PROBLEM 2.43
Two cables are tied together at C and are loaded as shown.
Determine the tension (a) in cable AC, (b) in cable BC.
SOLUTION
Free-Body Diagram
Force Triangle
Law of sines:
TAC
T
400 lb
= BC =
sin 60° sin 40° sin 80°
(a)
TAC =
400 lb
(sin 60°)
sin 80°
TAC = 352 lb
(b)
TBC =
400 lb
(sin 40°)
sin 80°
TBC = 261 lb
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
P
PROBLEM
2
2.48
Knowing that α = 20°, determine the tension (a) in cable
K
AC
C, (b) in rope BC.
TION
SOLUT
Freee-Body Diagraam
Force Trianggle
TAC
T
1200 lb
= BC =
sinn 110° sin 5° sin 65°
s
Law of sines:
(a )
TAC =
12000 lb
sin 110°
sin 65
6 °
TAC = 1244 lb
(b )
TBC =
1200 lb
sin 5°
sin 65
6 °
TBC = 115.4 lb
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 2.49
9
Two cables are tied togethher at C annd are loadded as show
wn.
Know
wing that P = 300 N, deetermine the tension in cables
c
AC and
a
BC.
TION
SOLUT
Free-B
Body Diagram
m
ΣFx = 0
− TCA sin 30 + TCB sin 30 − P coss 45° − 200N = 0
For P = 200N we have,
−0.5TCAA + 0.5TCB + 2112.13 − 200 = 0 (1)
ΣFy = 0
TCA coos30° − TCB cos30 − P sin 45 = 0
0.8
86603TCA + 0.886603TCB − 2112.13 = 0 (2))
multaneously gives,
Solvinng equations (1) and (2) sim
TCA = 134.6 N
TCB = 110.4 N
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PR
ROBLEM 2.67
A 600-lb
6
crate is
i supported by
b several roppeandd-pulley arranggements as shhown. Determine
for each arrangeement the tenssion in the rope.
(Seee the hint for Problem 2.66.)
TION
SOLUT
Free-Boody Diagram of Pulley
(a)
ΣFy = 0: 2T − (600 lb) = 0
T=
1
(600 lb)
2
T = 300 lb
(b)
ΣFy = 0: 2T − (600 lb) = 0
T=
1
(600 lb)
2
T = 300 lb
(c)
ΣFy = 0: 3T − (600 lb) = 0
1
T = ((600 lb)
3
T = 200 lb
(d )
ΣFy = 0: 3T − (600 lb) = 0
1
T = ((600 lb)
3
T = 200 lb
(e)
ΣFy = 0: 4T − (600 lb) = 0
T=
1
(600 lb)
4
T = 150.0 lb
Copyright © McGraw-Hill Education. Permission required for reproduction or display.