248
Ans:
4–21.
SOLUTION
F
18 in.
5 in.
30
In order to pull out the nail at B, the force
F
exerted on the
handle of the hammer must produce a clockwise moment of
500 lb # in. about point A. Determine the required magnitude
of force F.
249
Ans:
SOLUTION
=4 rad, r=10 mm
When u
=3 rad,
4–22.
Old clocks were constructed using a fusee B to drive the
gears and watch hands. The purpose of the fusee is to
increase the leverage developed by the mainspring A
as it uncoils and thereby loses some of its tension. The
mainspring can develop a torque (moment) Ts
=
k
u
,
where k
=
0.015 N
#
m
>
rad is the torsional stiffness and
u
is the angle of twist of the spring in radians. If the torque
Tf developed by the fusee is to remain constant as the
mainspring winds down, and
x=10
mm when
u
=
4 rad,
determine the required radius of the fusee when
u
=
3 rad. Tf
A
Ts
y
x
y
t
B
12 mm
x
250
Ans:
4–23.
The tower crane is used to hoist the 2-Mg load upward at
constant velocity.The 1.5-Mg jib BD, 0.5-Mg jib BC, and
6-Mg counterweight Chave centers of mass at G1,G2, and
G3,respectively. Determine the resultant moment produced
by the load and the weights of the tower crane jibs about
point Aand about point B.
SOLUTION
Since the moment arms of the weights and the load measured to points Aand Bare
C
BD
G2
G3
9.5m
7.5 m
4 m
G1
12.5 m
23 m
251
Ans:
*4–24.
The tower crane is used to hoist a 2-Mg load upward at con-
stant velocity.The 1.5-Mg jib BD and 0.5-Mg jib BC have
centers of mass at G1and G2,respectively. Determine the
required mass of the counterweight Cso that the resultant
moment produced by the load and the weight of the tower
crane jibs about point Ais zero.The center of mass for the
counterweight is located at G3.
SOLUTION
C
BD
G2
G3
A
9.5m
7.5 m
4 m
G1
12.5 m
23 m
252
4–25.
If the 1500-lb boom AB, the 200-lb cage BCD, and the 175-lb
man have centers of gravity located at points G1,G2and G3,
respectively, determine the resultant moment produced by
each weight about point A.
SOLUTION
Moment of the weight of boom AB about point A:
75
B
C
D
20 ft
10 ft
G1
G2
G3
1.75 ft
2.5 ft
A
253
Ans:
4–26.
If the 1500-lb boom AB, the 200-lb cage BCD, and the
175-lb man have centers of gravity located at points G1,G2
and G3,respectively, determine the resultant moment pro-
duced by all the weights about point A.
SOLUTION
Referring to Fig. a, the resultant moment of the weight about point Ais given by
75
B
C
D
20 ft
10 ft
G1
G2
G3
1.75 ft
2.5 ft
A
254
Ans:
SOLUTION
4–27.
Determine the moment of the force F about point O.
Express the result as a Cartesian vector. F fi {–6i + 4 j 8k} kN
4 m
3 m
6 m
2 m
1 m
Oy
z
P
A
255
SOLUTION
*4–28.
Determine the moment of the force F about point P.
Express the result as a Cartesian vector. F fi {–6i + 4 j 8k} kN
4 m
3 m
6 m
2 m
1 m
Oy
z
P
A
256
SOLUTION
Position Vector. The coordinates of point B are
B(8, 0.25, 1.5) ft
.
4–29.
The force
F={400i–100j–700k} lb
acts at the end of
the beam. Determine the moment of this force about
pointO.
1.5 ft
x
8 ft
0.25 ft
z
A
O
B
F
y
257
Ans:
SOLUTION
Position Vector. The coordinates of points A and B are
A (0, 0, 1.5) ft
and
B (8, 0.25, 1.5) ft
, respectively. Thus,
4–30.
The force
F={400i–100j–700k} lb
acts at the end of
the beam. Determine the moment of this force about
pointA.
1.5 ft
x
8 ft
0.25 ft
z
A
O
B
F
y
258
Ans:
SOLUTION
Position Vector. The coordinates of points A and P are
A (3, 3, –1) m
and
4–31.
Determine the moment of the force F about point P.
Express the result as a Cartesian vector.
2 m
1 m
3 m
3 m
3 m
2 m
A
O
P
x
y
z
259
Ans:
SOLUTION
*4–32.
The pipe assembly is subjected to the force of
F={600i+800j–500k} N
. Determine the moment of
this force about point A.
y
0.5 m
0.4 m
0.3 m
0.3 m
x
z
B
A
260
Ans:
SOLUTION
Position Vector. The coordinates of points B and C are
B (0.5, 0, 0) m
and
C (0.5, 0.7, –0.3) m,
respectively. Thus,
4–33.
The pipe assembly is subjected to the force of
F={600i+800j–500k} N
. Determine the moment of
this force about point B.
y
0.5 m
0.4 m
0.3 m
0.3 m
x
z
B
C
A
261
SOLUTION
Position Vectors And Force Vector. The coordinates of points A, B and C are
The Moment of Force F About Point A.
MA=rAB *F
OR
MA=rAC *F
4–34.
Determine the moment of the force of F
=
600 N about
point A.
4 m
4 m
z
x
y
6 m
6 m
A
C
B
F
45fi
262
SOLUTION
Position Vectors And Force Vector. The coordinates of points A, B and C are
Thus,
The Moment of Force F About Point A.
MA=rAB *F
OR
MA=rAC *F
The magnitude of
MA
is
4–35.
Determine the smallest force F that must be applied along
the rope in order to cause the curved rod, which has a radius
of 4 m, to fail at the support A. This requires a moment of
M=1500 N #m
to be developed at A.
4 m
4 m
z
x
y
6 m
6 m
A
C
B
F
45fi
263
Ans:
SOLUTION
Position And Force Vectors. The coordinates of point A are
A (0.4, 0.5, –0.3) m
.
Thus,
Moment of F About Point O. To produce zero moment about point O, the line of
action of F must pass through point O. Thus, F must directed from O to A (direction
defined by
uOA
). Thus,
OR F must directed from A to O (direction defined by
uAO
). Thus
*4–36.
Determine the coordinate direction angles
a,
b
,
g
of force
F, so that the moment of F about O is zero.
x
F
0.4 m
A
z
Oy
0.5 m
0.3 m
264
Ans:
SOLUTION
Position And Force Vectors. The coordinates of point A are
A (0.4, 0.5, –0.3) m
.
Thus
Moment of F About Point O.
MO=rOA *F
4–37.
Determine the moment of force F about point O. The force
has a magnitude of 800 N and coordinate direction angles of
a
=60°,
b
=120°,
g
=45°.
Express the result as a
Cartesian vector.
x
F
0.4 m
A
z
Oy
0.5 m
0.3 m
265
SOLUTION
Position Vectors And Force Vector. The coordinates of points A, C and D are
Moment of F About Point A.
MA=rAC *F
OR
MA=rAD *F
4–38.
Determine the moment of the force F about the door hinge
at A. Express the result as a Cartesian vector.
5 ft
1.5 ft
1.5 ft
7 ft 4 ft
z
C
A
BD
F fi 80 lb
45fi
SOLUTION
Position Vectors And Force Vector. The coordinates of points B, C and D are
B (–1.5, –3, 0) ft, C [0, –(3 +4 cos 45°), 4 sin 45°] ft and D (–5, 0, 0) ft,
respectively.
Thus,
Moment of F About Point B.
MB=rBC *F
or
MB=rBD *F
4–39.
Determine the moment of the force F about the door hinge
at B. Express the result as a Cartesian vector.
5 ft
1.5 ft
1.5 ft
3 ft
7 ft 4 ft
z
C
A
BD
F fi 80 lb
45fi
*4–40.
The curved rod has a radius of 5 ft. If a force of 60 lb acts at
its end as shown, determine the moment of this force about
point C.
SOLUTION
Position Vector and Force Vector:
Moment of Force About Point C: Applying Eq. 4–7, we have
FAB
5ft
5ft
60°
z
y
6ft
60 lb
A
C
B