBC
TAB = λ BATAB =
− ( 0.9 m ) i − ( 8.4 m ) j + ( 7.2 m ) k
TBC = λ BCTBC =
( 0.9 )2 + (8.4 )2 + ( 7.2 )2
m
( 777 N )
( 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 = λT2 = 

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
MO
i
j
k
=
0
9
1 N⋅m
−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
PROBLEM 3.23
An 8-lb force is applied to a wrench to tighten a showerhead. Knowing
that the centerline of the wrench is parallel to the x axis, determine the
moment of the force about A.
SOLUTION
Have
M A = rC/ A × F
where
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
and
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
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
M C = rA/C × FBA
where
rA/C = ( 0.96 m ) i − ( 0.12 m ) j + ( 0.72 m ) k
and
FBA = λ BA FBA


− ( 0.1 m ) i + (1.8 m ) j − ( 0.6 m ) k 
=
228 N )

(
2
2
2
( 0.1) + (1.8) + ( 0.6 ) m 

= − (12.0 N ) i + ( 216 N ) j − ( 72 N ) k
∴ MC
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
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
M A = rE/ A × TDE
(a) Have
rE/ A = ( 92 in.) j
where
TDE = λ DETDE
=
( 24 in.) i + (132 in.) j − (120 in.) k 360 lb
(
)
( 24 )2 + (132 )2 + (120 )2 in.
= ( 48 lb ) i + ( 264 lb ) j − ( 240 lb ) k
i
j
k
∴ M A = 0 92
0 lb ⋅ in. = − ( 22, 080 lb ⋅ in.) i − ( 4416 lb ⋅ in ) k
48 264 −240
or M A = − (1840 lb ⋅ ft ) i − ( 368 lb ⋅ ft ) k
M A = rG/ A × TCG
(b) Have
rG/ A = (108 in.) i + ( 92 in.) j
where
TCG = λ CGTCG =
− ( 24 in.) i + (132 in.) j − (120 in.) k
( 24 )2 + (132 )2 + (120 )2 in.
( 360 lb )
= − ( 48 lb ) i + ( 264 lb ) j − ( 240 lb ) k
i
j
k
∴ M A = 108 92
0 lb ⋅ in.
−48 264 −240
= − ( 22, 080 lb ⋅ in.) i + ( 25,920 lb ⋅ in.) j + ( 32,928 lb ⋅ in.) k
or M A = − (1840 lb ⋅ ft ) i + ( 2160 lb ⋅ ft ) j + ( 2740 lb ⋅ ft ) k
PROBLEM 3.26
The arms AB and BC of a desk lamp lie in a vertical plane that forms an
o
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.
SOLUTION
Have
M O = rC/O × FC
where
( 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
rC/O = ( 0.49118 m ) i + ( 0.25929 m ) j + ( 0.28358 m ) k
or
or
( FC ) x
= − ( 8 N ) cos 45° sin 20° = −1.93476 N
( FC ) y
= − ( 8 N ) sin 45° = −5.6569 N
( FC ) z
= ( 8 N ) cos 45° cos 20° = 5.3157 N
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
where
d = perpendicular distance from O to line AB.
Now
Μ O = rB/O × TBA
and
rB/O = ( 8.4 m ) j
TBA = λ BATAB =
− ( 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 =
and
i
j
k
0
8.4
0 N ⋅ m = ( 4233.6 N ⋅ m ) i + ( 529.2 N ⋅ m ) k
−63.0 −588 504
| MO | =
( 4233.6 )2 + ( 529.2 )2
= 4266.5 N ⋅ m
∴ 4266.5 N ⋅ m = ( 777 N ) d
or
d = 5.4911 m
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
| M O | = TBC d
where
d = perpendicular distance from O to line BC.
M O = rB/O × TBC
rB/O = 8.4 m j
TBC = λ 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
∴ MO
and
i
j
k
= 0
8.4
0 = (1008 N ⋅ m ) i − ( 4284 N ⋅ m ) k
510 −840 120
| 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
| M D | = FBAd
where
d = perpendicular distance from D to line AB.
M D = rA/D × FBA
rA/D = − ( 0.12 m ) j + ( 0.72 m ) k
FBA = λ BA FBA =
( − ( 0.1 m ) i + (1.8 m ) j − ( 0.6 m ) k ) ( 228 N )
( 0.1)2 + (1.8)2 + ( 0.6 )2
m
= − (12.0 N ) i + ( 216 N ) j − ( 72 N ) k
∴ MD
i
j
k
= 0
−0.12 0.72 N ⋅ m
−12.0 216 −72
= − (146.88 N ⋅ m ) i − ( 8.64 N ⋅ m ) j − (1.44 N ⋅ m ) k
and
|MD | =
(146.88)2 + (8.64 )2 + (1.44 )2
= 147.141 N ⋅ m
∴ 147.141 N ⋅ m = ( 228 N ) d
d = 0.64536 m
or d = 0.645 m
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
| M C | = FBAd
where
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 = λ BA FBA =
( − ( 0.1 m ) i + (1.8 m ) j − ( 0.6 ) k ) ( 228 N )
( 0.1)2 + (1.8)2 + ( 0.6 )2
m
= − (12.0 N ) i + ( 216 N ) j − ( 72 N ) k
i
j
k
∴ M C = 0.96 −0.12 0.72 N ⋅ m
−12.0 216 −72
= − (146.88 N ⋅ m ) i − ( 60.48 N ⋅ m ) j + ( 205.92 N ⋅ m ) k
and
| MC | =
(146.88)2 + ( 60.48)2 + ( 205.92 )2
= 260.07 N ⋅ m
∴ 260.07 N ⋅ m = ( 228 N ) d
d = 1.14064 m
or d = 1.141 m
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
M A = TDE d
where
d = perpendicular distance from A to line DE.
M A = rE/ A × TDE
rE/ A = ( 92 in.) j
TDE = λ DETDE =
( 24 in.) i + (132 in.) j − (120 in.) k 360 lb
(
)
( 24 )2 + (132 )2 + (120 )2 in.
= ( 48 lb ) i + ( 264 lb ) j − ( 240 lb ) k
i
j
k
∴ M A = 0 92
0 N⋅m
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 W
PROBLEM 3.32
In Problem 3.25, determine the perpendicular distance from point A to a
line drawn through points C and G.
Problem 3.25: The ramp ABCD is supported by cables at corners C
and D. The tension in each of the cables is 360 lb. Determine the moment
about A of the force exerted by (a) the cable at D, (b) the cable at C.
SOLUTION
Have
M A = TCG d
where
d = perpendicular distance from A to line CG.
M A = rG/ A × TCG
rG/ A = (108 in.) i + ( 92 in.) j
TCG = λ 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.
∴ 47,367 lb ⋅ in. = ( 360 lb ) d
d = 131.575 in.
or d = 10.96 ft W
PROBLEM 3.33
In Problem 3.25, determine the perpendicular distance from point B to a
line drawn through points D and E.
Problem 3.25: The ramp ABCD is supported by cables at corners C
and D. The tension in each of the cables is 360 lb. Determine the moment
about A of the force exerted by (a) the cable at D, (b) the cable at C.
SOLUTION
M B = TDE d
Have
where
d = perpendicular distance from B to line DE.
M B = rE/B × TDE
rE/B = − (108 in.) i + ( 92 in.) j
TDE = λ 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.
∴ 47,367 lb ⋅ in. = ( 360 lb ) d
d = 131.575 in.
or d = 10.96 ft W
PROBLEM 3.34
Determine the value of a which minimizes the perpendicular distance
from point C to a section of pipeline that passes through points A and B.
SOLUTION
Assuming a force F acts along AB,
M C = rA/C × F = F ( d )
where
d = perpendicular distance from C to line AB
F = λ 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
3−a 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
or
rA/C × F 2 = ( dF )
2
∴ ( 0.30154 + 0.50257a ) + ( 2.3693 − 0.57437a ) + ( 2.1108 ) = d 2
2
Setting
2
2
( )
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 W
PROBLEM 3.35
Given the vectors P = 7i − 2j + 5k, Q = −3i − 4j + 6k, and S = 8i +
j − 9k, compute the scalar products P ⋅ Q, P ⋅ S, and Q ⋅ S.
SOLUTION
P ⋅ Q = ( 7i − 2 j + 5k ) ⋅ ( −3i − 4 j + 6k )
= ( 7 )( −3) + ( −2 )( −4 ) + ( 5 )( 6 )
= 17
or P ⋅ Q = 17 W
P ⋅ S = ( 7i − 2 j + 5k ) ⋅ ( 8i + j − 9k )
= ( 7 )( 8 ) + ( −2 )(1) + ( 5 )( −9 )
=9
or P ⋅ S = 9 W
Q ⋅ S = ( −3i − 4 j + 6k ) ⋅ ( 8i + j − 9k )
= ( −3)( 8 ) + ( −4 )(1) + ( 6 )( −9 )
= −82
or Q ⋅ S = −82 W
PROBLEM 3.36
Form the scalar products B ⋅ C and B′ ⋅ C, where B = B′, and use the
results obtained to prove the identity
cos α cos β =
1
2
cos (α + β ) + 12 cos (α − β ) .
SOLUTION
By definition
B ⋅ C = BC cos (α − β )
where
B = B ( cos β ) i + ( sin β ) j
C = C ( cos α ) i + ( sin α ) j
∴ ( B cos β )( C cos α ) + ( B sin β )( C sin α ) = BC cos (α − β )
cos β cos α + sin β sin α = cos (α − β )
or
(1)
By definition
B′ ⋅ C = BC cos (α + β )
where
B′ = ( cos β ) i − ( sin β ) j
∴ ( B cos β )( C cos α ) + ( − B sin β )( C sin α ) = BC cos (α + β )
or
cos β cos α − sin β sin α = cos (α + β )
(2)
Adding Equations (1) and (2),
2 cos β cos α = cos (α − β ) + cos (α + β )
or cos α cos β =
1
1
cos (α + β ) + cos (α − β ) W
2
2
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 cosθ
or
( −1.95i − 2.40 j + 0.6k ) ⋅ ( −2.40 j + 1.80k ) = ( 3.15)( 3.0 ) cosθ
( −1.95)( 0 ) + ( −2.40 )( −2.40 ) + ( 0.6 )(1.8) = 9.45cosθ
∴ cosθ = 0.72381
and
θ = 43.630°
or θ = 43.6° W
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 =
and
( −2.4 )2 + (1.8)2
m = 3m
(1.2 )2 + ( −2.4 )2 + ( 0.3)2
m = 2.7 m
rC/ A = − ( 2.4 m ) j + (1.8 m ) k
rD/ A = (1.2 m ) i − ( 2.4 m ) j + ( 0.3 m ) k
By definition
rC/ A ⋅ rD/ A = rC/ A rD/ A cosθ
or
( −2.4 j + 1.8k ) ⋅ (1.2i − 2.4 j + 0.3k ) = ( 3)( 2.7 ) cosθ
( 0 )(1.2 ) + ( −2.4 )( −2.4 ) + (1.8)( 0.3) = 8.1cosθ
and
cosθ =
6.3
= 0.77778
8.1
θ = 38.942°
or θ = 38.9° W
PROBLEM 3.39
Steel framing members AB, BC, and CD are joined at B and C and are
braced using cables EF and EG. Knowing that E is at the midpoint of BC
and that the tension in cable EF is 330 N, determine (a) the angle
between EF and member BC, (b) the projection on BC of the force
exerted by cable EF at point E.
SOLUTION
λ BC ⋅ λ EF = (1)(1) cosθ
(a) By definition
where
λ BC =
λ EF =
∴
(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
( 7 ) 2 + ( 6 )2 + ( 6 ) 2 m
=
=
1
( −7i − 6 j + 6k )
11.0
(16i − 4.5j − 12k ) ⋅ ( −7i − 6 j + 6k )
20.5
1
(16i − 4.5j − 12k )
20.5
11.0
= cosθ
(16 )( −7 ) + ( −4.5)( −6 ) + ( −12 )( 6 ) = ( 20.5)(11.0 ) cosθ
and
 −157 
θ = cos −1 
 = 134.125°
 225.5 
or θ = 134.1° W
(b) By definition
(TEF )BC
= TEF cosθ
= ( 330 N ) cos134.125°
= −229.26 N
or (TEF ) BC = −230 N W
PROBLEM 3.40
Steel framing members AB, BC, and CD are joined at B and C and are
braced using cables EF and EG. Knowing that E is at the midpoint of BC
and that the tension in cable EG is 445 N, determine (a) the angle
between EG and member BC, (b) the projection on BC of the force
exerted by cable EG at point E.
SOLUTION
λ BC ⋅ λ EG = (1)(1) cosθ
(a) By definition
where
λ BC =
(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
λ 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
∴ λ BC ⋅ λ EG =
and
16 ( 8 ) + ( −4.5 )( −6 ) + ( −12 )( 4.875 )
= cosθ
( 20.5)(11.25)
 96.5 
θ = cos −1 
 = 64.967°
 228.06 
or θ = 65.0° W
(b) By definition
(TEG )BC
= TEG cosθ
= ( 445 N ) cos 64.967°
= 188.295 N
or (TEG ) BC = 188.3 N W
PROBLEM 3.41
Slider P can move along rod OA. An elastic cord PC is attached to the
slider and to the vertical member BC. Knowing that the distance from O
to P is 0.12 m and the tension in the cord is 30 N, determine (a) the angle
between the elastic cord and the rod OA, (b) the projection on OA of the
force exerted by cord PC at point P.
SOLUTION
λ OA ⋅ λ PC = (1)(1) cosθ
(a) By definition
λ OA =
where
=
( 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
λ 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 ) = cosθ
3
3 
3
θ = cos −1 ( 0.32445) = 71.068°
and
or θ = 71.1° W
(b)
(TPC )OA
= TPC cosθ = ( 30 N ) cos 71.068°
(TPC )OA
= 9.7334 N
or (TPC )OA = 9.73 N W
PROBLEM 3.42
Slider P can move along rod OA. An elastic cord PC is attached to the
slider and to the vertical member BC. Determine the distance from O to P
for which cord PC and rod OA are perpendicular.
SOLUTION
The requirement that member OA and the elastic cord PC be perpendicular implies that
λ OA ⋅ λ PC = 0
where
λ OA =
=
or
λ 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 = λ OAdOP =
x=
2
dOP ,
3
y =
(1)
dOP
( 2i + 2 j − k ) ,
3
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 W
PROBLEM 3.43
Determine the volume of the parallelepiped of Figure 3.25 when
(a) P = −(7 in.)i − (1 in.)j + (2 in.)k, Q = (3 in.)i − (2 in.)j + (4 in.)k, and
S = −(5 in.)i + (6 in.)j − (1 in.)k, (b) P = (1 in.)i + (2 in.)j − (1 in.)k,
Q = −(8 in.)i − (1 in.)j + (9 in.)k, and S = (2 in.)i + (3 in.)j + (1 in.)k.
SOLUTION
Volume of a parallelepiped is found using the mixed triple product.
(a)
Vol = P ⋅ ( Q × S )
−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 W
(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 W
PROBLEM 3.44
Given the vectors P = 4i − 2j + Pzk, Q = i + 3j − 5k, and S = −6i +
2j − k, determine the value of Pz for which the three vectors are coplanar.
SOLUTION
For the vectors to all be in the same plane, the mixed triple product is zero.
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 W
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
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
M A = rD/ A × TDE
where
rD/ A = ( 0.132 m ) j + ( 0.720 m ) k
Then
i
j
k
M A = 0 0.132 0.720 N ⋅ m
18.0 36.0 −36.0
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 W
PROBLEM 3.46
The 0.732 × 1.2-m -m lid ABCD of a storage bin is hinged along side AB
and is held open by looping cord DEC over a frictionless hook at E. If the
tension in the cord is 54 N, determine the moment about each of the
coordinate axes of the force exerted by the cord at C.
SOLUTION
z =
First note
( 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
M A = rE/ A × TCE
where
rE/ A = ( 0.360 m ) i + ( 0.852 m ) j
Then
i
j
k
M A = 0.360 0.852
0
N⋅m
−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 W
PROBLEM 3.47
A fence consists of wooden posts and a steel cable fastened to each post
and anchored in the ground at A and D. Knowing that the sum of the
moments about the z axis of the forces exerted by the cable on the posts
at B and C is −66 N · m, determine the magnitude TCD when TBA = 56 N.
SOLUTION
Based on
| M z | = k ⋅ ( rB ) y × TBA  + k ⋅ ( rC ) y × TCD 




where
M z = − ( 66 N ⋅ m ) k
( rB ) y
= ( rC ) y = (1 m ) j
TBA = λ 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 = λ CDTCD
=
( 2 m ) i − (1 m ) j − ( 2 m ) k T
=
1
TCD ( 2i − j − 2k )
3
3.0 m
CD
{
}
∴ − ( 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 W
PROBLEM 3.48
A fence consists of wooden posts and a steel cable fastened to each post
and anchored in the ground at A and D. Knowing that the sum of the
moments about the y axis of the forces exerted by the cable on the posts
at B and C is 212 N · m, determine the magnitude of TBA when
TCD = 33 N.
SOLUTION
Based on
| M y | = j ⋅ ( rB ) z × TBA + ( rC ) z × TCD 
where
M y = ( 212 N ⋅ m ) j
( rB ) z
= (8 m ) k
( rC ) z
= (2 m) k
TBA = λ BATBA
=
(1.5 m ) i − (1 m ) j − ( 3 m ) k T
=
TBA
(1.5i − j + 3k )
3.5
BA
3.5 m
TCD = λ 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
∴ TBA =
168
18.6667
or TBA = 49.0 N W
PROBLEM 3.49
To lift a heavy crate, a man uses a block and tackle attached to the bottom
of an I-beam at hook B. Knowing that the moments about the y and z axes
of the force exerted at B by portion AB of the rope are, respectively,
100 lb ⋅ ft and −400 lb ⋅ ft , determine the distance a.
SOLUTION
M O = rA/O × TBA
Based on
where
M O = M xi + M y j + M zk
= M xi + (100 lb ⋅ ft ) j − ( 400 lb ⋅ ft ) k
rA/O = ( 6 ft ) i + ( 4 ft ) j
TBA = λ BATBA
=
( 6 ft ) i − (12 ft ) j − ( a ) k T
BA
d BA
i j k
T
∴ M xi + 100 j − 400k = 6 4 0 BA
d
6 −12 −a BA
=
TBA
 − ( 4a ) i + ( 6a ) j − ( 96 ) k 
d BA 
100
d BA
6a
From j-coefficient:
100d AB = 6aTBA or TBA =
From k -coefficient:
−400d AB = −96TBA or TBA =
Equating Equations (1) and (2) yields
a=
400
d BA
96
(1)
(2)
100 ( 96 )
6 ( 400 )
or a = 4.00 ft W
PROBLEM 3.50
To lift a heavy crate, a man uses a block and tackle attached to the bottom
of an I-beam at hook B. Knowing that the man applies a 200-lb force to
end A of the rope and that the moment of that force about the y axis is
175 lb ⋅ ft , determine the distance a.
SOLUTION
(
| M y | = j ⋅ rA/O × TBA
Based on
)
rA/O = ( 6 ft ) i + ( 4 ft ) j
where
TBA = λ BATBA =
=
=
rA/B
d BA
TBA
( 6 ft ) i − (12 ft ) j − ( a ) k
d BA
( 200 lb )
200
( 6i − 12 j − ak )
d BA
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
180 + a 2 = 6.8571a
Squaring each side
180 + a 2 = 47.020a 2
Solving
a = 1.97771 ft
or a = 1.978 ft W
PROBLEM 3.51
A force P is applied to the lever of an arbor press. Knowing that P lies in
a plane parallel to the yz plane and that 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 φ and θ.
SOLUTION
Based on
M x = ( P cos φ ) ( 9 in.) sin θ  − ( P sin φ ) ( 9 in.) cosθ 
M y = − ( P cos φ )( 5 in.)
(2)
M z = − ( P sin φ )( 5 in.)
(3)
− ( P sin φ ) (5)
Equation (3) M z
:
=
Equation (2) M y
− ( P cos φ ) (5)
Then
or
(1)
tan φ =
−35
= 0.175
−200
φ = 9.9262°
or φ = 9.93° W
Substituting φ into Equation (2)
−200 lb ⋅ in. = − ( P cos 9.9262° ) (5 in.)
P = 40.608 lb
or P = 40.6 lb W
Then, from Equation (1)
230 lb ⋅ in. = ( 40.608 lb ) cos 9.9262° ( 9 in.) sin θ 
− ( 40.608 lb ) sin 9.9262°  ( 9 in.) cosθ 
or
0.98503sin θ − 0.172380cosθ = 0.62932
Solving numerically,
θ = 48.9° W
PROBLEM 3.52
A force P is applied to the lever of an arbor press. Knowing that P lies in
a plane parallel to the yz plane and that M y = −180 lb ⋅ in. and
M z = −30 lb ⋅ in., determine the moment M x of P about the x axis when
θ = 60°.
SOLUTION
Based on
M x = ( P cos φ ) ( 9 in.) sin θ  − ( P sin φ ) ( 9 in.) cosθ 
(1)
M y = − ( P cos φ )( 5 in.)
(2)
M z = − ( P sin φ )( 5 in.)
(3)
− ( P sin φ )( 5 )
Equation (3) M z
:
=
Equation (2) M y
− ( P cos φ )( 5 )
Then
−30
= tan φ
−180
or
∴ φ = 9.4623°
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. W
PROBLEM 3.53
The triangular plate ABC is supported by ball-and-socket joints at B and
D and is held in the position shown by cables AE and CF. If the force
exerted by cable AE at A is 220 lb, determine the moment of that force
about the line joining points D and B.
SOLUTION
(
M DB = λ DB ⋅ rA/D × TAE
Have
λ DB =
where
( 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 
TAE = λ AETAE = 
( 220 lb )
44 in.
= (180 lb ) i − (120 lb ) j + ( 40 lb ) k
∴ M DB
0.960 −0.280 0
8 lb ⋅ in.
= 0
−4
180 −120 40
= ( 0.960 ) ( −4 )( 40 ) − ( 8 )( −120 )  + ( −0.280 ) 8 (180 ) − 0 
= 364.8 lb ⋅ in.
or M DB = 365 lb ⋅ in. W
PROBLEM 3.54
The triangular plate ABC is supported by ball-and-socket joints at B and
D and is held in the position shown by cables AE and CF. If the force
exerted by cable CF at C is 132 lb, determine the moment of that force
about the line joining points D and B.
SOLUTION
(
M DB = λ DB ⋅ rC/D × TCF
Have
where
λ DB =
( 48 in.) i − (14 in.) j
50 in.
)
= 0.96i − 0.28 j
rC/D = ( 8 in.) j − (16 in.) k
TCF = λ CFTCF =
( 24 in.) i − ( 36 in.) j − (8 in.) k
44 in.
(132 lb )
= ( 72 lb ) i − (108 lb ) j − ( 24 lb ) k
∴ M DB
0.96 −0.28 0
8
= 0
−16 lb ⋅ in.
72 −108 −24
= 0.96 ( 8 )( −24 ) − ( −16 )( −108 )  + ( −0.28 ) ( −16 )( 72 ) − 0 
= −1520.64 lb ⋅ in.
or M DB = −1521 lb ⋅ in. W
PROBLEM 3.55
A mast is mounted on the roof of a house using bracket ABCD and is
guyed by cables EF, EG, and EH. Knowing that the force exerted by
cable EF at E is 66 N, determine the moment of that force about the line
joining points D and I.
SOLUTION
M DI = λ DI ⋅ rF /I × TEF 
Have
where
λ DI =
(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 = λ 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 
−1 0
0 1
2 −6 9
4
∴ M DI =
( 6 N )( 5.4 m ) 0
17
= 7.8582 ( 0 + 24 ) + ( −2 − 0 ) 
= 172.879 N ⋅ m
or M DI = 172.9 N ⋅ m W
PROBLEM 3.56
A mast is mounted on the roof of a house using bracket ABCD and is
guyed by cables EF, EG, and EH. Knowing that the force exerted by
cable EG at E is 61.5 N, determine the moment of that force about the
line joining points D and I.
SOLUTION
Have
M DI = λ DI ⋅ rG/I × TEG 
where
λ 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 = λ 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
(
M OA = λ OA ⋅ rC/O × P
Have
)
where
From triangle OBC
a
2
( OA) x
=
( OA) z
= ( OA ) x tan 30° =
( OA)2
Since
a 1 
a
=


2 3  2 3
= ( OA) x + ( OA) y + ( OAz )
2
2
2
 a 
2
a
a 2 =   + ( OA) y + 

2
 
2 3
or
∴
( OA) y
a2 −
=
2
2
a2 a2
2
−
=a
4
12
3
Then
rA/O =
a
2
a
i +a j+
k
2
3
2 3
and
λ OA =
1
i+
2
2
1
j+
k
3
2 3
P = λ BC P
=
=
( a sin 30° ) i − ( a cos30° ) k
a
(
P
i − 3k
2
)
rC/O = ai
( P)
PROBLEM 3.57 CONTINUED
∴ M OA
1
2
=
1
1
2
1
3 2 3
P
( a )  
0
0
2
0 − 3
=
aP  2 
−
 (1) − 3
2  3 
=
aP
2
(
)
M OA =
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,
JJJG JJJG
OA ⋅ BC = 0
where
From triangle OBC
and
Then
or
so that
a
2
( OA) x
=
( OA) z
= ( OA ) x tan 30° =
a 1 
a
=
2  3  2 3
JJJG  a 
 a 
∴ OA =   i + ( OA)y j + 
k
2
2 3
JJJG
BC = ( a sin 30° ) i − ( a cos 30° ) k
=
a
a 3
i−
k
2
2
=
a
i − 3k
2
(
)
a
 a  
a
=0
 i + ( OA ) y j + 
 k  ⋅ i − 3k
2
2
2
3

 

(
)
a2
a2
+ ( OA)y ( 0 ) −
=0
4
4
JJJG JJJG
∴ OA ⋅ BC = 0
JJJG
JJJG
OA is perpendicular to BC.
PROBLEM 3.58 CONTINUED
(b) Have M OA = Pd , with P acting along BC and d the
JJJG JJJG
perpendicular distance from OA to BC.
From the results of Problem 3.57,
M OA =
∴
or
Pa
2
Pa
= Pd
2
d =
a
2
PROBLEM 3.59
The 8-ft-wide portion ABCD of an inclined, cantilevered walkway is
partially supported by members EF and GH. Knowing that the
compressive force exerted by member EF on the walkway at F is 5400 lb,
determine the moment of that force about edge AD.
SOLUTION
(
M AD = λ AD ⋅ rE/ A × TEF
Having
λ AD =
where
( 24 ft ) i + ( 3 ft ) j
( 24 )2 + ( 3)2 ft
)
1
(8i + j)
65
=
rE/ A = ( 7 ft ) i − ( 3 ft ) j
TEF = λ EFTEF =
(8 ft − 7 ft ) i + 3 ft + ( 248 ) ( 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
= −18,456.4 lb ⋅ ft
or M AD = −18.46 kip ⋅ ft
PROBLEM 3.60
The 8-ft-wide portion ABCD of an inclined, cantilevered walkway is
partially supported by members EF and GH. Knowing that the
compressive force exerted by member GH on the walkway at H is
4800 lb, determine the moment of that force about edge AD.
SOLUTION
(
M AD = λ AD ⋅ rG/ A × TGH
Having
λ AD =
where
( 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 = λ GH TGH =
(16 ft − 20 ft ) i + 6 ft + ( 1624 ) ( 3 ft ) j + (8 ft ) k
( 4 )2 + ( 8 )2 + ( 8 )2
ft
( 4800 lb )
= 1600  − (1 lb ) i + ( 2 lb ) j + ( 2 lb ) k 
8
∴ M AD =
1 0
3200 lb ⋅ ft
−3 0 =
( −48 − 20 )
65
−1 2 2
(1600 lb )( 2 ft ) 10
65
= −26,989 lb ⋅ ft
or M AD = −27.0 kip ⋅ 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
F1 = F1λ1
First note that
F2 = F2λ 2
and
Let M1 = moment of F2 about the line of action of M1
and M 2 = moment of F1 about the line of action of M 2
Now, by definition
(
)
(
)
(
)
(
)
M1 = λ1 ⋅ rB/ A × F2 = λ1 ⋅ rB/ A × λ 2 F2
M 2 = λ 2 ⋅ rA/B × F1 = λ 2 ⋅ rA/B × λ1 F1
F1 = F2 = F
Since
and
(
rA/B = −rB/ A
)
M1 = λ1 ⋅ rB/ A × λ 2 F
(
)
M 2 = λ 2 ⋅ −rB/ A × λ1 F
Using Equation (3.39)
(
)
(
λ1 ⋅ rB/ A × λ 2 = λ 2 ⋅ −rB/ A × λ1
so that
(
)
)
M 2 = λ1 ⋅ rB/ A × λ 2 F
∴ M12 = M 21
PROBLEM 3.62
In Problem 3.53, determine the perpendicular distance between cable AE
and the line joining points D and B.
Problem 3.53: The triangular plate ABC is supported by ball-and-socket
joints at B and D and is held in the position shown by cables AE and CF.
If the force exerted by cable AE at A is 220 lb, determine the moment of
that force about the line joining points D and B.
SOLUTION
(
M DB = λ DB ⋅ rA/D × TAE
Have
where
λ DB =
( 48 in.) i − (14 in.) j
50 in.
)
= 0.96i − 0.28 j
rA/D = − ( 4 in.) j + ( 8 in.) k
TAE = λ AETAE
=
( 36 in.) i − ( 24 in.) j + (8 in.) k
44 in.
( 220 lb )
= (180 lb ) i − (120 lb ) j + ( 40 lb ) k
∴ M DB
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
TAE about line DB. The parallel component of TAE will be used to find
the perpendicular component.
PROBLEM 3.62 CONTINUED
Have
(TAE )parallel
= λ 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 )2parallel
=
( 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 ball-and-socket
joints at B and D and is held in the position shown by cables AE and CF.
If the force exerted by cable CF at C is 132 lb, determine the moment of
that force about the line joining points D and B.
SOLUTION
( M DB ) = λ DB ⋅ ( rC/D × TCF )
Have
λ DB =
where
( 48 in.) i − (14 in.) j
50 in.
= 0.96i − 0.28j
rC/D = ( 8 in.) j − (16 in.) k
TCF = λ CF TCF
=
( 24 in.) i − ( 36 in.) j − (8 in.) k
44 in.
(132 lb )
= ( 72 lb ) i − (108 lb ) j − ( 24 lb ) k
∴ M DB
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.
PROBLEM 3.63 CONTINUED
Have
(TCF )parallel
= λ 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 )2parallel
=
(132 )2 − ( 99.36 )2
= 86.900 lb
Then
M DB = (TCF )perp. ( d )
−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
(
M DI = λ DI ⋅ rF /I × TEF
Have
λ DI =
where
(1.6 m ) i − ( 0.4 m ) j
0.4 17 m
=
)
1
( 4i − j)
17
rF /I = ( 5.4 m ) k
TEF = λ 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 =
4 −1 0
0 1 = 172.879 N ⋅ m
2 −6 9
( 6 N )( 5.4 m ) 0
17
Only the perpendicular component of TEF contributes to the moment of
TEF about line DI. The parallel component of TEF will be used to find
the perpendicular component.
Have
(TEF )parallel
= λ DI ⋅ TEF
=
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 )2parallel
=
( 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
M DI = λ DI ⋅ rG/I × TEG 
Have
λ DI =
where
(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 = λ 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 ) = λ 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
=
( 61.5)
2
 42 
−

 17 
2
= 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
FBC = FBC j
Have
M OA = λ OA ⋅ rB/O × FBC
where
(
λ OA =
)
( 0.24 m ) i + ( 0.24 m ) j − ( 0.12 m ) k
0.36 m
2
2
1
i+ j− k
3
3
3
=
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 ) = λ OA ⋅ FBC =  i + j − k  ⋅ FBC j
3
3 
3
=
2
FBC
3
FBC = ( FBC )parallel + ( FBC )perp.
Since
( FBC )perp.
=
( FBC )2 − ( FBC )2parallel
=
( 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
M y = j ⋅ rD/ A × TDE
where
rD/ A = ( 0.132 j + 0.720k ) m
TDE = λ DETDE
=
( 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 y-axis. 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
( TDE ) = ( TDE )parallel + ( TDE )perp.
Since
(TDE )perp.
Then
=
2
(TDE )2 − (TDE )parallel
=
( 54 )2 − ( 36 )2
= 40.249 N
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 21-N forces,
(b) the perpendicular distance between the 12-N forces if the resultant of
the two couples is zero, (c) the value of α if the resultant couple is
1.8 N ⋅ m clockwise and d is 1.05 m.
SOLUTION
(a) Have
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 α )(12 N )
∴ sin α = 0.52381
and
α = 31.588°
or α = 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
M = Pd
(a) Have
 1 ft 
10 lb ⋅ ft = P (10 in.) 

 12 in. 
or
∴ P = 12 lb
d BC =
(b)
=
or Pmin = 12.00 lb
( BE )2 + ( EC )2
(10 in.)2 + ( 6 in.)2
= 11.6619 in.
M = Pd
Have
 1 ft 
10 lb ⋅ ft = P (11.6619 in.) 

 12 in. 
P = 10.2899 lb
d AC =
(c)
=
or P = 10.29 lb
( AD )2 + ( DC )2
(10 in.)2 + (16 in.)2
= 2 89 in.
M = Pd AC
Have
(
)
 1 ft 
10 lb ⋅ ft = P 2 89 in. 

 12 in. 
P = 6.3600 lb
or P = 6.36 lb
PROBLEM 3.70
Two 60-mm-diameter pegs are mounted on a steel plate at A and C, and
two rods are attached to the plate at B and D. A cord is passed around the
pegs and pulled as shown, while the rods exert on the plate 10-N forces as
indicated. (a) Determine the resulting couple acting on the plate when T =
36 N. (b) If only the cord is used, in what direction should it be pulled to
create the same couple with the minimum tension in the cord?
(c) Determine the value of that minimum tension.
SOLUTION
M = Σ ( Fd )
(a) Have
= ( 36 N )( 0.345 m ) − (10 N )( 0.380 m )
= 8.62 N ⋅ m
M = 8.62 N ⋅ m
(b)
M = Td = 8.62 N ⋅ m
Have
For T to be minimum, d must be maximum.
∴ Tmin must be perpendicular to line AC
tan θ =
0.380 m
= 1.33333
0.285 m
θ = 53.130°
and
or θ = 53.1°
M = Tmin d max
(c) Have
M = 8.62 N ⋅ m
where

d max = 

( 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 50-mm-diameter idler rollers
mounted on the plate as shown. Two flat belts pass around the rollers, and
rollers A and D will be adjusted so that the tension in each belt is 45 N.
Determine (a) the resultant couple acting on the plate if a = 0.2 m, (b) the
value of a so that the resultant couple acting on the plate is 54 N ⋅ m
clockwise.
SOLUTION
(a) Note when a = 0.2 m, rC/F is perpendicular to the inclined 45 N
forces.
Have
M = Σ ( 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
M = 54.0 N ⋅ m
(b)
M = Moment of couple due to horizontal forces at A and D
+ Moment of force-couple 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 =
−45
N
2
Fy =
−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
M = M1 + M 2
Based on
where
M1 = − ( 8 N ⋅ m ) j
M2 = − (6 N⋅m)k
∴ M = − (8 N ⋅ m ) j − ( 6 N ⋅ m ) k
M =
and
( 8 )2 + ( 6 ) 2
= 10 N ⋅ m
or M = 10.00 N ⋅ m
λ =
or
− (8 N ⋅ m ) j − ( 6 N ⋅ m ) k
M
=
= −0.8j − 0.6k
10 N ⋅ m
M
M = M λ = (10 N ⋅ m )( −0.8j − 0.6k )
cosθ x = 0
∴ θ x = 90°
cosθ y = −0.8
∴ θ y = 143.130°
cosθ z = −0.6
∴ θ z = 126.870°
or θ x = 90.0°, θ y = 143.1°, θ 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
M = M1 + M 2
where
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
M 2 = rD/ A × P2 E
Also,
rD/ A = ( 0.20 m ) j − ( 0.55 m ) k
P2 E = λ 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
M = ( 40 N ⋅ m ) i + ( 96 N ⋅ m ) j + ( 22.0 N ⋅ m ) i
and
+ ( 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 =
M x2 + M y2 + M z2 =
( 62.0 )2 + (148.8)2 + (19.2 )2
= 162.339 N ⋅ m
or M = 162.3 N ⋅ m
λ =
M
62.0i + 148.8j + 19.2k
=
M
162.339
= 0.38192i + 0.91660 j + 0.118271k
cosθ x = 0.38192
∴ θ x = 67.547°
or θ x = 67.5°
cosθ y = 0.91660
∴ θ y = 23.566°
or θ y = 23.6°
cosθ z = 0.118271
∴ θ z = 83.208°
or θ 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
M = M4 + M7
where
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
M 7 = rD/F × F7 D
Also,
rD/F = − ( 5 in.) i + ( 3 in.) j
F7 D = λ ED F7 D
=
=
− ( 5 in.) i + ( 3 in.) j + ( 7 in.) k
( 5 )2 + ( 3)2 + ( 7 )2
in.
( 7 lb )
7 lb
( −5i + 3j + 7k )
83
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
M = ( 40 lb ⋅ in.) j +  0.76835 ( 21i + 35j) lb ⋅ in.
and
= (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.
λ =
(16.1353 lb ⋅in.) i + ( 66.892 lb ⋅in.) j
M
=
M
68.811 lb ⋅ in.
= 0.23449i + 0.97212 j
cosθ x = 0.23449
∴ θ x = 76.438°
or θ x = 76.4°
cosθ y = 0.97212
∴ θ y = 13.5615°
or θ y = 13.56°
cosθ z = 0.0
∴ θ z = 90°
or θ 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
M = M 4 + M 7 + M5
Have
where
M 4 = rG/C × F4G
M 7 = rD/F × F7 D
i j k
= −10 0 0 lb ⋅ in. = ( 40 lb ⋅ in.) j
0 0 4
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 =
M x2 + M y2 + M z2 =
(18.8647 )2 + ( 66.892 )2 + ( 50 )2
= 85.618 lb ⋅ in.
or M = 85.6 lb ⋅ in.
λ =
M
−18.8647i + 66.892 j + 50k
=
= −0.22034i + 0.78129 j + 0.58399k
85.618
M
cosθ x = −0.22034
∴ θ x = 102.729°
or θ x = 102.7°
cosθ y = 0.78129
∴ θ y = 38.621°
or θ y = 38.6°
cosθ z = 0.58399
∴ θ z = 54.268°
or θ 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
M = M1 + M 2 + M P
Have
M1 = rC/B × P1C
where
M 2 = rD/ A × P2 E
i
j
k
= 0.96 −0.40 0 = ( 40 N ⋅ m ) i + ( 96 N ⋅ m ) j
0
0
−100
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.)
M P = rE/ A
i
j
k
× PE = 0.48 0.20 −1.10 = ( 231 N ⋅ m ) i + (100.8 N ⋅ m ) k
0 210
0
∴ 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 =
M x2 + M y2 + M z2 =
( 293)2 + (148.8)2 + (120 )2
= 349.84 N ⋅ m
or M = 350 N ⋅ m
λ =
293i + 148.8 j + 120k
M
=
= 0.83752i + 0.42533j + 0.34301k
349.84
M
cosθ x = 0.83752
∴ θ x = 33.121°
or θ x = 33.1°
cosθ y = 0.42533
∴ θ y = 64.828°
or θ y = 64.8°
cosθ z = 0.34301
∴ θ z = 69.940°
or θ 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
M = M1 + M 2 + M 3
Have
M1 = − (1.1 lb ⋅ ft )( cos 25° j + sin 25°k )
where
M 2 = − (1.1 lb ⋅ ft ) j
M 3 = − (1.3 lb ⋅ ft )( cos 20° j − sin 20°k )
∴ 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 =
M x2 + M y2 + M z2 =
( 0 )2 + ( 3.3185)2 + ( 0.020254 )2
= 3.3186 lb ⋅ ft
or M = 3.32 lb ⋅ ft
λ =
( 0 ) i − 3.3185j − 0.020254k
M
=
M
3.3186
= −0.99997 j − 0.0061032k
cosθ x = 0
∴ θ x = 90°
cosθ y = −0.99997
∴ θ y = 179.555°
cosθ z = −0.0061032
∴ θ z = 90.349°
or θ x = 90.0°
or θ y = 179.6°
or θ 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
force-couple system (a) at A, (b) at B.
SOLUTION
(a) Based on
ΣF : FA = T = 1000 N
or FA = 1000 N
20°
ΣM A : M A = (T sin 50° )( dA )
= (1000 N ) sin 50° ( 2.25 m )
= 1723.60 N ⋅ m
or M A = 1724 N ⋅ m
(b) Based on
ΣF : FB = T = 1000 N
or FB = 1000 N
20°
ΣMB : M B = (T sin 50° )( d B )
= (1000 N ) sin 50° (1.25 m )
= 957.56 N ⋅ m
or M B = 958 N ⋅ m
PROBLEM 3.79
The 20-lb horizontal force P acts on a bell crank as shown. (a) Replace P
with an equivalent force-couple system at B. (b) Find the two vertical
forces at C and D which are equivalent to the couple found in part a.
SOLUTION
(a) Based on
ΣF : PB = P = 20 lb
or PB = 20 lb
ΣM : 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,
ΣM : M D = PC d
100 lb ⋅ in. = PC ( 4 in.)
∴ PC = 25 lb
or PC = 25 lb
ΣFy : 0 = PD − PC
∴ PD = 25 lb
or PD = 25 lb
PROBLEM 3.80
A 700-N force P is applied at point A of a structural member. Replace P
with (a) an equivalent force-couple system at C, (b) an equivalent system
consisting of a vertical force at B and a second force at D.
SOLUTION
ΣF : PC = P = 700 N
(a) Based on
or PC = 700 N
60°
ΣM C : M C = − Px dCy + Py dCx
Px = ( 700 N ) cos60° = 350 N
where
Py = ( 700 N ) sin 60° = 606.22 N
dCx = 1.6 m
dCy = 1.1 m
∴ M C = − ( 350 N )(1.1 m ) + ( 606.22 N )(1.6 m )
= −385 N ⋅ m + 969.95 N ⋅ m
= 584.95 N ⋅ m
or M C = 585 N ⋅ m
ΣFx : PDx = P cos 60°
(b) Based on
= ( 700 N ) cos 60°
= 350 N
ΣM D :
( P cos 60° )( d DA ) =
PB ( d DB )
( 700 N ) cos 60° ( 0.6 m ) = PB ( 2.4 m )
PB = 87.5 N
or PB = 87.5 N
PROBLEM 3.80 CONTINUED
ΣFy : P sin 60° = PB + PDy
( 700 N ) sin 60° = 87.5 N + PDy
PDy = 518.72 N
PD =
=
( PDx )2 + ( PDy )
2
( 350 )2 + ( 518.72 )2
= 625.76 N
 PDy 
−1  518.72 
 = tan 
 = 55.991°
 350 
 PDx 
θ = tan −1 
or PD = 626 N
56.0°
PROBLEM 3.81
A landscaper tries to plumb a tree by applying a 240-N force as shown.
Two helpers then attempt to plumb the same tree, with one pulling at B
and the other pushing with a parallel force at C. Determine these two
forces so that they are equivalent to the single 240-N force shown in the
figure.
SOLUTION
Based on
ΣFx : − ( 240 N ) cos30° = − FB cosα − FC cosα
− ( FB + FC ) cos α = − ( 240 N ) cos 30°
or
ΣFy :
( 240 N ) sin 30° =
( FB + FC ) sin α
or
(1)
FB sin α + FC sin α
= ( 240 N ) sin 30°
(2)
From
Equation (2)
: tan α = tan 30°
Equation (1)
∴ α = 30°
Based on
ΣM 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),
30°
− (100 N + FC ) cos30° = −240cos30°
FC = 140 N
or FC = 140.0 N
30°
PROBLEM 3.82
A landscaper tries to plumb a tree by applying a 240-N force as shown.
(a) Replace that force with an equivalent force-couple system at C. (b)
Two helpers attempt to plumb the same tree, with one applying a
horizontal force at C and the other pulling at B. Determine these two
forces if they are to be equivalent to the single force of part a.
SOLUTION
ΣFx : − ( 240 N ) cos30° = − FC cos30°
(a) Based on
∴ FC = 240 N
or FC = 240 N
ΣM C : ( 240 N ) cos10° ( d A ) = M C
30°
d A = 0.25 m
∴ M C = 59.088 N ⋅ m
or M C = 59.1 N ⋅ m
ΣFy :
(b) Based on
( 240 N ) sin 30° =
FB sin α
FB sin α = 120
or
(1)
ΣM 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
ΣFx : − ( 240 N ) cos30° = −146.722 N − FB cosα
FB cosα = 61.124
(2)
From
Equation (1)
:
Equation (2)
tan α =
α = 63.007°
From Equation (1),
FB =
120
= 1.96323
61.124
or α = 63.0°
120
= 134.670 N
sin 63.007°
or FB = 134.7 N
63.0°
PROBLEM 3.83
A dirigible is tethered by a cable attached to its cabin at B. If the tension
in the cable is 250 lb, replace the force exerted by the cable at B with an
equivalent system formed by two parallel forces applied at A and C.
SOLUTION
Require the equivalent forces acting at A and C be parallel and at an angle
of α with the vertical.
Then for equivalence,
ΣFx :
( 250 lb ) sin 30° =
FA sin α + FB sin α
ΣFy : − ( 250 lb ) cos 30° = − FA cos α − FB cos α
(1)
(2)
Dividing Equation (1) by Equation (2),
( 250 lb ) sin 30°
− ( 250 lb ) cos 30°
=
( FA + FB ) sin α
− ( FA + FB ) cos α
Simplifying yields α = 30°
Based on
ΣM C : ( 250 lb ) cos 30° (12 ft ) = ( FA cos 30° )( 32 ft )
∴ FA = 93.75 lb
or FA = 93.8 lb
60°
Based on
ΣM A : − ( 250 lb ) cos 30° ( 20 ft ) = ( FC cos 30° ) ( 32 ft )
∴ FC = 156.25 lb
or FC = 156.3 lb
60°
PROBLEM 3.84
Three workers trying to move a 3 × 3 × 4-ft crate apply to the crate the
three horizontal forces shown. (a) If P = 60 lb, replace the three forces
with an equivalent force-couple system at A. (b) Replace the force-couple
system of part a with a single force, and determine where it should be
applied to side AB. (c) Determine the magnitude of P so that the three
forces can be replaced with a single equivalent force applied at B.
SOLUTION
(a) Based on
ΣFz : − 50 lb + 50 lb + 60 lb = FA
FA = 60 lb
or FA = ( 60.0 lb ) k
Based on
ΣM A :
( 50 lb )( 2 ft ) − ( 50 lb )( 0.6 ft ) = M A
M A = 70 lb ⋅ ft
(a)
or M A = ( 70.0 lb ⋅ ft ) j
(b) Based on
ΣFz : − 50 lb + 50 lb + 60 lb = F
F = 60 lb
or F = ( 60.0 lb ) k
Based on
ΣM A : 70 lb ⋅ ft = 60 lb (x )
(b)
x = 1.16667 ft
or x = 1.167 ft from A along AB
(c) Based on
ΣMB : − ( 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 600-N forces are
reversed.
SOLUTION
(a)
ΣFy : FC + FD + FE = F
Have
F = −800 N + 600 N − 600 N
F = −800 N
Have
or F = 800 N
ΣM G : FC ( d − 1.5 m ) − FD ( 2 m ) = 0
(800 N )( d
− 1.5 m ) − ( 600 N )( 2 m ) = 0
d =
1200 + 1200
800
d = 3m
or d = 3.00 m
(b)
Changing directions of the two 600 N forces only changes sign of the couple.
∴ F = −800 N
and
or F = 800 N
ΣM G : FC ( d − 1.5 m ) + FD ( 2 m ) = 0
(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 force-couple system at A.
(b) Determine the single force which is equivalent to the force-couple
system obtained in part a, and specify its point of application on a line
drawn through points A and D.
SOLUTION
(a) Have
ΣF : FB + FC + FD = FA
FB = −FD
Since
∴ FA = FC = 110 N
20°
or FA = 110.0 N
Have
20.0°
ΣM 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
ΣF : FA = FE
or FE = 110.0 N
20.0°
ΣM : M A = [ FE cos 20°] ( a )
∴ 3.2545 N ⋅ m = (110 N ) cos 20° ( a )
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 force-couple.
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 z-direction.
Based on
ΣMB : − 3.60 N ⋅ m = (2.5 N )cos 25°  ( a )
a = −1.590 m
F′ = ( 2.5 N )( cos 25°i + 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 α = 40°,
specify the magnitude and the line of action of the equivalent force.
(b) Specify the value of α if the line of action of the equivalent force is
to intersect line CD 12 in. to the right of D.
SOLUTION
ΣFx : − ( 3 lb ) sin 40° + ( 3 lb ) sin 40° = Fx
(a) Have
∴ Fx = 0
ΣFy : − ( 3 lb ) cos 40° − 10 lb + ( 3 lb ) cos 40° = Fy
Have
∴ Fy = −10 lb
or F = 10.00 lb
Note: The two 3-lb forces form a couple
ΣM A : rC/ A × PC + rB/ A × PB = rX / A × F
and
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
ΣM 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 α cos α 0
0 −1 0
0 −1 0
k : 48cos α + 30sin α − 160 = −120
24cosα = 20 − 15sin α
PROBLEM 3.88 CONTINUED
Squaring both sides of the equation, and
using the identity cos 2 α = 1 − sin 2 α , results in
sin 2 α − 0.74906sin α − 0.21973 = 0
Using quadratic formula
sin α = 0.97453
sinα = − 0.22547
so that
α = 77.0°
and
α = −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, α = 30°
Then,
ΣM 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 270-lb force P is applied to the end of a
cantilever beam. Replace P with an equivalent force-couple system at G.
SOLUTION
Have
ΣF : − ( 270 lb ) i = F
∴ F = − ( 270 lb ) i
Also, have
ΣM 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 I-beam
to lift a large cylindrical tank. Knowing that the tension in rope AB is
324 N, replace the force exerted at A by rope AB with an equivalent
force-couple system at E.
SOLUTION
ΣF : TAB = F
Have
where
TAB = λ ABTAB
=
( 0.75 m ) i − ( 6.0 m ) j + ( 3.0 m ) k
6.75 m
( 324 N )
∴ TAB = 36 N ( i − 8j + 4k )
so that
F = ( 36.0 N ) i − ( 288 N ) j + (144.0 N ) k
Have
ΣM 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 I-beam
to lift a large cylindrical tank. Knowing that the tension in rope CD is
366 N, replace the force exerted at C by rope CD with an equivalent
force-couple system at O.
SOLUTION
ΣF : TCD = F
Have
where
TCD = λ 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
F = − (18.00 N ) i − ( 336 N ) j + (144.0 N ) k
Have
ΣM O : rC/O × TCD = M
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 45-lb force directed along line AB.
Replace that force with an equivalent force-couple system at C.
SOLUTION
Have
ΣF : PAB = FC
where
PAB = λ AB PAB
=
( 2.0 in.) i + ( 38 in.) j − ( 24 in.) k
44.989 in.
( 45 lb )
or FC = ( 2.00 lb ) i + ( 38.0 lb ) j − ( 24.0 lb ) k
Have
ΣM C : rB/C × PAB = M C
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 25-lb force acting in a vertical plane parallel to the yz plane is applied
to the 8-in.-long horizontal handle AB of a socket wrench. Replace the
force with an equivalent force-couple system at the origin O of the
coordinate system.
SOLUTION
Have
ΣF : 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
ΣM 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 315-N force F and 70-N · m couple M are applied to corner A of the
block shown. Replace the given force-couple system with an equivalent
force-couple system at corner D.
SOLUTION
Have
ΣF : F = FD
= λ 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
ΣM D : M + rI /D × F = M D
where
M = λ 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
W = −Wj
and
M = M1 + M 2 = − ( M 2 cos 25° ) i + ( M1 − M 2 sin 25° ) k
Note that since W and M are perpendicular, it follows that they can be
replaced with a single equivalent force.
F =W
(a) Have
or
F = −Wj = − ( 2.4 N ) j
or F = − ( 2.40 N ) j
(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 20-lb 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
force-couple system ( F2 , M 2 ) at corner B and if ( M 2 ) z = 0, determine
(a) the distance d, (b) F2 and M 2.
SOLUTION
ΣM Bz : M 2 z = 0
(a) Have
(
)
k ⋅ rH /B × F1 + M1z = 0
where
(1)
rH /B = ( 31 in.) i − ( 2 in.) j
F1 = λ EH F1
=
=
( 6 in.) i + ( 6 in.) j − ( 7 in.) k
11.0 in.
( 20 lb )
20 lb
( 6i + 6 j − 7k )
11.0
M1z = k ⋅ M1
M1 = λ EJ M1
=
−di + ( 3 in.) j − ( 7 in.) k
d 2 + 58 in.
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
( 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
d = 5.3955 in.
From which
or d = 5.40 in.
(b)
F2 = F1 =
20 lb
( 6i + 6 j − 7k )
11.0
= (10.9091i + 10.9091j − 12.7273k ) lb
or F2 = (10.91 lb ) i + (10.91 lb ) j − (12.73 lb ) k
M 2 = rH /B × F1 + M1
i j k
20 lb ⋅ in. ( −5.3955 ) i + 3j − 7k
= 31 −2 0
+
( 480 lb ⋅ in.)
11.0
9.3333
6 6 −7
= ( 25.455i + 394.55 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 4-ft-long beam is subjected to a variety of loadings. (a) Replace each
loading with an equivalent force-couple system at end A of the beam.
(b) Which of the loadings are equivalent?
SOLUTION
(a)
ΣFy : − 200 lb − 100 lb = Ra
(a) Have
or R a = 300 lb
and
ΣM A : 900 lb ⋅ ft − (100 lb )( 4 ft ) = M a
or M a = 500 lb ⋅ ft
ΣFy : − 300 lb = Rb
(b) Have
or R b = 300 lb
ΣM A : − 450 lb ⋅ ft = M b
and
or M b = 450 lb ⋅ ft
ΣFy : 150 lb − 450 lb = Rc
(c) Have
or R c = 300 lb
and
ΣM A : 2250 lb ⋅ ft − ( 450 lb )( 4 ft ) = M c
or M c = 450 lb ⋅ ft
ΣFy : − 200 lb + 400 lb = Rd
(d) Have
or R d = 200 lb
and
ΣM A :
( 400 lb )( 4 ft ) − 1150 lb ⋅ ft
= Md
or M d = 450 lb ⋅ ft
(e) Have
ΣFy : − 200 lb − 100 lb = Re
or R e = 300 lb
and
ΣM 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
ΣFy : − 400 lb + 100 lb = R f
or R f = 300 lb
and
ΣM A : −150 lb ⋅ ft + 150 lb ⋅ ft + (100 lb )( 4 ft ) = M f
or M f = 400 lb ⋅ ft
(g) Have
ΣFy : −100 lb − 400 lb = Rg
or R g = 500 lb
and
ΣM A : 100 lb ⋅ ft + 2000 lb ⋅ ft − ( 400 lb )( 4 ft ) = M g
or M g = 500 lb ⋅ ft
(h) Have
ΣFy : −150 lb − 150 lb = Rh
or R h = 300 lb
and
ΣM 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 4-ft-long beam is loaded as shown. Determine the loading of Problem
3.98 which is equivalent to this loading.
SOLUTION
ΣFy : −100 lb − 200 lb = R
Have
or R = 300 lb
and
ΣM 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 force-couples at A
case
R
(a)
300 lb
500 lb ⋅ ft
(b)
300 lb
450 lb ⋅ ft
(c)
300 lb
450 lb ⋅ ft
(d )
200 lb
450 lb ⋅ ft
(e)
300 lb
100 lb ⋅ ft
(f )
300 lb
400 lb ⋅ ft
(g )
500 lb
500 lb ⋅ ft
(h)
300 lb
450 lb ⋅ ft
M
PROBLEM 3.100
Determine the single equivalent force and the distance from point A to its
line of action for the beam and loading of (a) Problem 3.98b,
(b) Problem 3.98d, (c) Problem 3.98e.
Problem 3.98: A 4-ft-long beam is subjected to a variety of loadings.
(a) Replace each loading with an equivalent force-couple system at end A
of the beam. (b) Which of the loadings are equivalent?
SOLUTION
(a)
For equivalent single force at distance d from A
ΣFy : − 300 lb = R
Have
or R = 300 lb
ΣM C :
and
( 300 lb )( d ) − 450 lb ⋅ ft
=0
or d = 1.500 ft
(b)
ΣFy : − 200 lb + 400 lb = R
Have
or R = 200 lb
and
ΣM C :
( 200 lb )( d ) + ( 400 lb )( 4 − d ) − 1150 lb ⋅ ft
=0
or d = 2.25 ft
(c)
Have
ΣFy : − 200 lb − 100 lb = R
or R = 300 lb
and
ΣM 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 force-couple systems act at the corners of a metal block,
which has been machined into the shape shown. Determine which of
these systems is equivalent to a force 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 force-couple system at A for each of the five force-couple
systems will be determined. Each will then be compared to the given
force-couple system to determine if they are equivalent.
Force-couple system at B
ΣF :
Have
F = (10 N ) j
or
and
(10 N ) j = F
(
)
ΣM A : ΣM B + rB/ A × F = M
( 4 N ⋅ m ) i + ( 2 N ⋅ m ) k + ( 0.2 m ) i × (10 N ) j = M
M = ( 4 N⋅m) i + ( 4 N⋅m)k
Comparing to given force-couple system at A,
Is Not Equivalent
Force-couple system at C
Have
or
and
ΣF :
(10 N ) j = F
F = (10 N ) j
(
)
ΣM 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 force-couple system at A,
Is Not Equivalent
PROBLEM 3.101 CONTINUED
Force-couple system at E
ΣF :
Have
(10 N ) j = F
F = (10 N ) j
or
(
)
ΣM A : M E + rE/ A × F = M
and
( 6 N ⋅ m ) i + ( 0.4 m ) i − ( 0.08 m ) j × (10 N ) j = M
M = (6 N⋅m) i + ( 4 N⋅m)k
Comparing to given force-couple system at A,
Is Equivalent
Force-couple system at G
ΣF :
Have
(10 N ) i + (10 N ) j = F
F = (10 N ) i + (10 N ) j
or
F has two force components
∴ force-couple system at G
Is Not Equivalent
Force-couple system at I
(10 N ) j = F
Have
ΣF :
or
F = (10 N ) j
and
(
)
ΣM A : ΣM 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 force-couple system at A,
Is Not Equivalent
PROBLEM 3.102
The masses of two children sitting at ends A and B of a seesaw are 38 kg
and 29 kg, respectively. Where should a third child sit so that the
resultant of the weights of the three children will pass through C if she
has a mass of (a) 27 kg, (b) 24 kg.
SOLUTION
First
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,
ΣM 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,
ΣM 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
WA = WB = m A g = (1.8 kg ) g
First
WC = mC g = (1.6 kg ) g
d = 0.75 m
(a)
Have
R = WA + WB + WC
R = (1.8 + 1.8 + 1.6 ) kg  g
R = ( 5.2 g ) N
or
Have
ΣM 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
D=
(b)
L
= 1.25 m
2
Have
ΣM 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
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 = 1.1 m
(a)
ΣF : −WC − WD − WE = R
Have
∴ R = −800 N − 700 N − 540 N
R = 2040 N
(a)
R = 2040 N
or
Have
ΣM 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)
L
ΣM A = − R  
2
From
∴
− ( 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
ΣFx : − ( 90 N ) sin 30° + (125 N ) cos 40° = Rx
or Rx = 50.756 N
ΣFy : − ( 90 N ) cos30° − 200 N − (125 N ) sin 40° = Ry
or Ry = −358.29 N
and
( 50.756 )2 + ( −358.29 )2
R=
Then
tan θ =
Ry
Rx
=
−358.29
= −7.0591
50.756
= 361.87 N
∴ θ = −81.937°
or R = 362 N
81.9°
Also
ΣM A : M − ( 90 N ) sin 35° ( 0.6 m ) − ( 200 N ) cos 25°  ( 0.85 m ) − (125 N ) sin 65°  (1.25 m ) = 0
∴ M = 326.66 N ⋅ m
or M = 327 N ⋅ m
PROBLEM 3.106
To test the strength of a 25 × 20-in. 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
ΣF : − ( 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
R = − (18.95 lb ) i + ( 41.3 lb ) j
or
R=
Rx2 + Ry2 =
(18.9461)2 + ( 41.297 )2
= 45.436 lb
 Ry 
−1  41.297 
 = tan 
 = −65.355°
 −18.9461 
 Rx 
θ x = tan −1 
or R = 45.4 lb
(b) Have
65.4°
ΣM 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
M B = rB × R
Since
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
or x = 4.63 in.
For
x = 0,
y =
191.246
= 10.0942 in.
18.9461
or y = 10.09 in.
PROBLEM 3.107
Solve Problem 3.106 assuming that P = 28 lb.
Problem 3.106: To test the strength of a 25 × 20-in. 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
ΣF : − ( 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
R = − (18.95 lb ) i + ( 51.3 lb ) j
or
R=
Rx2 + Ry2 =
(18.9461)2 + ( 51.297 )2
= 54.684 lb
 Ry 
−1  51.297 
 = tan 
 = −69.729°
 −18.9461 
 Rx 
θ x = tan −1 
or R = 54.7 lb
(b) Have
69.7°
ΣM B = M B
 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
M B = rB × R
Since
i
j
k
∴ ( 221.246 lb ⋅ in.) k =
x
y
0 = ( 51.297 x + 18.9461y ) k
−18.9461 51.297 0
For
y = 0,
x=
221.246
= 4.3130 in.
51.297
or x = 4.31 in.
For
x = 0,
y =
221.246
= 11.6776 in.
18.9461
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 α = 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 α so that the line of action of the
resultant passes through fold EF.
SOLUTION
Position the origin for the coordinate system along the centerline of the
sheet metal at the intersection with line EF.
ΣF = R
(a) Have
R = −  2.6 j − 5.25 j − 10.5 ( cos 45°i + sin 45° j) − 3.2i  kN
∴ R = − (10.6246 kN ) i − (15.2746 kN ) j
R=
Rx2 + Ry2 =
(10.6246 )2 + (15.2746 )2
= 18.6064 kN
 Ry 
−1  −15.2746 
 = tan 
 = 55.179°
 −10.6246 
 Rx 
θ = tan −1 
or R = 18.61 kN
55.2°
M EF = ΣM EF
Have
where
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,
M EF = dRy = d ( −15.2746 kN )
Have
∴d =
15.4903 N ⋅ m
= −1.01412 × 10−3 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 α + 40 mm 
∴ (128 N ⋅ m ) sin α = 106 N ⋅ m
sin α = 0.828125
α = 55.907°
or α = 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 α so that the resultant of the applied forces is parallel to the
10.5 N force, (b) the corresponding resultant of the applied forces and the
point of intersection of its line of action with a line drawn through points
A and B.
SOLUTION
(a) For the resultant force, R, to be parallel to the 10.5 kN force,
α =φ
∴ tan α = tan φ =
Ry
Rx
where
Rx = −3.2 kN − (10.5 kN ) sin α
Ry = −2.6 kN − 5.25 kN − (10.5 kN ) cos α
∴ tan α =
3.2 + 10.5 sinα
7.85 + 10.5cos α
tan α =
and
3.2
= 0.40764
7.85
α = 22.178°
or α = 22.2°
α = 22.178°
(b) From
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=
Rx2 + Ry2 =
( 7.1636 )2 + (17.5732 )2
= 18.9770 kN
R = 18.98 kN
or
67.8°
Then
M EF = ΣM EF
where
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,
M EF = dRy = d ( −17.5732 )
Have
∴d =
57.682 N ⋅ m
= −3.2824 × 10−3 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
R = ΣF
Have
R = ( 240 N )( cos 70°i − sin 70° j) − (160 N ) j
+ ( 300 N )( − cos 40°i − sin 40° j) − (180 N ) j
∴ R = − (147.728 N ) i − ( 758.36 N ) j
R=
Rx2 + Ry2 =
(147.728)2 + ( 758.36 )2
= 772.62 N
 Ry 
−1  −758.36 
 = tan 
 = 78.977°
 −147.728 
 Rx 
θ = tan −1 
or R = 773 N
79.0°
ΣM A = dRy
Have
where
ΣM A = − [ 240 N cos 70°] ( 6 m ) − [ 240 N sin 70°] ( 4 m )
− (160 N )(12 m ) +
[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 α = 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
P = 5 lb,
(a)
α = 40°
R = ΣF
Have
= ( 5 lb )( cos 40°i + sin 40° j) − ( 3 lb ) i − ( 2 lb ) j
∴ R = ( 0.83022 lb ) i + (1.21394 lb ) j
R=
Rx2 + Ry2 =
( 0.83022 )2 + (1.21394 )2
= 1.47069 lb
 Ry 
−1  1.21394 
 = tan 
 = 55.632°
 0.83022 
 Rx 
θ = tan −1 
or R = 1.471 lb
55.6°
MB = ΣMB = dRy
(b) From
where
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
M B = rD/B × R
(c) From
− ( 23.211 lb ⋅ in.) k = ( −d1 cos 50°i + 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
ΣFx = 0
Based on
P cos α − 3 lb = 0
∴ P cos α = 3 lb
(1)
ΣFy = 0
and
P sin α − 2 lb = 0
∴ P sin α = 2 lb
(2)
Dividing Equation (2) by Equation (1),
tan α =
2
3
∴ α = 33.690°
Substituting into Equation (1),
P=
3 lb
= 3.6056 lb
cos 33.690°
P = 3.61 lb
or
(a) Based on
33.7°
ΣM 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
ΣM 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 RB = 0 for
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 force-couple at A due to belts on pulley A
ΣF : −120 N − 160 N = RA
Have
∴ R A = 280 N
ΣM A : − 40 N ( 0.02 m ) = M A
Have
∴ M A = 0.8 N ⋅ m
Equivalent force-couple at B due to belts on pulley B
ΣF :
Have
( 210 N + 150 N )
∴ R B = 360 N
25° = R B
25°
ΣM B : − 60 N ( 0.015 m ) = M B
Have
∴ M B = 0.9 N ⋅ m
Equivalent force-couple at F
Have
ΣF : R F = ( −280 N ) j + ( 360 N )( cos 25°i + sin 25° j)
= ( 326.27 N ) i − (127.857 N ) j
R = RF =
2
2
RFx
+ RFy
=
( 326.27 )2 + (127.857 )2
= 350.43 N
 RFy 
−1  −127.857 
 = tan 
 = −21.399°
 326.27 
 RFx 
θ = tan −1 
or R F = R = 350 N
21.4°
PROBLEM 3.113 CONTINUED
Have
ΣM F : M F = − ( 280 N )( 0.06 m ) − 0.80 N ⋅ m
− ( 360 N ) cos 25° ( 0.010 m )
+ ( 360 N ) sin 25° ( 0.120 m ) − 0.90 N ⋅ m
M F = − ( 3.5056 N ⋅ m ) k
To determine where a single resultant force will intersect line FE,
M F = dR y
∴ d =
MF
−3.5056 N ⋅ m
=
= 0.027418 m = 27.418 mm
Ry
−127.857 N
or d = 27.4 mm
PROBLEM 3.114
As follower AB rolls along the surface of member C, it exerts a constant
force F perpendicular to the surface. (a) Replace F with an equivalent
force-couple system at the point D obtained by drawing the perpendicular
from the point of contact to the x axis (b) For 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 α = −  
 dx 
−1
=
a2  1 
 
2b  x 
For equivalence
ΣF : F = R
ΣM D :
( F cosα )( y A ) = M D
where
cos α =
2bx
(a )
2
2
+ ( 2bx )
∴ MD =
2
,

x2 
y A = b 1 − 2 
a 


x3 
2 Fb 2  x − 2 
a 

a 4 + 4b 2 x 2
Therefore, the equivalent force-couple system at D is
R = F
 a2 
tan −1 

 2bx 

x3 
2Fb 2  x − 2 
a 

M =
a 4 + 4b 2 x 2
PROBLEM 3.114 CONTINUED
dM
=0
dx
(b) To maximize M, the value of x must satisfy
where, for a = 1 m, b = 2 m
M =
∴
dM
= 8F
dx
)
1 + 16 x 2
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
2
2
)
−
1
2
 = 0
3
32 x 4 + 3x 2 − 1 = 0
or
x2 =
(
8F x − x 3
−3 ± 9 − 4 ( 32 )( −1)
2 ( 32 )
= 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 3-in.-diameter cylindrical sheet
metal container, the insertion tool exerts the forces shown on the
enclosure. Each of the forces is parallel to one of the coordinate axes.
Replace these forces with an equivalent force-couple system at C.
SOLUTION
For equivalence
Σ F: FA + FB + FC + FD = R C
R C = − ( 5 lb ) j − ( 3 lb ) j − ( 4 lb ) k − ( 7 lb ) i
∴ R C = ( −7 lb ) i − ( 8 lb ) j − ( 4 lb ) k
Also for equivalence
ΣM C : rA′/C × FA + rB′/C × FB + rD′/C × FD = M C
or
MC
i j
k
i
j
k
i
j
k
= 0 0 −1.5 in. + 1 in. 0 −1.5 in. + 0 1.5 in. 1.5 in.
0 5 lb
0
0 −3 lb
0
−7 lb
0
0
= ( −7.50 lb ⋅ in. − 0 ) i  + ( 0 − 4.50 lb ⋅ in.) i + ( −3.0 lb ⋅ in. − 0 ) k 
+ (10.5 lb ⋅ in. − 0 ) j + ( 0 + 10.5 lb ⋅ in.) k 
or M C = − (12.0 lb ⋅ in.) i + (10.5 lb ⋅ in.) j + ( 7.5 lb ⋅ in.) k
PROBLEM 3.116
Two 300-mm-diameter pulleys are mounted on line shaft AD. The belts B
and C lie in vertical planes parallel to the yz plane. Replace the belt forces
shown with an equivalent force-couple system at A.
SOLUTION
Equivalent force-couple at each pulley
Pulley B
R B = ( 290 N )( − cos 20° j + sin 20°k ) − 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 − cos10°k )
= − (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 N⋅m
0 −702.51 99.186
i
j
k
+ 0.90
0
0
N⋅m
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 force-couple
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
ΣF : A + B = C
Have
Fx : Ax + Bx = −40 N
or
∴ Bx = − ( Ax + 40 N )
(1)
ΣFy : Ay + By = 0
Ay = − By
or
(2)
ΣFz : 10 N + Bz = 20 N
Bz = 10 N
or
ΣM C : rB/C × B + rA/C × A = M C
Have
∴
or
(3)
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
( 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
(4)
j - coefficient
− 0.05Bx + 0.2 Ax = 4 N ⋅ m
(5)
k - coefficient
0.2 By + 0.2 Ay = 0
(6)
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 force-couple system at A
consisting of the force R = ( 3.9 lb ) i + Ry j − (1.1 lb ) k and the couple
M RA = M xi + (1.5 lb ⋅ ft ) j − (1.1 lb ⋅ ft ) k. . (b) Find the corresponding
values of Ry and M x .
SOLUTION
ΣF : B + C = R
Have
ΣFx : Bx + Cx = 3.9 lb
or
Bx = 3.9 lb − Cx
(1)
ΣFy : C y = Ry
(2)
ΣFz : C z = −1.1 lb
(3)
ΣM A : rB/ A × B + rC/ A × C + M B = M RA
Have
i j k
i
j
k
1
1
4 0 2.0 + ( 2 lb ⋅ ft ) i = M xi + (1.5 lb ⋅ ft ) j − (1.1 lb ⋅ ft ) k
∴
x 0 4.5 +
12
12
Bx 0 0
C x C y −1.1
( 2 − 0.166667C y ) i + ( 0.375Bx + 0.166667Cx + 0.36667 ) j + ( 0.33333C y ) k
= M xi + (1.5 ) j − (1.1) k
From
i - coefficient
2 − 0.166667C y = M x
(4)
j - coefficient
0.375Bx + 0.166667Cx + 0.36667 = 1.5
(5)
k - coefficient
0.33333C y = −1.1
(6)
or
C y = −3.3 lb
(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
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 10-lb force at
F lies in a plane parallel to the yz plane, determine (a) the angle α the
force at F should form with the horizontal if duct AB is not to tend to
rotate about the vertical, (b) the force-couple system at B equivalent to
the given force system when this condition is satisfied.
SOLUTION
(a) Duct AB will not have a tendency to rotate about the vertical or y-axis if:
(
)
R
M By
= j ⋅ ΣM RB = j ⋅ rF /B × FF + rE/B × FE = 0
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 α ) j + ( cos α ) k 
FE = − ( 5 lb ) k
∴
ΣM RB
i
j
k
i
j k
= (10 lb ) 45 in. −23 in. 28 in. + ( 5 lb )( 2 in.) 27 −17 14
0
sin α cos α
0 0 −1
= ( −230 cos α − 280sin α + 170 ) i − ( 450 cos α − 270 ) j + ( 450sin α ) k  lb ⋅ in.
Thus,
R
M By
= −450 cos α + 270 = 0
cos α = 0.60
α = 53.130°
or α = 53.1°
PROBLEM 3.119 CONTINUED
(b) R = FE + FF
where
FE = − ( 5 lb ) k
FF = (10 lb )( sin 53.130° j + cos 53.130°k ) = ( 8 lb ) j + ( 6 lb ) k
∴ R = ( 8 lb ) j + (1 lb ) k
and
M = ΣM RB = −  230 ( 0.6 ) + 280 ( 0.8 ) − 170  i −  450 ( 0.6 ) − 270  j +  450 ( 0.8 ) k
= − (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 10-lb force at
F lies in a plane parallel to the yz plane and that α = 60°, (a) replace the
given force system with an equivalent force-couple system at C, (b)
determine whether duct CD will tend to rotate clockwise or
counterclockwise relative to elbow C, as viewed from D to C.
SOLUTION
R = ΣF = FF + FE
(a) Have
FF = 10 lb ( sin 60° ) j + ( cos 60° ) k  = ( 8.6603 lb ) j + ( 5.0 lb ) k
where
FE = − ( 5 lb ) k
∴ R = ( 8.6603 lb ) j
or R = ( 8.66 lb ) j
Have
M CR = Σ ( r × F ) = rF /C × FF + rE/C × FE
where
rF /C = ( 9 in.) i − ( 2 in.) j
rE/C = (18 in.) i − (13 in.) j
∴
M CR
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
or M CR = ( 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
M CD
= λ DC ⋅ M CR
where
λ DC =
Then
− (18 in.) i + ( 4 in.) j
2 85 in.
R
M CD
=
=
1
( −9i + 2 j)
85
1
( −9i + 2 j) ⋅ ( 55i + 45j + 77.942k ) lb ⋅ in.
85
= ( −53.690 + 9.7619 ) lb ⋅ in.
= −43.928 lb ⋅ in.
Since λ DC ⋅ M CR < 0, duct DC tends to rotate clockwise relative to elbow C as viewed from D to C.
PROBLEM 3.121
The head-and-motor assembly of a radial drill press was originally
positioned with arm AB parallel to the z axis and the axis of the chuck
and bit parallel to the y axis. The assembly was then rotated 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 force-couple system at the center O of the base of the vertical
column.
SOLUTION
R =F
Have
= ( 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
M O = rB/O × F + M C
Have
where
rB/O = ( 0.280 m ) sin 25°  i + ( 0.300 m ) j + ( 0.280 m ) cos 25°  k
= ( 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 force-couple system shown, where R = 300 N and
M = 37.5 N ⋅ m. Replace this force-couple system with an equivalent
force-couple system at C.
SOLUTION
 − ( 0.2 m ) i + (1.4 m ) j − ( 0.5 m ) k 
R C = R = ( 300 N ) λ AB = 300 N 

1.50 m


From
R C = − ( 40.0 N ) i + ( 280 N ) j − (100 N ) k
M C = rA/C × R + M
From
where
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 ) λ BA = ( 37.5 N ⋅ m ) 

1.50 m


= ( 5.0 N ⋅ m ) i − ( 35.0 N ⋅ m ) j + (12.5 N ⋅ m ) k
∴ MC
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
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 × 15-ft raft. If the weights of the
children at points A, B, and C are 85 lb, 60 lb, and 90 lb, respectively,
determine the magnitude and the point of application of the resultant of
the three weights.
SOLUTION
Have
ΣF : FA + FB + FC = R
− ( 85 lb ) j − ( 60 lb ) j − ( 90 lb ) j = R
− ( 235 lb ) j = R
Have
or R = 235 lb
ΣM 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
ΣM 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 × 15-ft raft. The weights of the
children at points A, B, and C are 85 lb, 60 lb, and 90 lb, respectively. If a
fourth child of weight 95 lb climbs onto the raft, determine where she
should stand if the other children remain in the positions shown and the
line of action of the resultant of the four weights is to pass through the
center of the raft.
SOLUTION
Have
ΣF : FA + FB + FC + FD = R
− ( 85 lb ) j − ( 60 lb ) j − ( 90 lb ) j − ( 95 lb ) j = R
∴ R = − ( 330 lb ) j
Have
ΣM x : FA ( z A ) + FB ( z B ) + FC ( zC ) + FD ( z D ) = R ( z H )
(85 lb )( 9 ft ) + ( 60 lb )(1.5 ft ) + ( 90 lb )(14.25 ft ) + ( 95 lb )( zD ) = ( 330 lb )( 7.5 ft )
∴ z D = 3.5523 ft
Have
or z D = 3.55 ft
ΣM z : FA ( x A ) + FB ( xB ) + FC ( xC ) + FD ( xD ) = R ( xH )
(85 lb )( 3 ft ) + ( 60 lb )( 4.5 ft ) + ( 90 lb )(14.25 ft ) + ( 95 lb )( xD ) = ( 330 lb )( 7.5 ft )
∴ xD = 7.0263 ft
or xD = 7.03 ft
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
ΣF : FA + FB + FC + FD = R
− ( 580 kN ) j − ( 2350 kN ) j − ( 330 kN ) j − (140 kN ) j = R
∴ R = − ( 3400 kN ) j
Have
R = 3400 kN
ΣM 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
ΣM 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 580-kN force is shoveled so that the this load acts at E, determine
a and b knowing that the point of application of the resultant of the four
loads is then at B.
SOLUTION
Have
ΣF : FB + FC + FD + FE = R
− ( 2350 kN ) j − ( 330 kN ) j − (140 kN ) j − ( 580 kN ) j = R
∴ R = − ( 3400 kN ) j
Have
ΣM 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
ΣM 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 × 4-m flatbed trailer with two
0.6 × 0.6 × 0.6-m boxes and one 0.6 × 0.6 × 1.2-m box. Each of the
boxes at the rear of the trailer is positioned so that it is aligned with both
the back and a side of the trailer. Determine the smallest load the students
should place in a second 0.6 × 0.6 × 1.2-m box and where on the trailer
they should secure it, without any part of the box overhanging the sides
of the trailer, if each box is uniformly loaded and the line of action of the
resultant of the weights of the four boxes is to pass through the point of
intersection of the centerlines of the trailer and the axle. (Hint: Keep in
mind that the box may be placed either on its side or on its end.)
SOLUTION
For the smallest weight on the trailer so that the resultant force of the four weights acts over the axle at the
intersection with the center line of the trailer, the added 0.6 × 0.6 × 1.2-m box should be placed adjacent to
one of the edges of the trailer with the 0.6 × 0.6-m side on the bottom. The edges to be considered are based
on the location of the resultant for the three given weights.
ΣF : − ( 200 N ) j − ( 400 N ) j − (180 N ) j = R
Have
∴ R = − ( 780 N ) j
Have
ΣM z :
( 200 N )( 0.3 m ) + ( 400 N )(1.7 m ) + (180 N )(1.7 m ) = ( 780 N )( x )
∴ x = 1.34103 m
Have
ΣM x :
( 200 N )( 0.3 m ) + ( 400 N )( 0.6 m ) + (180 N )( 2.4 m ) = ( 780 N )( z )
∴ z = 0.93846 m
From the statement of the problem, it is known that the resultant of R from the original loading and the
lightest load W passes through G, the point of intersection of the two center lines. Thus, ΣM G = 0.
Further, since the lightest load W is to be as small as possible, the fourth box should be placed as far from G
as possible without the box overhanging the trailer. These two requirements imply
( 0.3 m ≤
x ≤ 1 m ) (1.8 m ≤ z ≤ 3.7 m )
PROBLEM 3.127 CONTINUED
Let x = 0.3 m,
ΣM Gz :
( 200 N )( 0.7 m ) − ( 400 N )( 0.7 m ) − (180 N )( 0.7 m ) + W ( 0.7 m ) = 0
∴ W = 380 N
ΣM Gx : − ( 200 N )(1.5 m ) − ( 400 N )(1.2 m ) + (180 N )( 0.6 m ) + ( 380 N )( z − 1.8 m ) = 0
∴ z = 3.5684 m < 3.7 m
Let z = 3.7 m,
∴ acceptable
ΣM Gx : − ( 200 N )(1.5 m ) − ( 400 N )(1.2 m ) + (180 N )( 0.6 m ) + W (1.7 m ) = 0
∴ W = 395.29 N > 380 N
Since the weight W found for x = 0.3 m is less than W found for z = 3.7 m, x = 0.3 m results in the
smallest weight W.
or W = 380 N at
( 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 × 4-m flatbed trailer with
two 0.6 × 0.6 × 0.6-m boxes and one 0.6 × 0.6 × 1.2-m box. Each of the
boxes at the rear of the trailer is positioned so that it is aligned with both
the back and a side of the trailer. Determine the smallest load the students
should place in a second 0.6 × 0.6 × 1.2-m box and where on the trailer
they should secure it, without any part of the box overhanging the sides
of the trailer, if each box is uniformly loaded and the line of action of the
resultant of the weights of the four boxes is to pass through the point of
intersection of the centerlines of the trailer and the axle. (Hint: Keep in
mind that the box may be placed either on its side or on its end.)
SOLUTION
For the largest additional weight on the trailer with the box having at least one side coinsiding with the side of
the trailer, the box must be as close as possible to point G. For x = 0.6 m, with a small side of the box
touching the z-axis, satisfies this condition.
Let x = 0.6 m,
ΣM Gz :
( 200 N )( 0.7 m ) − ( 400 N )( 0.7 m ) − (180 N )( 0.7 m ) + W ( 0.4 m ) = 0
∴ W = 665 N
and
ΣM GX : − ( 200 N )(1.5 m ) − ( 400 N )(1.2 m ) + (180 N )( 0.6 m ) + ( 665 N )( z − 1.8 m ) = 0
∴ z = 2.8105 m
(2 m <
z < 4 m)
∴ acceptable
or W = 665 N at
( 0.6 m, 0, 2.81 m )
PROBLEM 3.129
A block of wood is acted upon by three forces of the same magnitude P
and having the directions shown. Replace the three forces with an
equivalent wrench and determine (a) the magnitude and direction of the
resultant R, (b) the pitch of the wrench, (c) the point where the axis of the
wrench intersects the xy plane.
SOLUTION
First, reduce the given force system to a force-couple at the origin.
Have
ΣF : Pi − Pi − Pk = R
∴ R = − Pk
Have
ΣM O : − P ( 3a ) k − P ( 3a ) j + P ( −ai + 3aj) = M OR
∴ M OR = Pa ( −i − 3k )
Then let vectors ( R, M1 ) represent the components of the wrench, where their directions are the same.
R = − Pk
(a)
or Magnitude of R = P
Direction of R : θ x = 90°, θ y = 90°, θ z = −180°
(b) Have
M1 = λ R ⋅ M OR
= −k ⋅  Pa ( −i − 3k ) 
= 3Pa
and pitch
p=
M1 3Pa
=
= 3a
R
P
or p = 3a
PROBLEM 3.129 CONTINUED
(c) Have
M OR = M1 + M 2
∴ M 2 = M OR − M1 = Pa ( −i − 3k ) − ( −3Pak ) = − Pai
Require
M 2 = rQ/O × R
− Pai = ( xi + yj) × ( − P ) k = Pxj − Pyi
From
i : − Pa = − Py
or
y =a
j: x = 0
∴ The axis of the wrench is parallel to the z-axis and intersects the xy plane at x = 0, y = a
PROBLEM 3.130
A piece of sheet metal is bent into the shape shown and is acted upon by
three forces. Replace the three forces with an equivalent wrench and
determine (a) the magnitude and direction of the resultant R, (b) the pitch
of the wrench, (c) the point where the axis of the wrench intersects the
yz plane.
SOLUTION
First, reduce the given force system to a force-couple system at the origin.
( 2P ) i − ( P ) j + ( P ) j = R
ΣF :
Have
∴ R = ( 2P ) i
ΣM O : Σ ( rO × F ) = M OR
Have
M OR
i j k
i j k
= Pa 2 2 2.5 + 0 0 4 = Pa ( −1.5i + 5j − 6k )
2 −1 0
0 1 0
R = 2 Pi
(a)
or Magnitude of R = 2 P
Direction of R : θ x = 0°, θ y = −90°, θ z = 90°
(b) Have
M1 = λ R ⋅ M OR
λR =
R
R
= i ⋅ ( −1.5Pai + 5Paj − 6 Pak )
= −1.5Pa
and pitch
p=
M1
−1.5Pa
=
= −0.75a
R
2P
or p = −0.75a
PROBLEM 3.130 CONTINUED
M OR = M1 + M 2
(c) Have
∴ M 2 = M OR − M1 = ( 5Pa ) j − ( 6Pa ) k
Require
M 2 = rQ/O × R
( 5Pa ) j − ( 6Pa ) k = ( yj + zk ) × ( 2Pi ) = − ( 2Py ) k + ( 2Pz ) j
From
i : 5Pa = 2 Pz
∴ z = 2.5a
From
k : − 6 Pa = −2 Py
∴ y = 3a
∴ The axis of the wrench is parallel to the x-axis and intersects the yz-plane 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 force-couple at the origin.
ΣF : − (10 N ) j − (11 N ) j = R
Have
∴ R = − ( 21 N ) j
ΣM O : Σ ( rO × F ) + ΣM C = M OR
Have
M OR
i j
k
i j
k
= 0 0 0.5 N ⋅ m + 0 0 −0.375 N ⋅ m − (12 N ⋅ m ) j
0 −10 0
0 −11
0
= ( 0.875 N ⋅ m ) i − (12 N ⋅ m ) j
R = − ( 21 N ) j
(a)
(b) Have
M1 = λ R ⋅ M OR
λR =
or R = − ( 21 N ) j
R
R
= ( − j) ⋅ ( 0.875 N ⋅ m ) i − (12 N ⋅ m ) j
= 12 N ⋅ m
and pitch
p=
and
M1 = − (12 N ⋅ m ) j
M1 12 N ⋅ m
=
= 0.57143 m
R
21 N
or p = 0.571 m
PROBLEM 3.131 CONTINUED
M OR = M1 + M 2
(c) Have
∴ M 2 = M OR − M1 = ( 0.875 N ⋅ m ) i
M 2 = rQ/O × R
Require
∴
( 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 y-axis and intersects the xz-plane 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 force-couple system.
Have
Have
ΣF : − ( 6 lb ) i − ( 4.5 lb ) j = R
R = 7.5 lb
ΣM O : ∑ ( rO × F ) + ∑ M C = M OR
M OR = −6 lb ( 8 in.) j − (160 lb ⋅ in.) i − ( 72 lb ⋅ in.) j
= − (160 lb ⋅ in.) i − (120 lb ⋅ in.) j
M OR = 200 lb ⋅ in.
R = − ( 6 lb ) i − ( 4.5 lb ) j
(a)
(b) Have
M1 = λ R ⋅ M OR
λ =
R
R
= ( −0.8i − 0.6 j) ⋅  − (160 lb ⋅ in.) i − (120 lb ⋅ in.) j
= 200 lb ⋅ in.
M1 = 200 lb ⋅ in. ( −0.8i − 0.6j)
and
Pitch
p=
M1
200 lb ⋅ in.
=
= 26.667 in.
R
7.50 lb
or p = 26.7 in.
(c) From above note that
M1 = M OR
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 force-couple at the origin.
Have
ΣF : − ( 20 lb ) k − ( 21 lb ) j = − ( 21 lb ) j − ( 20 lb ) k = R
R = 29 lb
ΣM O : ∑ ( rO × F ) + ∑ M C = M OR
and
i j k
i j k
20 lb ( 4 in.) 4 3 0 + 21 lb ( 4 in.) 6 0 1 + ( −300 j − 320k ) lb ⋅ in. = M OR
0 0 −1
0 −1 0
∴ M OR = − (156 lb ⋅ in.) i + ( 20 lb ⋅ in.) j − ( 824 lb ⋅ in.) k
R = − ( 21 lb ) j − ( 20 lb ) k
(a)
(b) Have
M1 = λ R ⋅ M OR
=−
λR =
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
M1 = M1λ R = − ( 401.03 lb ⋅ in.) j − ( 381.93 lb ⋅ in.) k
and
Then pitch
p=
M1 553.80 lb ⋅ in.
=
= 19.0964 in.
R
29 lb
or p = 19.10 in.
M OR = M1 + M 2
(c) Have
∴ M 2 = M OR − 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
M 2 = rQ/O × R
Require
( −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 xz-plane 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 force-couple at the origin at B.
15 
 8
ΣF : − ( 79.2 lb ) k − ( 51 lb )  i +
j = R
17 
 17
(a) Have
∴ R = − ( 24.0 lb ) i − ( 45.0 lb ) j − ( 79.2 lb ) k
R = 94.2 lb
and
ΣM B : rA/B × FA + M A + M B = M RB
Have
M RB
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 RB = (1248 lb ⋅ in.) i − ( 630 lb ⋅ in.) j − ( 660 lb ⋅ in.) k
(b) Have
M1 = λ R ⋅ M OR
=
λR =
R
R
−24.0i − 45.0 j − 79.2k
⋅ (1248 lb ⋅ in.) i − ( 630 lb ⋅ in.) j − ( 660 lb ⋅ in.) k 
94.2
= 537.89 lb ⋅ in.
PROBLEM 3.134 CONTINUED
M1 = M1λ R
and
= − (137.044 lb ⋅ in.) i − ( 256.96 lb ⋅ in.) j − ( 452.24 lb ⋅ in.) k
Then pitch
p=
M1 537.89 lb ⋅ in.
=
= 5.7101 in.
R
94.2 lb
or p = 5.71 in.
M RB = M1 + M 2
(c) Have
∴ M 2 = M RB − M1 = (1248i − 630 j − 660k ) − ( −137.044i − 256.96 j − 452.24k )
= (1385.04 lb ⋅ in.) i − ( 373.04 lb ⋅ in.) j − ( 207.76 lb ⋅ in.) k
M 2 = rQ/B × R
Require
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
−207.76 = −45x
∴ z = 30.779 in.
∴ x = 4.6169 in.
∴ The axis of the wrench intersects the xz-plane at
x = 4.62 in., z = 30.8 in.
PROBLEM 3.135
A flagpole is guyed by three cables. If the tensions in the cables have the
same magnitude P, replace the forces exerted on the pole with an
equivalent wrench and determine (a) the resultant force R, (b) the pitch of
the wrench, (c) the point where the axis of the wrench intersects the xz
plane.
SOLUTION
(a) First reduce the given force system to a force-couple at the origin.
ΣF : Pλ BA + Pλ DC + Pλ DE = R
Have
 4
3  3
4   −9
4
12  
R = P  j − k  +  i − j  +  i − j +
k
5
5
5
5
25
5
25  
 
 

∴ R =
R=
3P
25
( 2 )2 + ( 20 )2 + (1)2
=
3P
( 2i − 20 j − k )
25
27 5
P
25
ΣM : Σ ( rO × P ) = M OR
Have
−4 P
3P 
4P 
4P
12 P 
 3P
 −9 P
j−
k  + ( 20a ) j × 
i−
j  + ( 20a ) j × 
i−
j+
k  = M OR
5 
5 
5
25 
 5
 5
 25
( 24a ) j × 
∴ M OR =
M1 = λ R ⋅ M OR
(b) Have
where
24 Pa
( −i − k )
5
λR =
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 =
and pitch
p=
M1 = M 1λ R =
(c)
Then
1
9 5
( 2i − 20 j − k ) ⋅
24 Pa
−8Pa
( −i − k ) =
5
15 5
M1
−8Pa  25  −8a
=

=
R
81
15 5  27 5 P 
or p = −0.0988a
−8Pa  1 
8Pa
( −2i + 20 j + k )

 ( 2i − 20 j − k ) =
675
15 5  9 5 
M 2 = M OR − M1 =
24Pa
8Pa
8Pa
( −i − k ) −
( −2i + 20 j + k ) =
( −403i − 20 j − 406k )
5
675
675
M 2 = rQ/O × R
Require
 8Pa 
 3P 

 ( −403i − 20 j − 406k ) = ( xi + zk ) × 
 ( 2i − 20 j − k )
675


 25 
 3P 
=
  20 zi + ( x + 2 z ) j − 20 xk 
 25 
From i:
8 ( −403)
Pa
 3P 
= 20 z 

675
 25 
∴ z = −1.99012a
From k:
8 ( −406 )
Pa
 3P 
= −20 x 

675
 25 
∴ x = 2.0049a
∴ The axis of the wrench intersects the xz-plane at
x = 2.00a, z = −1.990a
PROBLEM 3.136
Determine whether the force-and-couple system shown can be reduced to
a single equivalent force R. If it can, determine R and the point where the
line of action of R intersects the yz plane. If it cannot be so reduced,
replace the given system with an equivalent wrench and determine its
resultant, its pitch, and the point where its axis intersects the yz plane.
SOLUTION
First, reduce the given force system to a force-couple at D.
Have
ΣF : FDA + FED = FDAλ DA + FEDλ ED = R
where
 − ( 0.300 m ) i + ( 0.225 m ) j + ( 0.200 m ) k 
FDA = 136 N 

0.425 m


= − ( 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
ΣM D : M A = M D
 − ( 0.150 m ) i − ( 0.150 m ) j + ( 0.450 m ) k  16 N ⋅ m
M RD = (16 N ⋅ m ) 
( −i − j + 3k )
 =
0.150 11 m
11


PROBLEM 3.136 CONTINUED
The force-couple at D can be replaced by a single force if R is perpendicular to M RD . To be perpendicular,
R ⋅ M RD = 0.
Have
R ⋅ M RD = ( −168i + 72 j − 32k ) ⋅
=
16
( −i − j + 3k )
11
128
( 21 − 9 − 12 )
11
=0
∴ Force-couple can be reduced to a single equivalent force.
To determine the coordinates where the equivalent single force intersects the yz-plane, M RD = rQ/D × R
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
j
k
( y − 0.075) z m
9
−4
or
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:
From k:
−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
∴ line of action of R intersects the yz-plane at
y = 290 mm, z = −28.4 mm
PROBLEM 3.137
Determine whether the force-and-couple system shown can be reduced to
a single equivalent force R. If it can, determine R and the point where the
line of action of R intersects the yz plane. If it cannot be so reduced,
replace the given system with an equivalent wrench and determine its
resultant, its pitch, and the point where its axis intersects the yz plane.
SOLUTION
First, reduce the given force system to a force-couple at the origin.
ΣF : FA + FG = R
Have
 ( 4 in.) i + ( 6 in.) j − (12 in.) k 
∴ R = (10 lb ) k + 14 lb 
 = ( 4 lb ) i + ( 6 lb ) j − ( 2 lb ) k
14 in.


R=
and
Have
56 lb
ΣM O : ∑ ( rO × F ) + ∑ M C = M OR
{
}
M OR = (12 in.) j × (10 lb ) k  + (16 in.) i × ( 4 lb ) i + ( 6 lb ) j − (12 lb ) k 
 (16 in.) i − (12 in.) j 
 ( 4 in.) i − (12 in.) j + ( 6 in.) k 
+ ( 84 lb ⋅ in.) 
 + ( 120 lb ⋅ in.) 

20 in.
14 in.




∴ M 0R = ( 221.49 lb ⋅ in.) i + ( 38.743 lb ⋅ in.) j + (147.429 lb ⋅ in.) k
= (18.4572 lb ⋅ ft ) i + ( 3.2286 lb ⋅ ft ) j + (12.2858 lb ⋅ ft ) k
PROBLEM 3.137 CONTINUED
The force-couple at O can be replaced by a single force if the direction of R is perpendicular to M OR .
To be perpendicular R ⋅ M OR = 0
Have
R ⋅ M OR = ( 4i + 6 j − 2k ) ⋅ (18.4572i + 3.2286 j + 12.2858k ) = 0?
= 73.829 + 19.3716 − 24.572
≠0
∴ System cannot be reduced to a single equivalent force.
To reduce to an equivalent wrench, the moment component along the line of action of P is found.
M1 = λ R ⋅ M OR
λR =
R
R
 ( 4i + 6 j − 2k ) 
=
 ⋅ (18.4572i + 3.2286 j + 12.2858k )
56


= 9.1709 lb ⋅ ft
M1 = M1λ R = ( 9.1709 lb ⋅ ft )( 0.53452i + 0.80178 j − 0.26726k )
and
And pitch
p=
M1 9.1709 lb ⋅ ft
=
= 1.22551 ft
R
56 lb
or p = 1.226 ft
Have
M 2 = M OR − M1 = (18.4572i + 3.2286 j + 12.2858k ) − ( 9.1709 )( 0.53452i + 0.80178 j − 0.26726k )
= (13.5552 lb ⋅ ft ) i − ( 4.1244 lb ⋅ ft ) j + (14.7368 lb ⋅ ft ) k
Require
M 2 = rQ/O × R
(13.5552i − 4.1244 j + 14.7368k ) = ( yj + zk ) × ( 4i + 6 j − 2k )
= − ( 2 y + 6z ) i + ( 4z ) j − ( 4 y ) k
From j:
−4.1244 = 4z
From k:
14.7368 = −4 y
or
or
z = −1.0311 ft
y = −3.6842 ft
∴ line of action of the wrench intersects the yz plane at
y = −3.68 ft, z = 1.031 ft
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
ΣFx : Ax + Bx = 0
(1)
ΣFz : Az + Bz = R
(2)
ΣM x : Az ( a ) + Bz ( a + b ) = 0
(3)
ΣM z : − Ax ( a ) − Bx ( a + b ) = M
(4)
Bx = − Ax
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
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


M
B = −
 b

a 
i −  Rk

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 x-axis intercept of B.
The known components of the wrench can be expressed as
R = Rxi + Ry j + Rzk
M = M xi + M y j + M zk
and
while the unknown forces A and B can be expressed as
A = Axi + Ay j + Azk
and
B = Bxi + Bzk
Since the position vector of point P is given, it follows that the scalar components (x, y, z) of the position
vector rP are also known.
Then, for equivalence of the two systems
ΣFx : Rx = Ax + Bx
(1)
ΣFy : Ry = Ay
(2)
ΣFz : Rz = Az + Bz
(3)
ΣM x : M x = yAz − zAy
(4)
ΣM y : M y = zAx − xAz − bBz
(5)
ΣM z : M z = xAy − yAx
(6)
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.
Ay = Ry
From Equation (2)
Equation (6)
1
Ax =   xRy − M z
 y
)
Equation (1)
1
Bx = Rx −   xRy − M z
 y
(
)
Equation (4)
1
Az =   M x + zRy
 y
)
Equation (3)
1
Bz = Rz −   M x + zRy
 y
)
Equation (5)
(
(
(
b=
( xM x + yM y + zM z )
( M x − yRz + zRy )
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
(1)
∑ Fy : R = Ay + By
(2)
∑ Fz : 0 = Az + Bz
(3)
∑ M x : 0 = − zBy
(4)
∑ M y : M = −aAz − xBz + zBx
(5)
∑ M z : 0 = aAy + xBy
(6)
Since A and B are made perpendicular,
A⋅B = 0
There are eight unknowns:
or
Ax Bx + Ay By + Az Bz = 0
Ax , Ay , Az , Bx , By , Bz , x, z
But only seven independent equations. Therefore, there exists an infinite number of solutions.
(7)
PROBLEM 3.140 CONTINUED
0 = − zBy
Next consider Equation (4):
If By = 0, Equation (7) becomes
Ax Bx + Az Bz = 0
Ax2 + Az2 = 0
Using Equations (1) and (3) this equation becomes
Since the components of A must be real, a nontrivial solution is not possible. Thus, it is required that
By ≠ 0, so that from Equation (4), z = 0.
To obtain one possible solution, arbitrarily let Ax = 0.
(Note: Setting Ay , Az , or Bz equal to zero results in unacceptable solutions.)
The defining equations then become.
0 = Bx
Then
(1)′
R = Ay + By
(2)
0 = Az + Bz
(3)
M = −aAz − xBz
(5)′
0 = aAy + xBy
(6)
Ay By + Az Bz = 0
(7)′
Equation (2) can be written
Ay = R − By
Equation (3) can be written
Bz = − Az
Equation (6) can be written
x=−
aAy
By
Substituting into Equation (5)′,
 R − By 
M = −aAz −  −a
 ( − Az )

By 

or
Az = −
M
By
aR
Substituting into Equation (7)′,
M
 M

By 
By  = 0
( R − By ) By +  − aR
 aR 
(8)
PROBLEM 3.140 CONTINUED
By =
or
a 2 R3
a R2 + M 2
2
Then from Equations (2), (8), and (3)
Ay = R −
Az = −
Bz =
a 2 R3
RM 2
=
a2R2 + M 2
a2R2 + M 2

M 
a 2 R3
aR 2 M
=
−
 2 2

aR  a R + M 2 
a2R2 + M 2
aR 2 M
a2R2 + M 2
In summary
A=
RM
( Mj − aRk )
a R2 + M 2
B=
aR 2
( aRj + Mk )
a2R2 + M 2
2
Which shows that it is possible to replace a wrench with two perpendicular forces, one of which is applied at a
given point.
Lastly, if R > 0 and M > 0, it follows from the equations found for A and B that Ay > 0 and By > 0.
From Equation (6), x < 0 (assuming a > 0). Then, as a consequence of letting Ax = 0, force A lies in a plane
parallel to the yz plane and to the right of the origin, while force B lies in a plane parallel to the yz plane but to
the left of the origin, as shown in the figure below.
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
λ A = λxi + λ y j + λzk
it follows that force A can be expressed as
(
A = Aλ A = A λxi + λ y j + λz k
)
Force B can be expressed as
B = Bxi + By j + Bzk
Next, observe that since the axis of the wrench and the prescribed line of action AA′ are known, it follows
that the distance a can be determined. In the following solution, it is assumed that a is known.
Then, for equivalence
(
ΣFx : 0 = Aλx + Bx
(1)
ΣFy : R = Aλ y + By
(2)
ΣFz : 0 = Aλz + Bz
(3)
ΣM x : 0 = − zBy
(4)
ΣM y : M = −aAλz + zBx − xBz
(5)
ΣM z : 0 = aAλ y + xBy
(6)
)
Since there are six unknowns A, Bx , By , Bz , x, z and six independent equations, it will be possible to
obtain a solution.
PROBLEM 3.141 CONTINUED
Case 1: Let z = 0 to satisfy Equation (4)
Aλ y = R − By
Now Equation (2)
Bz = − Aλz
Equation (3)
x=−
Equation (6)
aAλ y
By
 a 
= −
 R − By
 By 


(
)
Substitution into Equation (5)
  a 

M = −aAλz −  − 
 R − By ( − Aλz ) 
  By 

(
∴ A=−
1 M 
λz  aR 
)
By
Substitution into Equation (2)
R=−
1 M 
B λ + By
λz  aR  y y
∴ By =
Then
A=−
λz aR 2
λz aR − λ y M
MR
R
=
aR
λz aR − λ y M
λy −
λz
M
Bx = − Aλx =
λx MR
λz aR − λ y M
Bz = − Aλz =
λz MR
λz aR − λ y M
In summary
A=
B=
and
P
λA
aR
λy −
λz
M
R
( λ Mi + λz aRj + λz Mk )
λz aR − λ y M x


 λz aR − λ y M
R 
x = a 1 −
 = a 1 − R 

By 
λz aR 2




 
 
or x =
Note that for this case, the lines of action of both A and B intersect the x axis.
λy M
λz R
PROBLEM 3.141 CONTINUED
Case 2: Let By = 0 to satisfy Equation (4)
A=
Now Equation (2)
R
λy
Equation (1)
λ
Bx = − R  x
 λy





Equation (3)
λ
Bz = − R  z
 λy





aAλ y = 0
Equation (6)
which requires a = 0
Substitution into Equation (5)
 λ
M = z −R  x
  λ y
  λ 

 − x −R  z 

  λ y  

or
This last expression is the equation for the line of action of force B.
In summary
 R
A=
 λy

 R
B=
 λy


λ A



 ( −λ x i − λ z k )


Assuming that λx , λ y , λz > 0, the equivalent force system is as shown below.
Note that the component of A in the xz plane is parallel to B.
M 
λz x − λx z =   λ y
 R
PROBLEM 3.142
A worker tries to move a rock by applying a 360-N force to a steel bar as
shown. (a) Replace that force with an equivalent force-couple system at
D. (b) Two workers attempt to move the same rock by applying a vertical
force at A and another force at D. Determine these two forces if they are
to be equivalent to the single force of part a.
SOLUTION
(a) Have
ΣF : 360 N ( − sin 40°i − cos 40° j) = − ( 231.40 N ) i − ( 275.78 N ) j = F
or F = 360 N
ΣM D : rB/D × R = M
Have
where
50°
rB/D = − ( 0.65 m ) cos 30°  i + ( 0.65 m ) sin 30°  j
= − ( 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
or M = 230 N ⋅ m
ΣM D : M = rA/D × FA
rA/D = − (1.05 m ) cos 30° i + (1.05 m ) sin 30°  j
= − ( 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
( 0.90933FA ) k
or
= 230.45k
∴ FA = 253.42 N
or FA = 253 N
ΣF : F = FA + FD
Have
− ( 231.40 N ) i − ( 275.78 N ) j = − ( 253.42 N ) j + FD ( − cosθ i − sin θ j)
From
i : 231.40 N = FD cosθ
(1)
j: 22.36 N = FD sin θ
(2)
Equation (2) divided by Equation (1)
tan θ = 0.096629
∴ θ = 5.5193°
or
θ = 5.52°
Substitution into Equation (1)
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 360-N force to a steel bar as
shown. If two workers attempt to move the same rock by applying a force
at A and a parallel force at C, determine these two forces so that they will
be equivalent to the single 360-N force shown in the figure.
SOLUTION
ΣF : R = FA + FC
Have
− ( 360 N ) sin 40° i − ( 360 N ) cos 40° j = − ( FA + FC ) sin θ  i − ( FA + FC ) cosθ  j
From
i:
( 360 N ) sin 40° = ( FA + FC ) sin θ
(1)
j:
( 360 N ) cos 40° = ( FA + FC ) cosθ
(2)
Dividing Equation (1) by Equation (2),
tan 40° = tan θ
∴ θ = 40°
Substituting θ = 40° into Equation (1),
FA + FC = 360 N
Have
where
ΣM C : rB/C × R = rA/C × FA
rB/C = ( 0.35 m )( −cos30°i + sin 30° j) = − ( 0.30311 m ) i + ( 0.175 m ) j
(3)
PROBLEM 3.143 CONTINUED
R = ( 360 N )( −sin40°i − cos 40° j) = − ( 231.40 N ) i − ( 275.78 N ) j
rA/C = ( 0.75 m )( −cos30°i + sin 30 j) = − ( 0.64952 m ) i + ( 0.375 m ) j
FA = FA ( − sin 40°i − 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
or
FA = 168.0 N
Substituting into Equation (3),
FC = 360 − 168.002 = 191.998 N
or
FC = 192.0 N
or FA = 168.0 N
50°
FC = 192.0 N
50°
PROBLEM 3.144
A force and a couple are applied as shown to the end of a cantilever
beam. (a) Replace this system with a single force F applied at point C,
and determine the distance d from C to a line drawn through points D and
E. (b) Solve part a if the directions of the two 360-N forces are reversed.
SOLUTION
(a)
(a) Have
ΣF : F = ( 360 N ) j − ( 360 N ) j − ( 600 N ) k
or F = − ( 600 N ) k
and
ΣM D :
( 360 N )( 0.15 m ) = ( 600 N )( d )
∴ d = 0.09 m
or d = 90.0 mm below ED
F = − ( 600 N ) k
(b) Have from part a
(b)
and
ΣM 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
(
)
W = mg = 80 kg 9.81 m/s 2 = 784.8 N
(a) By definition
ΣM E : M E = ( 784.8 N )( 0.25 m )
Have
∴ 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
ΣM E : 196.2 N ⋅ m = FB ( 0.98615 m )
∴ FB = 198.954 N
and
 0.85 m 
θ = tan −1 
 = 59.534°
 0.5 m 
or FB = 199.0 N
59.5°
PROBLEM 3.146
A crate of mass 80 kg is held in the position shown. Determine (a) the
moment produced by the weight W of the crate about E, (b) the smallest
force applied at A which creates a moment of equal magnitude and
opposite sense about E, (c) the magnitude, sense, and point of application
on the bottom of the crate of the smallest vertical force which creates a
moment of equal magnitude and opposite sense about E.
SOLUTION
(
)
W = mg = 80 kg 9.81 m/s 2 = 784.8 N
(a) By definition
ΣM E : M E = ( 784.8 N )( 0.25 m )
Have
∴ 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
ΣM E : 196.2 N ⋅ m = FA ( 0.61033 m )
∴ FA = 321.47 N
and
 0.35 m 
θ = tan −1 
 = 34.992°
 0.5 m 
or FA = 321 N
35.0°
(c) The smallest force acting on the bottom of the crate resulting in a
moment about E will be located at the point on the bottom of the
crate farthest from E and acting perpendicular to line CED. The
sense of the force will be such as to produce a counterclockwise
moment about E. A force acting vertically upward at D satisfies
these conditions.
PROBLEM 3.146 CONTINUED
Have
ΣM 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 x-axis due to the two cable forces can be found
using the z-components of each force acting at their intersection with the
xy-plane (A and D). The x-components of the forces are parallel to the xaxis, 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.
ΣM x :
Have
(TAB ) z
where
(TAB ) z ( y A ) + (TDE ) z ( yD ) = M x
= k ⋅ TAB = k ⋅ (TABλ AB )

 −i − 12 j + 12k  
= k ⋅  255 lb 
  = 180 lb
17



(TDE ) z
= k ⋅ TDE = k ⋅ (TDE λ 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 x-axis due to the two cable forces can be found
using the z components of each force acting at the intersection with the xy
plane (A and D). The x components of the forces are parallel to the x axis,
and the y components of the forces intersect the x axis. Therefore, neither
the x or y components produce a moment about the x axis.
ΣM x :
Have
(TAB ) z
where
(TAB ) z ( y A ) + (TDE ) z ( yD ) = M x
= k ⋅ TAB = k ⋅ (TABλ AB )

 −i − 12 j + 12k  
= k ⋅ 306 lb 
  = 216 lb
17



(TDE ) z
= k ⋅ TDE = k ⋅ (TDE λ 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 force-couple
system with an equivalent force-couple system at A knowing that
R = 21.2 lb and M = 13.25 lb ⋅ ft.
SOLUTION
ΣF : R = R A = Rλ BC
Have
λ BC =
where
∴ RA =
( 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
ΣM A : rC/ A × R + M = M A
Have
where
rC/ A = ( 42 in.) i + ( 48 in.) k =
1
( 42i + 48k ) ft
12
= ( 3.5 ft ) i + ( 4.0 ft ) k
R = ( 8.40 lb ) i − (19.20 lb ) j − ( 3.20 lb ) k
M = −λ BC M
=
−42i + 96 j + 16k
(13.25 lb ⋅ ft )
106
= − ( 5.25 lb ⋅ ft ) i + (12 lb ⋅ ft ) j + ( 2 lb ⋅ ft ) k
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 60-N forces are applied to a lever as shown. Determine the
moment of the couple formed by the two forces (a) by resolving each
force into horizontal and vertical components and adding the moments of
the two resulting couples, (b) by using the perpendicular distance
between the two forces, (c) by summing the moments of the two forces
about point A.
SOLUTION
(a) Have
where
ΣM 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
ΣM A : Σ ( 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 32-lb motor is mounted on the floor. Find the resultant of the weight
and the forces exerted on the belt, and determine where the line of action
of the resultant intersects the floor.
SOLUTION
ΣF :
Have
( 60 lb ) i − ( 32 lb ) j + (140 lb )( cos 30°i + sin 30°j) = R
∴ R = (181.244 lb ) i + ( 38.0 lb ) j
or R = 185.2 lb
11.84°
ΣM O : ΣM O = xRy
Have
∴
− (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 θ = 25°, M x = −61 lb ⋅ ft, and
M z = −43 lb ⋅ ft, determine θ and d.
SOLUTION
Have
ΣM O : rA/O × F = M O
where
rA/O = − ( 4 in.) i + (11 in.) j − ( d ) k
F = F ( cosθ cos φ i − sin θ j + cosθ sin φ k )
F = 70 lb, θ = 25°
For
F = ( 70 lb ) ( 0.90631cos φ ) i − 0.42262 j + ( 0.90631sin φ ) k 
∴ MO
i
j
k
= ( 70 lb )
−4
11
−d
in.
−0.90631cos φ −0.42262 0.90631sin φ
= ( 70 lb ) ( 9.9694sin φ − 0.42262d ) i + ( −0.90631d cos φ + 3.6252sin φ ) j
+ (1.69048 − 9.9694 cos φ ) k  in.
and
M x = ( 70 lb )( 9.9694sin φ − 0.42262d ) in. = − ( 61 lb ⋅ ft )(12 in./ft )
(1)
M y = ( 70 lb )( −0.90631d cos φ + 3.6252sin φ ) in.
(2)
M z = ( 70 lb )(1.69048 − 9.9694cos φ ) in. = −43 lb ⋅ ft (12 in./ft )
(3)
PROBLEM 3.152 CONTINUED
From Equation (3)
 634.33 
φ = cos −1 
 = 24.636°
 697.86 
or φ = 24.6°
From Equation (1)
 1022.90 
d =
 = 34.577 in.
 29.583 
or d = 34.6 in.
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
ΣM O : rA/O × F = M O
where
rA/O = − ( 4 in.) i + (11 in.) j − ( 27 in.) k
F = F ( cosθ cos φ i − sin θ j + cosθ sin φ k )
∴ MO
i
j
k
= F
−4
11
−27 lb ⋅ in.
cosθ cos φ − sin θ cosθ sin φ
= F (11cosθ sin φ − 27sin θ ) i + ( −27 cosθ cos φ + 4cosθ sin φ ) j
+ ( 4sin θ − 11cosθ cos φ ) k  ( lb ⋅ in.)
and
M x = F (11cosθ sin φ − 27sin θ )( lb ⋅ in.)
(1)
M y = F ( −27 cosθ cos φ + 4cosθ sin φ )( lb ⋅ in.)
(2)
M z = F ( 4sin θ − 11cosθ cos φ )( lb ⋅ in.)
(3)
Now, Equation (1)
cosθ sin φ =
1  Mx

+ 27sin θ 

11  F

(4)
and
cosθ cos φ =
1
Mz 
 4sin θ −

F 
11 
(5)
Equation (3)
Substituting Equations (4) and (5) into Equation (2),
1
1 M
M 

  
M y = F −27   4sin θ − z   + 4   x + 27sin θ   
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