PROBLEM 3.2
A 20–lb force is applied to the control rod AB as shown. Knowing that the length
of the rod is 9 in. and that the moment of the force about B is 120 lb ∙ in. clockwise,
determine the value of α.
SOLUTION
Free–Body Diagram of Rod AB:
25
αθ
=−°
(20 lb)cosQ
θ
=
and
( )(9 in.)
B
MQ=
Therefore,
120 lb-in. (20 lb)(cos )(9 in.)
120 lb-in.
cos 180 lb-in.
θ
θ
=
=
or
48.190
θ
= °
Therefore,
23.2
α
= °
consent of McGraw–Hill Education.
PROBLEM 3.3
A 300-N force P is applied at Point A of the bell crank shown.
(a) Compute the moment of the force P about O by resolving
it into horizontal and vertical components. (b) Using the
result of part (a), determine the perpendicular distance from O
to the line of action of P.
SOLUTION
(0.2 m)cos40
0.153209 m
(0.2 m)sin 40
0.128558 m
x
y
= °
=
= °
=
/
(0.153209 m) (0.128558 m)
AO
∴= +r ij
(a)
(300 N)sin30
150 N
(300 N)cos30
259.81 N
x
y
F
F
= °
=
= °
=
(150 N) (259.81 N)= +Fi j
/
(0.153209 0.128558 ) m (150 259.81 ) N
(39.805 19.2837 ) N m
(20.521 N m)
O AO
= ×
= + ×+
=−⋅
= ⋅
Mr F
i j ij
kk
k
20.5 N m
O
= ⋅M
(b)
O
M Fd=
20.521 N m (300 N)( )
0.068403 m
d
d
⋅=
=
68.4 mmd=
PROBLEM 3.4
A 400–N force P is applied at Point A of the bell crank shown.
(a) Compute the moment of the force P about O by resolving it
into components along line OA and in a direction perpendicular
to that line. (b) Determine the magnitude and direction of the
smallest force Q applied at B that has the same moment as P
about O.
SOLUTION
PROBLEM 3.5
A 300–N force is applied at A as shown. Determine
(a) the moment of the 300–N force about D, (b) the
smallest force applied at B that creates the same moment
about D.
SOLUTION
(a)
(300 N)cos 25
271.89 N
(300 N)sin 25
126.785 N
(271.89 N) (126.785 N)
x
y
F
F
= °
=
= °
=
= +Fi j
(0.1m) (0.2 m)
[ (0.1m) (0.2 m) ] [(271.89 N) (126.785 N) ]
(12.6785 N m) (54.378 N m)
(41.700 N m)
D
D
DA==−−
= ×
=−− × +
=− ⋅+ ⋅
= ⋅
r ij
M rF
M ij i j
kk
k
FFF
41.7 N m
D= ⋅M
DB
FFF
at 45°
()
41.700 N m (0.28284 m)
D
Q DB
Q
=
⋅=
MFFF
147.4 NQ=
45.0°
PROBLEM 3.6
A 300–N force is applied at A as shown. Determine
(a) the moment of the 300–N force about D, (b) the
magnitude and sense of the horizontal force applied at C
that creates the same moment about D, (c) the smallest
force applied at C that creates the same moment about D.
SOLUTION
(a) See solution to prob. 3.5 for figure and analysis leading to the determination of MD
41.7 N m
D
= ⋅M
(b) Since C is horizontal
C=Ci
(0.2 m) (0.125 m)
(0.125 m)
41.7 N m (0.125 m)( )
333.60 N
D
DC
CC
C
C
= = −
=×=
⋅=
=
r ij
Mri k
C
334 NC=
22
0.125 m
tan 0.2 m
32.0
( ); (0.2 m) (0.125 m)
0.23585 m
D
C DC DC
a
a
=
= °
= = +
=
M
41.70 N m (0.23585 m)C⋅=
176.8 N=C
58.0°
consent of McGraw–Hill Education.
PROBLEM 3.7
The tailgate of a car is supported by the hydraulic lift BC. If
the lift exerts a 125–lb force directed along its centerline on
the ball and socket at B, determine the moment of the force
about A.
SOLUTION
First note
22
(12.0 in.) (2.33 in.)
12.2241in.
CB
d= +
=
Then
12.0 in.
cos 12.2241in.
2.33 in.
sin 12.2241in.
θ
θ
=
=
and
cos sin
125 lb [(12.0 in.) (2.33 in.) ]
12.2241in.
CB CB CB
FF
θθ
= −
= −
F ij
ij
Now
/A B A CB
= ×Mr F
where
/(15.3 in.) (12.0 in. 2.33 in.)
(15.3 in.) (14.33 in.)
BA= −+
= −
ri j
ij
Then
125 lb
[(15.3 in.) (14.33 in.) ] (12.0 2.33 )
12.2241in.
(1393.87 lb in.)
A
=−× −
= ⋅
M i j ij
k
(116.156 lb ft)= ⋅ k
or
116.2 lb ft
A
= ⋅M
consent of McGraw–Hill Education.
PROBLEM 3.8
The tailgate of a car is supported by the hydraulic lift BC. If
the lift exerts a 125–lb force directed along its centerline on
the ball and socket at B, determine the moment of the force
about A.
SOLUTION
First note
22
(17.2 in.) (7.62 in.)
18.8123 in.
CB
d= +
=
Then
17.2 in.
cos 18.8123 in.
7.62 in.
sin 18.8123 in.
θ
θ
=
=
and
( cos ) ( sin )
125 lb [(17.2 in.) (7.62 in.) ]
18.8123 in.
CB CB CB
FF
θθ
= −
= +
F ij
ij
Now
/A B A CB
= ×Mr F
where
/(20.5 in.) (4.38 in.)
BA= −
r ij
Then
125 lb
[(20.5 in.) (4.38 in.) ] (17.2 7.62 )
18.8123 in.
A
=−× −M i j ij
(1538.53 lb in.)
(128.2 lb ft)
= ⋅
= ⋅
k
k
or
128.2 lb ft
A
= ⋅M
consent of McGraw–Hill Education.
PROBLEM 3.9
It is known that the connecting rod AB exerts on the crank BC a 500–lb force
directed down and to the left along the centerline of AB. Determine the moment of
the force about C.
SOLUTION
Using (a):
11
() ()
7 24
(2.24 in.) 500 lb (1.68 in.) 500 lb
25 25
1120 lb in.
C AB x AB y
M yF xF= +
  
= ×+ ×
  
  
= ⋅
(a)
1.120 kip in.
C
= ⋅M
Using (b):
2()
7
(8 in.) 500 lb
25
1120 lb in.
C AB x
M yF=

= ×


= ⋅
(b)
1.120 kip in.
C= ⋅M
PROBLEM 3.10
It is known that the connecting rod AB exerts on the crank BC a 500–lb force directed
down and to the left along the centerline of AB. Determine the moment of the force
about C.
SOLUTION
Using (a):
11
() ()
7 24
(2.24 in.) 500 lb (1.68 in.) 500 lb
25 25
492.8 lb in.
C AB x AB y
M yF xF=−+
  
=− ×+ ×
  
  
=+⋅
(a)
493 lb in.
C
= ⋅M
Using (b):
2()
7
(3.52 in.) 500 lb
25
492.8 lb in.
C AB x
M yF=

= ×


=+⋅
(b)
493 lb in.
C
= ⋅M
PROBLEM 3.2
A 20–lb force is applied to the control rod AB as shown. Knowing that the length
of the rod is 9 in. and that the moment of the force about B is 120 lb ∙ in. clockwise,
determine the value of α.
SOLUTION
Free–Body Diagram of Rod AB:
25
αθ
=−°
(20 lb)cosQ
θ
=
and
( )(9 in.)
B
MQ=
Therefore,
120 lb-in. (20 lb)(cos )(9 in.)
120 lb-in.
cos 180 lb-in.
θ
θ
=
=
or
48.190
θ
= °
Therefore,
23.2
α
= °
consent of McGraw–Hill Education.
PROBLEM 3.3
A 300-N force P is applied at Point A of the bell crank shown.
(a) Compute the moment of the force P about O by resolving
it into horizontal and vertical components. (b) Using the
result of part (a), determine the perpendicular distance from O
to the line of action of P.
SOLUTION
(0.2 m)cos40
0.153209 m
(0.2 m)sin 40
0.128558 m
x
y
= °
=
= °
=
/
(0.153209 m) (0.128558 m)
AO
∴= +r ij
(a)
(300 N)sin30
150 N
(300 N)cos30
259.81 N
x
y
F
F
= °
=
= °
=
(150 N) (259.81 N)= +Fi j
/
(0.153209 0.128558 ) m (150 259.81 ) N
(39.805 19.2837 ) N m
(20.521 N m)
O AO
= ×
= + ×+
=−⋅
= ⋅
Mr F
i j ij
kk
k
20.5 N m
O
= ⋅M
(b)
O
M Fd=
20.521 N m (300 N)( )
0.068403 m
d
d
⋅=
=
68.4 mmd=
PROBLEM 3.4
A 400–N force P is applied at Point A of the bell crank shown.
(a) Compute the moment of the force P about O by resolving it
into components along line OA and in a direction perpendicular
to that line. (b) Determine the magnitude and direction of the
smallest force Q applied at B that has the same moment as P
about O.
SOLUTION
PROBLEM 3.5
A 300–N force is applied at A as shown. Determine
(a) the moment of the 300–N force about D, (b) the
smallest force applied at B that creates the same moment
about D.
SOLUTION
(a)
(300 N)cos 25
271.89 N
(300 N)sin 25
126.785 N
(271.89 N) (126.785 N)
x
y
F
F
= °
=
= °
=
= +Fi j
(0.1m) (0.2 m)
[ (0.1m) (0.2 m) ] [(271.89 N) (126.785 N) ]
(12.6785 N m) (54.378 N m)
(41.700 N m)
D
D
DA==−−
= ×
=−− × +
=− ⋅+ ⋅
= ⋅
r ij
M rF
M ij i j
kk
k
FFF
41.7 N m
D= ⋅M
DB
FFF
at 45°
()
41.700 N m (0.28284 m)
D
Q DB
Q
=
⋅=
MFFF
147.4 NQ=
45.0°
PROBLEM 3.6
A 300–N force is applied at A as shown. Determine
(a) the moment of the 300–N force about D, (b) the
magnitude and sense of the horizontal force applied at C
that creates the same moment about D, (c) the smallest
force applied at C that creates the same moment about D.
SOLUTION
(a) See solution to prob. 3.5 for figure and analysis leading to the determination of MD
41.7 N m
D
= ⋅M
(b) Since C is horizontal
C=Ci
(0.2 m) (0.125 m)
(0.125 m)
41.7 N m (0.125 m)( )
333.60 N
D
DC
CC
C
C
= = −
=×=
⋅=
=
r ij
Mri k
C
334 NC=
22
0.125 m
tan 0.2 m
32.0
( ); (0.2 m) (0.125 m)
0.23585 m
D
C DC DC
a
a
=
= °
= = +
=
M
41.70 N m (0.23585 m)C⋅=
176.8 N=C
58.0°
consent of McGraw–Hill Education.
PROBLEM 3.7
The tailgate of a car is supported by the hydraulic lift BC. If
the lift exerts a 125–lb force directed along its centerline on
the ball and socket at B, determine the moment of the force
about A.
SOLUTION
First note
22
(12.0 in.) (2.33 in.)
12.2241in.
CB
d= +
=
Then
12.0 in.
cos 12.2241in.
2.33 in.
sin 12.2241in.
θ
θ
=
=
and
cos sin
125 lb [(12.0 in.) (2.33 in.) ]
12.2241in.
CB CB CB
FF
θθ
= −
= −
F ij
ij
Now
/A B A CB
= ×Mr F
where
/(15.3 in.) (12.0 in. 2.33 in.)
(15.3 in.) (14.33 in.)
BA= −+
= −
ri j
ij
Then
125 lb
[(15.3 in.) (14.33 in.) ] (12.0 2.33 )
12.2241in.
(1393.87 lb in.)
A
=−× −
= ⋅
M i j ij
k
(116.156 lb ft)= ⋅ k
or
116.2 lb ft
A
= ⋅M
consent of McGraw–Hill Education.
PROBLEM 3.8
The tailgate of a car is supported by the hydraulic lift BC. If
the lift exerts a 125–lb force directed along its centerline on
the ball and socket at B, determine the moment of the force
about A.
SOLUTION
First note
22
(17.2 in.) (7.62 in.)
18.8123 in.
CB
d= +
=
Then
17.2 in.
cos 18.8123 in.
7.62 in.
sin 18.8123 in.
θ
θ
=
=
and
( cos ) ( sin )
125 lb [(17.2 in.) (7.62 in.) ]
18.8123 in.
CB CB CB
FF
θθ
= −
= +
F ij
ij
Now
/A B A CB
= ×Mr F
where
/(20.5 in.) (4.38 in.)
BA= −
r ij
Then
125 lb
[(20.5 in.) (4.38 in.) ] (17.2 7.62 )
18.8123 in.
A
=−× −M i j ij
(1538.53 lb in.)
(128.2 lb ft)
= ⋅
= ⋅
k
k
or
128.2 lb ft
A
= ⋅M
consent of McGraw–Hill Education.
PROBLEM 3.9
It is known that the connecting rod AB exerts on the crank BC a 500–lb force
directed down and to the left along the centerline of AB. Determine the moment of
the force about C.
SOLUTION
Using (a):
11
() ()
7 24
(2.24 in.) 500 lb (1.68 in.) 500 lb
25 25
1120 lb in.
C AB x AB y
M yF xF= +
  
= ×+ ×
  
  
= ⋅
(a)
1.120 kip in.
C
= ⋅M
Using (b):
2()
7
(8 in.) 500 lb
25
1120 lb in.
C AB x
M yF=

= ×


= ⋅
(b)
1.120 kip in.
C= ⋅M
PROBLEM 3.10
It is known that the connecting rod AB exerts on the crank BC a 500–lb force directed
down and to the left along the centerline of AB. Determine the moment of the force
about C.
SOLUTION
Using (a):
11
() ()
7 24
(2.24 in.) 500 lb (1.68 in.) 500 lb
25 25
492.8 lb in.
C AB x AB y
M yF xF=−+
  
=− ×+ ×
  
  
=+⋅
(a)
493 lb in.
C
= ⋅M
Using (b):
2()
7
(3.52 in.) 500 lb
25
492.8 lb in.
C AB x
M yF=

= ×


=+⋅
(b)
493 lb in.
C
= ⋅M