PROBLEM 6.126
Solve Problem 6.125 when (a)
(b)
6 .
0,
PROBLEM 6.125 The control rod CE passes
through a horizontal hole in the body of the
toggle system shown. Knowing that link BD
is 250 mm long, determine the force Q
required to hold the system in equilibrium
when
20 .
SOLUTION
We note that BD is a two-force member.
(a)
Free body: Member ABC
0:
BD
Since
MC
35 mm
;
250 mm
250 mm, sin
8.048
0: (100 N)(350 mm) FBD sin (200 mm) 0
FBD 1250 N
Fx
0: FBD cos
Cx
0
(1250 N)(cos 8.048 ) Cx
0
Cx 1237.7 N
Member CE:
Fx
0: (1237.7 N) Q 0
Q 1238 N
Q 1237.7 N
(b)
Free body: Member ABC
6
Dimensions in mm
Since
BD
250 mm,
sin
1 14.094
mm
250 mm
3.232
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PROBLEM 6.126 (Continued)
MC
0: ( FBD sin )198.90 ( FBD cos )20.906 (100 N)348.08 0
FBD [198.90sin 3.232
20.906cos3.232 ] 34808
FBD
Fx
0: FBD cos
Cx
0
(1084.8 N)cos3.232
Cx
0
Cx
Member DE:
Fx
1084.8 N
1083.1 N
0: Q Cx
Q 1083.1 N
Q 1083 N
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PROBLEM 6.131
Arm ABC is connected by pins to a collar at B and to crank CD at
C. Neglecting the effect of friction, determine the couple M
required to hold the system in equilibrium when
0.
SOLUTION
Free body: Member ABC
Fx
0: Cx
240 N 0
Cx
MC
240 N
0: (240 N)(500 mm) B(160 mm) 0
B
Fy
0: C y
750 N
Cy
750 N
0
750 N
Free body: Member CD
MD
0: M
(750 N)(300 mm) (240 N)(125 mm) 0
M
195 103 N mm
M 195.0 kN m
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PROBLEM 6.142
A log weighing 800 lb is lifted by a pair of tongs as shown.
Determine the forces exerted at E and F on tong DEF.
SOLUTION
FBD AB:
By symmetry:
Ay
and
By
400 lb
Ax
Bx
6
(400 lb)
5
480 lb
D
Note:
so
FBD DEF:
MF
B
Dx
480 lb
Dy
400 lb
(10.5 in.)(400 lb) (15.5 in.)(480 lb) (12 in.)Ex
Ex
Fx
0:
970 lb
480 lb 970 lb Fx
Fx
Fy
0: 400 lb Fy
Fy
E 970 lb
0
490 lb
0
400 lb
F
633 lb
39.2
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0