PROBLEM 3.70
Two 80-N forces are applied as shown to the corners B and D of
a rectangular plate. (a) Determine the moment of the couple
formed by the two forces by resolving each force into horizontal
and vertical components and adding the moments of the two
resulting couples. (b) Use the result obtained to determine the
perpendicular distance between lines BE and DF.
SOLUTION
(a)
Resolving forces into components:
P (80 N)sin 50
61.284 N
Q (80 N)cos50
M
(51.423 N)(0.5 m) (61.284 N)(0.3 m)
7.3263 N m
M
(b)
51.423 N
7.33 N m
Distance between lines BE and DF
M
or
Fd
7.3263 N m=(80 N)d
d
0.091579 m
d
91.6 mm
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PROBLEM 3.72
Four 1 12 -in.-diameter pegs are attached to a board as shown.
Two strings are passed around the pegs and pulled with the
forces indicated. (a) Determine the resultant couple acting on
the board. (b) If only one string is used, around which pegs
should it pass and in what directions should it be pulled to
create the same couple with the minimum tension in the string?
(c) What is the value of that minimum tension?
SOLUTION
M
(a)
(60 lb)(10.5 in.) (40 lb)(13.5 in.)
630 lb in. 540 lb in.
M
(b)
1170 lb in.
With only one string, pegs A and D, or B and C should be used. We
have
tan
9
12
36.9
90
53.1
Direction of forces:
(c)
With pegs A and D:
53.1
With pegs B and C:
53.1
The distance between the centers of the two pegs is
122
92
15 in.
Therefore, the perpendicular distance d between the forces is
d
15 in. 2
3
in.
4
16.5 in.
We must have
M
Fd 1170 lb in. F (16.5 in.)
F
70.9 lb
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PROBLEM 3.82
The tension in the cable attached to the end
C of an adjustable boom ABC is 560 lb.
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
F : FA
T
or
560 lb
FA
M A: M A
560 lb
20.0°
(T sin 50 )(d A )
(560 lb)sin 50 (18 ft)
7721.7 lb ft
or
(b)
MA
Based on
F : FB
T
or
560 lb
FB
M B: MB
7720 lb ft
560 lb
20.0°
(T sin 50 )(d B )
(560 lb)sin 50°(10 ft)
4289.8 lb ft
or
MB
4290 lb ft
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