PROBLEM 3.
76
Knowing that P = 210 N, replace the three couples with a single
equivalent couple, specifying its magnitude and the direction of its axis.
SOLUTION
Have
1 2 P
= + + M M M M
where ( ) ( )
1 / 1
0.96 0.40 0 40 N m 96 N m
0 0 100
C B C
= = = +
i j k
M r P i j
( ) ( ) ( )
2 / 2
0 0.20 0.55 22.0 N m 52.8 N m 19.2 N m
96 0 110
D A E
= = = + +
i j k
M r P i j k
(See Solution to Problem 3.73.)
( ) ( )
/
0.48 0.20 1.10 231 N m 100.8 N m
0 210 0
P E A E
= = = +
i j k
M r P i k
( ) ( ) ( ) 40 22 231 96 52.8 19.2 100.8 N m ( = + + + + + +
M i j k
( ) ( ) ( ) 293 N m 148.8 N m 120 N m = + + i j k
( ) ( ) ( )
2 2 2
2 2 2
293 148.8 120 349.84 N m
x y z
M M M = + + = + + = M
or 350 N m M =
293 148.8 120
0.83752 0.42533 0.34301
349.84
+ +
= = = + +
M i j k
i j k
M
cos 0.83752 33.121
x x
= = or 33.1
x
=
cos 0.42533 64.828
y y
= = or 64.8
y
=
cos 0.34301 69.940
z z
= = or 69.9
z
=
PROBLEM 3.77
In a manufacturing operation, three holes are drilled simultaneously in a
workpiece. Knowing that the holes are perpendicular to the surfaces of
the workpiece, replace the couples applied to the drills with a single
equivalent couple, specifying its magnitude and the direction of its axis.
SOLUTION
Have
1 2 3
= + + M M M M
where ( )( )
1
1.1 lb ft cos 25 sin 25 = + M j k
( )
2
1.1 lb ft = M j
( )( )
3
1.3 lb ft cos 20 sin 20 = M j k
( ) ( ) 0.99694 1.1 1.22160 0.46488 0.44463 = + + M j k
( ) ( ) 3.3185 lb ft 0.020254 lb ft = j k
and ( ) ( ) ( )
2 2 2
2 2 2
0 3.3185 0.020254
x y z
M M M = + + = + + M
3.3186 lb ft =
or 3.32 lb ft M =
( ) 0 3.3185 0.020254
3.3186
= =
i j k M
M
0.99997 0.0061032 = j k
cos 0
x
= 90
x
= or 90.0
x
=
cos 0.99997
y
= 179.555
y
= or 179.6
y
=
cos 0.0061032 90.349
z z
= = or 90.3
z
=
PROBLEM 3.78
The tension in the cable attached to the end C of an adjustable boom ABC
is 1000 N. Replace the force exerted by the cable at C with an equivalent
force-couple system (a) at A, (b) at B.
SOLUTION
(a) Based on : 1000 N
A
F F T = =
or 1000 N
A
= F 20
( )( ) : sin50
A A A
M M T d =
( ) ( ) 1000 N sin50 2.25 m =
1723.60 N m =
or 1724 N m
A
= M
(b) Based on : 1000 N
B
F F T = =
or 1000 N
B
= F 20
( )( ) : sin50
B B B
M M T d =
( ) ( ) 1000 N sin50 1.25 m =
957.56 N m =
or 958 N m
B
= 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 : 20 lb
B
F P P = =
or 20 lb
B
= P
:
B B
M M Pd =
( ) 20 lb 5 in. =
100 lb in. =
or 100 lb in.
B
= M
(b) If the two vertical forces are to be equivalent to ,
B
M they must be a
couple. Further, the sense of the moment of this couple must be
counterclockwise.
Then, with
C
P and
D
P acting as shown,
:
D C
M M P d =
( ) 100 lb in. 4 in.
C
P =
25 lb
C
P =
or 25 lb
C
= P
: 0
y D C
F P P =
25 lb
D
P =
or 25 lb
D
= P
PROBLEM 3.80
A 700-N force P is applied at point A of a structural member. Replace P
with (a) an equivalent force-couple system at C, (b) an equivalent system
consisting of a vertical force at B and a second force at D.
SOLUTION
(a) Based on : 700 N
C
F P P = =
or 700 N
C
= P 60
:
C C x Cy y Cx
M M P d P d = +
where ( ) 700 N cos60 350 N
x
P = =
( ) 700 N sin60 606.22 N
y
P = =
1.6 m
Cx
d =
1.1 m
Cy
d =
( )( ) ( )( ) 350 N 1.1 m 606.22 N 1.6 m
C
M = +
385 N m 969.95 N m = +
584.95 N m =
or 585 N m
C
= M
(b) Based on : cos60
x Dx
F P P =
( ) 700 N cos60 =
350 N =
( )( ) ( ) : cos 60
D DA B DB
M P d P d =
( ) ( ) ( ) 700 N cos60 0.6 m 2.4 m
B
P ( =
87.5 N
B
P =
or 87.5 N
B
= P
PROBLEM 3.80 CONTINUED
: sin60
y B Dy
F P P P = +
( ) 700 N sin60 87.5 N
Dy
P = +
518.72 N
Dy
P =
( ) ( )
2
2
D Dx Dy
P P P = +
( ) ( )
2 2
350 518.72 625.76 N = + =
1 1
518.72
tan tan 55.991
350
Dy
Dx
P
P
| |
| |
= = =
|
|
\ .
\ .
or 626 N
D
P = 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
( ) : 240 N cos30 cos cos
x B C
F F F =
or ( ) ( ) cos 240 N cos30
B C
F F + = (1)
( ) : 240 N sin30 sin sin
y B C
F F F = +
or ( ) ( ) sin 240 N sin30
B C
F F + = (2)
From
( )
Equation (2)
: tan tan30
Equation 1
=
30 =
Based on
( ) ( ) ( ) ( )( ) : 240 N cos 30 20 0.25 m cos10 0.60 m
C B
M F ( =
100 N
B
F =
or 100.0 N
B
= F 30
From Equation (1), ( ) 100 N cos30 240cos30
C
F + =
140 N
C
F =
or 140.0 N
C
= F 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
(a) Based on ( ) : 240 N cos30 cos30
x C
F F =
240 N
C
F =
or 240 N
C
= F 30
( ) ( ) : 240 N cos10 0.25 m
C A C A
M d M d ( = =
59.088 N m
C
M =
or 59.1 N m
C
= M
(b) Based on ( ) : 240 N sin30 sin
y B
F F =
or sin 120
B
F = (1)
( ) ( ) ( ) : 59.088 N m 240 N cos10 cos 20
B C C C
M d F d ( =
( ) ( ) ( ) 59.088 N m 240 N cos10 0.60 m 0.60 m cos 20
C
F ( ( =
0.56382 82.724
C
F =
146.722 N
C
F =
or 146.7 N
C
= F
and ( ) : 240 N cos30 146.722 N cos
x B
F F =
cos 61.124
B
F = (2)
From
Equation (1)
:
Equation (2)
120
tan 1.96323
61.124
= =
63.007 = or 63.0 =
From Equation (1),
120
134.670 N
sin63.007
B
F = =
or 134.7 N
B
= F 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 with the vertical.
Then for equivalence,
( ) : 250 lb sin30 sin sin
x A B
F F F = + (1)
( ) : 250 lb cos30 cos cos
y A B
F F F = (2)
Dividing Equation (1) by Equation (2),
( )
( )
( )
( )
250 lb sin30 sin
250 lb cos30 cos
A B
A B
F F
F F
+
=
+
Simplifying yields 30 =
Based on
( ) ( ) ( )( ) : 250 lb cos30 12 ft cos30 32 ft
C A
M F ( =
93.75 lb
A
F =
or 93.8 lb
A
= F 60
Based on
( ) ( ) ( )( ) : 250 lb cos30 20 ft cos30 32 ft
A C
M F ( =
156.25 lb
C
F =
or 156.3 lb
C
= F 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 60 lb, P = 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)
(b)
(c)
(a) Based on
: 50 lb 50 lb 60 lb
z A
F F + + =
60 lb
A
F =
( ) or 60.0 lb
A
= F k
Based on
( )( ) ( )( ) : 50 lb 2 ft 50 lb 0.6 ft
A A
M M =
70 lb ft
A
M =
( ) or 70.0 lb ft
A
= M j
(b) Based on
: 50 lb 50 lb 60 lb
z
F F + + =
60 lb F =
( ) or 60.0 lb = F k
Based on
( ) : 70 lb ft 60 lb
A
M x =
1.16667 ft x =
or 1.167 ft from along x A AB =
(c) Based on
( ) ( ) ( ) ( ) ( ) ( ) : 50 lb 1 ft 50 lb 2.4 ft 3 ft 0
B
M P R + =
70
23.333 lb
3
P = =
or 23.3 lb P =
