PRO
OBLEM 2.1
1
Twoo forces are applied
a
as shoown to a hoook. Determinee graphically the
magnnitude and dirrection of theiir resultant using (a) the paarallelogram laaw,
(b) thhe triangle rulle.
TION
SOLUT
(a)
Parallelogram law:
l
(b)
T
Triangle
rule:
W measure:
We
R = 1391 kN, α = 477.8°
R = 1391 N
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47.8°
PROBLEM 2.3
Two structural members B and C are bolted to bracket A. Knowing that
both members are in tension and that P = 10 kN and Q = 15 kN,
determine graphically the magnitude and direction of the resultant force
exerted on the bracket using (a) the parallelogram law, (b) the triangle
rule.
SOLUTION
(a)
Parallelogram law:
(b)
Triangle rule:
R = 20.1 kN,
We measure:
α = 21.2°
R = 20.1 kN
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21.2°
PROBLEM 2.7
A telephone cable is clamped at A to the pole AB. Knowing that the
tension in the right-hand portion of the cable is T2 = 1000 lb, determine
by trigonometry (a) the required tension T1 in the left-hand portion if
the resultant R of the forces exerted by the cable at A is to be vertical,
(b) the corresponding magnitude of R.
SOLUTION
Using the triangle rule and the law of sines:
(a)
75° + 40° + β = 180°
β = 180° − 75° − 40°
= 65°
(b)
T1
1000 lb
=
sin 75° sin 65°
T1 = 938 lb
1000 lb
R
=
sin 75° sin 40°
R = 665 lb
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P
PROBLEM
2.11
A steel tank is to be positionned in an excavvation. Knowiing
thhat α = 20°, determine
d
by trigonometry (a) the requirred
m
magnitude
of the force P if
i the resultannt R of the two
t
fo
forces
applied at A is to be vertical,
v
(b) thhe correspondiing
m
magnitude
of R.
R
SOLUT
TION
Using thhe triangle rulee and the law of sines:
(a)
β + 50° + 60° = 180°
β = 180° − 50° − 60°
= 70°
(b)
4425 lb
P
=
ssin 70° sin 60°
P = 392 lb
4 lb
425
R
=
ssin 70° sin 50°
R = 346 lb
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PROBL
LEM 2.22
Determinne the x and y components of each of the forces shown..
TION
SOLUT
Computte the followin
ng distances:
OA = (600) 2 + (800) 2
= 1000 mm
m
OB = (560) 2 + (900) 2
= 1060 mm
m
OC = (480) 2 + (900) 2
= 1020 mm
m
F
800-N Force:
F
424-N Force:
F
408-N Force:
Fx = +(800 N)
N
800
1000
Fx = +640 N
Fy = +(800 N)
N
600
1000
Fy = +480 N
Fx = −(424 N)
N
560
1060
Fx = −224 N
Fy = −(424 N)
N
900
1060
Fy = −360 N
Fx = +(408 N)
N
480
1020
Fx = +192.0 N
Fy = −(408 N)
N
900
1020
Fy = −360 N
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PROBLE
EM 2.31
Determine the
t resultant of
o the three forrces of Problem
m 2.21.
PROBLEM
M 2.21 Determ
mine the x andd y componennts of each of the
forces show
wn.
TION
SOLUT
Componnents of the fo
orces were deteermined in Prooblem 2.21:
Force
F
x Comp. (lb)
y Compp. (lb)
29
2 lb
+21.0
+200.0
50
5 lb
–14.00
+48.0
51
5 lb
+24.0
–45.0
Rx = +31.0
Ry = + 23.0
R = Rx i + Ry j
= (31.0 lb)i + (23.0 lb) j
tann α =
Ry
Rx
23.0
31.0
α = 36.573°
=
R=
23.0 lbb
sin (36.5773°)
= 38.601 lbb
R = 38.66 lb
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36.6°
PR
ROBLEM 2.36
Knoowing that the tension in roppe AC is 365 N,
N determine the
resuultant of the thhree forces exeerted at point C of post BC.
TION
SOLUT
Determiine force comp
ponents:
Cable foorce AC:
500-N Force:
F
200-N Force:
F
and
960
= −2440 N
1460
1100
Fy = −(365 N)
= −2775 N
N
1460
Fx = −(365 N)
N
24
= 480 N
25
7
Fy = (500 N))
= 140 N
25
Fx = (500 N))
4
= 160 N
5
3
Fy = −(200 N)
N = −120 N
5
Fx = (200 N))
Rx = ΣFx = −240 N + 480 N + 160 N = 400
4 N
Ry = ΣFy = −275 N + 140 N − 120 N = −255 N
R = Rx2 + Ry2
= (400 N)
N 2 + (−255 N
N) 2
= 474.37 N
Further:
255
400
α = 32.5°
tan α =
R = 4744 N
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32.5°
PROBLEM 2.37
7
Knowiing that α = 40°, determiine the resultant of the thhree
forces shown.
SOLUT
TION
60-lb Foorce:
Fx = (60 llb) cos 20° = 56.382 lb
Fy = (60 llb)sin 20° = 200.521 lb
80-lb Foorce:
Fx = (80 lb)
l cos 60° = 40.000 lb
Fy = (80 lb)sin
l
60° = 699.282 lb
120-lb Force:
F
Fx = (1200 lb) cos30° = 103.923
1
lb
Fy = −(1220 lb)sin 30° = −60.000 lb
and
Rx = ΣFx = 200.305 lb
Ry = ΣFy = 29.803 lb
(
lb)2
R = (2000.305 lb)2 + (29.803
= 202..510 lb
Further:
tan α =
29.803
200.305
α = tan −1
= 8.46°
29.803
200.305
R = 2033 lb
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8.46°
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
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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
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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
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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
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