Unlock document.
This document is partially blurred.
Unlock all pages and 1 million more documents.
Get Access
PROBLEM 4.76
Solve Problem 4.75, assuming that the 170-N force applied at B is horizontal
and directed to the left.
PROBLEM 4.75
Rod AB is supported by a pin and bracket at A and rests
against a frictionless peg at C. Determine the reactions at A and C when a
170-N vertical force is applied at B.
PROBLEM 4.77
Member ABC is supported by a pin and bracket at B and by an
inextensible cord attached at A and C and passing over a
frictionless pulley at D. The tension may be assumed to be the
same in portions AD and CD of the cord. For the loading shown
and neglecting the size of the pulley, determine the tension in the
cord and the reaction at B.
Reaction at B must pass through D.
7 in.
tan 12 in.
30.256
7 in.
tan 24 in.
16.26
Force Triangle
Law of sines:
72 lb
sin59.744 sin13.996 sin106.26
(sin13.996 ) ( 72 lb)(sin 59.744°)
(0.24185) ( 72)(0.86378)
TT B
TT
TT
100.00 lbT
100.0 lbT
sin 106.26°
(100 lb) sin59.744
111.14 lb
B
111.1 lbB
30.3
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.78
Using the method of Section 4.7, solve Problem 4.22.
PROBLEM 4.22
A lever AB is hinged at C and attached to a
control cable at A. If the lever is subjected to a 500-N horizontal
force at B, determine (a) the tension in the cable, (b) the
reaction at C.
Reaction at C must pass through E, where
F
AD
and 500-N force intersect.
Since 250 mm,AC CD triangle
ACD
is isosceles.
We have 90 30 120C
and
1(180 120 ) 30
2
AD
Dimensions in mm
On the other hand, from triangle BCF:
( )sin30 200 sin 30 100 mm
250 100 150 mm
CF BC
FD CD CF
From triangle EFD, and since 30 :D
( )tan30 150tan30 86.60 mmEF FD
From triangle EFC:
100 mm
tan 86.60 mm
49.11
CF
EF
Force triangle
Law of sines
500 N
sin49.11 sin60 sin 70.89°
400 N, 458 N
AD
AD
FC
FC
(a)
400 N
AD
F
(b) 458 N
C
49.1°
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.79
Knowing that
30, determine the reaction (a) at B, (b) at C.
Reaction at C must pass through D where force
P
and reaction at B intersect.
In CDE: (3 1)
tan
31
36.2
R
R
Force Triangle
Law of sines: sin 23.8 sin126.2 sin30
2.00
1.239
PBC
BP
CP
(a)
2PB
60.0°
(b)
1.239PC
36.2°
PROBLEM 4.80
Knowing that
60, determine the reaction (a) at B, (b) at C.
In CDE:
3
tan
1
13
22.9
R
Force Triangle
Law of sines: sin52.9 sin67.1 sin60
1.155
1.086
PBC
BP
CP
(a)
1.155PB
30.0°
(b)
1.086PC
22.9°
PROBLEM 4.81
Determine the reactions at A and B when 50°.
150 mm
tan 933.01mm
9.1333
AB
BD
Force Triangle
Law of sines:
100 N
sin9.1333° sin15 sin155.867
163.1 N; 257.6 N
AB
AB
163.1 N
A
74.1° 258 N
B
65.0°
PROBLEM 4.82
Determine the reactions at A and B when
80.
PROBLEM 4.82 (Continued)
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.83
Rod AB is bent into the shape of an arc of circle and is lodged between two
pegs D and E. It supports a load
P
at end B. Neglecting friction and the weight
of the rod, determine the distance c corresponding to equilibrium when a 20
mm and R 100 mm.
PROBLEM 4.84
A slender rod of length L is attached to collars that can slide freely along
the guides shown. Knowing that the rod is in equilibrium, derive an
expression for the angle
in terms of the angle
.
