CHAPTER 2
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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
47.8°
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PROBLEM 2.2
2
Twoo forces are applied as shown
s
to a bracket
b
support. Determiine
grapphically the magnitude and directioon of their resultant
r
usiing
(a) the paralleloogram law, (b)
( the trianggle rule.
TION
SOLUT
(a)
Parallelogram law:
l
(b)
T
Triangle
rule:
W measure:
We
R = 9006 lb,
α = 26.6°
2
R = 9066 lb
26.6°
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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:
We measure:
R = 20.1 kN,
α = 21.2°
R = 20.1 kN
21.2°
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PROBLEM 2.4
Two structural members B and C are bolted to bracket A. Knowing that
both members are in tension and that P = 6 kips and Q = 4 kips,
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:
We measure:
R = 8.03 kips, α = 3.8°
R = 8.03 kips
3.8°
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PROBLEM 2.5
A stake is being pulled out of the ground by means of two ropes as shown.
Knowing that α = 30°, determine by trigonometry (a) the magnitude of the
force P so that the resultant force exerted on the stake is vertical, (b) the
corresponding magnitude of the resultant.
SOLUTION
Using the triangle rule and the law of sines:
(a)
(b)
120 N
P
=
sin 30° sin 25°
P = 101.4 N
30° + β + 25° = 180°
β = 180° − 25° − 30°
= 125°
120 N
R
=
sin 30° sin125°
R = 196.6 N
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PROBLEM 2.6
A telephone cable is clamped at A to the pole AB. Knowing that the
tension in the left-hand portion of the cable is T1 = 800 lb, determine
by trigonometry (a) the required tension T2 in the right-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° + α = 180°
α = 180° − 75° − 40°
= 65°
(b)
T2
800 lb
=
sin 65° sin 75°
T2 = 853 lb
800 lb
R
=
sin 65° sin 40°
R = 567 lb
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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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PROBLEM 2.8
A disabled automobile is pulled by means of two
ropes as shown. The tension in rope AB is 2.2 kN,
and the angle α is 25°. Knowing that the resultant
of the two forces applied at A is directed along the
axis of the automobile, determine by trigonometry
(a) the tension in rope AC, (b) the magnitude of the
resultant of the two forces applied at A.
SOLUTION
Using the law of sines:
TAC
2.2 kN
R
=
=
sin 30° sin125° sin 25D
TAC = 2.603 kN
R = 4.264 kN
(a)
TAC = 2.60 kN
(b)
R = 4.26 kN
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PROBLEM 2.9
A disabled automobile is pulled by means of two
ropes as shown. Knowing that the tension in rope
AB is 3 kN, determine by trigonometry the tension
in rope AC and the value of α so that the resultant
force exerted at A is a 4.8-kN force directed along
the axis of the automobile.
SOLUTION
Using the law of cosines:
TAC 2 = (3 kN)2 + (4.8 kN)2 − 2(3 kN)(4.8 kN) cos 30°
TAC = 2.6643 kN
Using the law of sines:
sin α
sin 30°
=
3 kN 2.6643 kN
α = 34.3°
TAC = 2.66 kN
34.3°
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PROBLEM
M 2.10
Two forces are
a applied as shown to a hoook support. Knowing
K
that the
magnitude of
o P is 35 N, determine
d
by trigonometry (a) the requirred
angle α if thee resultant R of
o the two forcces applied to the support iss to
be horizontal, (b) the correesponding maggnitude of R.
TION
SOLUT
Using thhe triangle rulee and law of sines:
(a)
siin α sin 25°
=
5 N
50
35 N
s α = 0.603774
sin
α = 37.1388°
(b)
α = 37.1°
α + β + 25° = 180°
β = 180° − 25° − 37.138°
= 117.8622°
R
35 N
=
sin1177.862° sin 25°
2
R = 73.2 N
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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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P
PROBLEM
2.12
A steel tank is to be positionned in an excavvation. Knowiing
thhat the maggnitude of P is 500 lb,, determine by
trrigonometry (a) the required angle α if thhe resultant R of
thhe two forces applied at A is to be vertical,
v
(b) the
c
corresponding
magnitude off R.
SOLUT
TION
Using thhe triangle rulee and the law of sines:
(a)
(α + 330°) + 60° + β = 180°
β = 180° − (α + 30°) − 60°
β = 90° − α
sin (90° − α ) sin 60°
=
500 lb
425 lb
90° − α = 47.402°
(b)
R
500 lb
=
sin (42.598° + 300°) sin 60°
α = 42.6°
R = 551 lb
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P
PROBLEM
2.13
A steel tank is to be positioned
p
in an excavation.
D
Determine
byy trigonometrry (a) the magnitude and
a
d
direction
of thee smallest forcce P for whichh the resultantt R
o the two forces
of
f
appliedd at A is vertical,
v
(b) the
c
corresponding
magnitude off R.
TION
SOLUT
The smaallest force P will
w be perpenndicular to R.
(a)
P = (425 lb) co
os 30°
(b)
R = (425 lb)sin
n 30°
P = 368 lb
R = 21
13 lb
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P
PROBLEM
2.14
For the hook support
F
s
of Proob. 2.10, deterrmine by trigoonometry (a) the
m
magnitude
andd direction of the
t smallest force
fo
P for whhich the resulttant
R of the twoo forces appliied to the suupport is horrizontal, (b) the
corresponding magnitude off R.
SOLUT
TION
The smaallest force P will
w be perpenndicular to R.
(a)
P = (50 N)sin 25°
(b)
R = (50 N) cos 25°
P = 21.1 N
R = 45.3 N
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PROBLEM 2.15
For the hook support shown, determine by trigonometry the
magnitude and direction of the resultant of the two forces applied
to the support.
SOLUTION
Using the law of cosines:
R 2 = (200 lb) 2 + (300 lb) 2
− 2(200 lb)(300 lb) cos (45D + 65°)
R = 413.57 lb
Using the law of sines:
sin α
sin (45D + 65°)
=
300 lb
413.57 lb
α = 42.972°
β = 90D + 25D − 42.972°
R = 414 lb
72.0°
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PROBLEM 2.16
Solve Prob. 2.1 by trigonometry.
PROBLEM 2.1
Two forces are applied as shown to a hook. Determine graphically the
magnitude and direction of their resultant using (a) the parallelogram law,
(b) the triangle rule.
SOLUTION
Using the law of cosines:
R 2 = (900 N) 2 + (600 N )2
− 2(900 N )(600 N) cos (135°)
R = 1390.57N
Using the law of sines:
sin(α − 30D ) sin (135°)
=
600N
1390.57N
D
α − 30 = 17.7642°
α = 47.764D
R = 1391N
47.8°
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PROBLEM 2.17
2
Soolve Problem 2.4
2 by trigonoometry.
PR
ROBLEM 2.44 Two structuural members B and C are bolted
b
to bracket
A.. Knowing thaat both membbers are in tennsion and thatt P = 6 kips and
a
Q = 4 kips, deetermine graphhically the maagnitude and direction of the
resultant force exerted on thhe bracket usinng (a) the parrallelogram laaw,
r
(bb) the triangle rule.
TION
SOLUT
Using thhe force triang
gle and the law
ws of cosines and
a sines:
We have:
γ = 1800° − (50° + 25°)
= 1055°
Then
R 2 = (4 kips) 2 + (6 kipps) 2 − 2(4 kips)(6 kips) cos1105°
= 64.423 kips 2
R = 8.00264 kips
And
4 kipps
8.02644 kips
=
sin(25° + α )
sin105°
sin(25° + α ) = 0.481337
25° + α = 28.7755°
α = 3.775°
R = 8.03 kips
k
3.8°
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PROBLEM 2.18
For the stake of Prob. 2.5, knowing that the tension in one rope is 120 N,
determine by trigonometry the magnitude and direction of the force P so
that the resultant is a vertical force of 160 N.
PROBLEM 2.5 A stake is being pulled out of the ground by means of two
ropes as shown. Knowing that α = 30°, determine by trigonometry (a) the
magnitude of the force P so that the resultant force exerted on the stake is
vertical, (b) the corresponding magnitude of the resultant.
SOLUTION
Using the laws of cosines and sines:
P 2 = (120 N) 2 + (160 N) 2 − 2(120 N)(160 N) cos 25°
P = 72.096 N
And
sin α
sin 25°
=
120 N 72.096 N
sin α = 0.70343
α = 44.703°
P = 72.1 N
44.7°
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PROBLEM 2.19
Two forces P and Q are applied
a
to the lid of a storagge bin as show
wn.
Knowing thhat P = 48 N and Q = 60 N,
N determine by trigonomeetry
the magnituude and directiion of the resuultant of the tw
wo forces.
SOLUT
TION
Using thhe force triang
gle and the law
ws of cosines and
a sines:
We havee
γ = 180° − (220° + 10°)
= 150°
Then
and
R 2 = (48 N) 2 + (60 N) 2
− 2(48 N)(60
N
N) cos1550°
R = 104.366 N
48 N 104.366 N
=
sin150°
sin α
sinn α = 0.22996
α = 13.2947°
Hence:
φ = 180° − α − 80°
= 180° − 133.2947° − 80°
= 86.705°
R = 104.4 N
86.7°°
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P
PROBLEM
M 2.20
Two forces P and Q are appplied to the lid
T
l of a storagge bin as show
wn.
K
Knowing
that P = 60 N andd Q = 48 N, determine
d
by trigonometry the
m
magnitude
andd direction of the resultant of
o the two forcces.
SOLUT
TION
Using thhe force triang
gle and the law
ws of cosines and
a sines:
We havee
γ = 180° − (20° + 10°)
= 150°
Then
and
R 2 = (60 N)2 + (448 N) 2
− 2(60 N)(488 N) cos 150°
R = 104.366 N
60 N 104.366 N
=
sin α
sin150°
sin α = 0.28745
α = 16.7054°
Hence:
φ = 180° − α − 180°
= 180° − 16.7054° − 80°
= 83.2295°
R = 104.44 N
83.3°
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PROB
BLEM 2.21
Determiine the x and y components of each of thee forces shownn.
TION
SOLUT
Computte the followin
ng distances:
OA = (84) 2 + (80) 2
= 116 in..
OB = (28)2 + (96) 2
= 100 in..
OC = (48)2 + (90) 2
= 102 in..
29-lb Foorce:
50-lb Foorce:
51-lb Foorce:
Fx = +(29 lbb)
84
116
Fx = +21.0 lb
Fy = +(29 lbb)
80
116
Fy = +20.0 lb
Fx = −(50 lbb)
28
100
Fx = −14.00 lb
Fy = +(50 lbb)
96
100
Fy = + 48.0 lb
Fx = +(51 lbb)
48
102
Fx = +24.0 lb
Fy = −(51 lbb)
90
102
Fy = −45.0 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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PROBLEM 2.23
Determine the x and y components of each of the forces shown.
SOLUTION
80-N Force:
120-N Force:
150-N Force:
Fx = +(80 N)cos 40°
Fx = 61.3 N
Fy = +(80 N)sin 40°
Fy = 51.4 N
Fx = +(120 N)cos70°
Fx = 41.0 N
Fy = +(120 N)sin 70°
Fy = 112.8 N
Fx = −(150 N)cos35°
Fx = −122. 9 N
Fy = +(150 N)sin 35°
Fy = 86.0 N
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PROBLEM 2.24
Determine the x and y components of each of the forces shown.
SOLUTION
40-lb Force:
50-lb Force:
60-lb Force:
Fx = +(40 lb)cos60°
Fx = 20.0 lb
Fy = −(40 lb)sin 60°
Fy = −34.6 lb
Fx = −(50 lb)sin 50°
Fx = −38.3 lb
Fy = −(50 lb)cos50°
Fy = −32.1 lb
Fx = +(60 lb)cos 25°
Fx = 54.4 lb
Fy = +(60 lb)sin 25°
Fy = 25.4 lb
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PROBLEM 2.25
Member BC exerts on member AC a force P directed along line BC. Knowing
that P must have a 325-N horizontal component, determine (a) the magnitude
of the force P, (b) its vertical component.
SOLUTION
BC = (650 mm) 2 + (720 mm) 2
= 970 mm
⎛ 650 ⎞
Px = P ⎜
⎟
⎝ 970 ⎠
(a)
or
⎛ 970 ⎞
P = Px ⎜
⎟
⎝ 650 ⎠
⎛ 970 ⎞
= 325 N ⎜
⎟
⎝ 650 ⎠
= 485 N
P = 485 N
(b)
⎛ 720 ⎞
Py = P ⎜
⎟
⎝ 970 ⎠
⎛ 720 ⎞
= 485 N ⎜
⎟
⎝ 970 ⎠
= 360 N
Py = 970 N
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PRO
OBLEM 2.26
6
Membber BD exertss on memberr ABC a forcce P directed along line BD.
B
Know
wing that P muust have a 300--lb horizontal component, determine
d
(a) the
t
magniitude of the foorce P, (b) its vertical
v
compoonent.
TION
SOLUT
P sin 35° = 300 lb
(a)
P=
(b)
V
Vertical
compo
onent
300 lb
sin 35°
P = 523 lb
Pv = P cos35°
= (523 lb) cos 355°
Pv = 428 lb
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PR
ROBLEM 2.27
2
Thhe hydraulic cyylinder BC exxerts on membber AB a forcee P
dirrected along liine BC. Know
wing that P muust have a 6000-N
com
mponent perppendicular to member
m
AB, determine
d
(a) the
maagnitude of thee force P, (b) its componentt along line AB
B.
TION
SOLUT
180° = 45° + α + 90° + 300°
α = 180° − 45° − 90° − 30°
= 15°
(a)
Px
P
P
P= x
coss α
6000 N
=
coss15°
= 6211.17 N
cos α =
P = 621 N
(b)
tan α =
Py
Px
Py = Px tan
t α
= (6000 N) tan15°
= 1600.770 N
Py = 160.8 N
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PRO
OBLEM 2.2
28
Cablee AC exerts on
o beam AB a force P directed along linne AC. Knowiing
that P must have a 350-lb verticcal componentt, determine (aa) the magnituude
of thee force P, (b) its horizontal component.
TION
SOLUT
(a)
P=
=
Py
cos 55°
350 lb
cos 55°
= 610.21 lb
(b)
P = 610 lb
Px = P sin 555°
= (610.211 lb)sin 55°
= 499.85 lb
Px = 500 lb
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P
PROBLEM
2
2.29
Thhe hydraulic cylinder BD exerts
e
on mem
mber ABC a force P directed
allong line BD
D. Knowing that P mustt have a 750-N component
peerpendicular too member ABC, determine (a) the magnitude of the force
P, (b) its compoonent parallel to ABC.
TION
SOLUT
(a)
750 N = P sinn 20°
P = 21922.9 N
(b)
P = 2190 N
PABC = P coss 20°
= (2192.9 N) cos 20°°
PABC
= 2060 N
A
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PROBLEM 2.30
The guy wire BD exerts on the telephone pole AC a force P directed along
BD. Knowing that P must have a 720-N component perpendicular to the pole
AC, determine (a) the magnitude of the force P, (b) its component along line
AC.
SOLUTION
(a)
37
Px
12
37
=
(720 N)
12
= 2220 N
P=
P = 2.22 kN
(b)
35
Px
12
35
= (720 N)
12
= 2100 N
Py =
Py = 2.10 kN
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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
36.6°
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PROBLE
EM 2.32
Determine the resultant of
o the three forrces of Probleem 2.23.
M 2.23 Determ
mine the x andd y componennts of each of the
PROBLEM
forces show
wn.
TION
SOLUT
Componnents of the fo
orces were dettermined in Prroblem 2.23:
Force
y Com
mp. (N)
x Comp. (N)
80 N
+61.3
+
+51.4
120 N
+41.0
+1112.8
150 N
–122.9
+
+86.0
Rx = −20.6
Ry = + 250.2
2
R = Rx i + R y j
= (−20..6 N)i + (250.22 N)j
tan α =
Ry
Rx
250.22 N
20.66 N
tan α = 12.14456
α = 85.2993°
tan α =
R=
2500.2 N
sin 855.293°
R = 251 N
85.3°°
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PROBL
LEM 2.33
Determine the resultantt of the three forces
f
of Problem 2.24.
PROBLE
EM 2.24 Deteermine the x and y componnents of eachh of
the forcess shown.
TION
SOLUT
Force
x Comp. (lb)
y Comp. (lb)
40
4 lb
+20.00
–34.664
50
5 lb
–38.30
–32.114
60
6 lb
+54.38
+25.336
Rx = +36.08
Ry = −41.442
R = Rx i + Ry j
= (+36.08 lb)i + (−41.42 lbb)j
tan α =
Ry
Rx
41.42 lb
36.08 lb
tan α = 1.14800
tan α =
α = 48.942°
R=
41.42 lb
sin 48.942°
R = 54.99 lb
48.9°
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PROBL
LEM 2.34
Determinee the resultantt of the three forces
f
of Probllem 2.22.
PROBLE
EM 2.22 Deterrmine the x annd y componennts of each of the
forces shoown.
SOLUT
TION
Componnents of the fo
orces were determined in Problem 2.22:
Force
x Comp. (N)
(
y Comp. (N)
80
00 lb
+640
+4800
42
24 lb
–224
–3600
40
08 lb
+192
–3600
Rx = +608
Ry = −2400
R = Rx i + Ry j
= (608 lb)i + (−240 lb) j
Ry
tann α =
Rx
240
608
α = 21.541°
240 N
R=
sin(21.5441°)
= 653.65 N
=
R = 6544 N
21.5°
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PRO
OBLEM 2.35
5
Know
wing that α = 35°, determiine the resulttant of the thhree
forcess shown.
SOLUT
TION
Fx = +(100 N) cos35° = +81.915 N
100-N Force:
F
Fy = −(100 N)sin 35° = −57.358 N
Fx = +(150 N) cos 65° = +63.393 N
150-N Force:
F
Fy = −(150 N)sin 65° = −135.946 N
Fx = −(200 N) cos35° = −163.830 N
200-N Force:
F
Fy = −(200 N)sin 35° = −114.715 N
Force
x Com
mp. (N)
N)
y Comp. (N
100 N
+81.915
−57.3558
150 N
+63.393
−135.9446
200 N
−163.830
−114.7115
Rx = −18.522
Ry = −308.022
R = Rx i + Ry j
= (−18.5222 N)i + (−3088.02 N) j
taan α =
Ry
Rx
308.02
18.522
α = 86.559°
=
R=
308.02 N
sin 86.559
R = 3099 N
86.6°
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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
32.5°
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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
8.46°
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PRO
OBLEM 2.3
38
Know
wing that α = 75°, determ
mine the resulttant of the three
forcees shown.
TION
SOLUT
60-lb Foorce:
Fx = (60 lb) coos 20° = 56.3882 lb
Fy = (60 lb)siin 20° = 20.521 lb
80-lb Foorce:
Fx = (80 lb) coos 95° = −6.97725 lb
Fy = (80 lb)sin 95° = 79.6966 lb
120-lb Force:
F
Fx = (120 lb) cos
c 5° = 119.5443 lb
Fy = (120 lb)ssin 5° = 10.4599 lb
Then
Rx = ΣFx = 1668.953 lb
Ry = ΣFy = 1110.676 lb
and
R = (168.9553 lb)2 + (110..676 lb)2
= 201.976 lb
110.676
168.953
tann α = 0.65507
α = 33.228°
tann α =
R = 2022 lb
33.2°
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PROBLEM 2.39
For the collar of Problem 2.35, determine (a) the required value
of α if the resultant of the three forces shown is to be vertical,
(b) the corresponding magnitude of the resultant.
SOLUTION
Rx = ΣFx
= (100 N) cos α + (150 N) cos (α + 30°) − (200 N) cos α
Rx = −(100 N) cos α + (150 N) cos (α + 30°)
(1)
Ry = ΣFy
= −(100 N)sin α − (150 N)sin (α + 30°) − (200 N)sin α
Ry = −(300 N)sin α − (150 N)sin (α + 30°)
(a)
(2)
For R to be vertical, we must have Rx = 0. We make Rx = 0 in Eq. (1):
−100 cos α + 150 cos (α + 30°) = 0
−100 cos α + 150 (cos α cos 30° − sin α sin 30°) = 0
29.904 cos α = 75sin α
29.904
75
= 0.39872
α = 21.738°
tan α =
(b)
α = 21.7°
Substituting for α in Eq. (2):
Ry = −300sin 21.738° − 150sin 51.738°
= −228.89 N
R = | Ry | = 228.89 N
R = 229 N
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PROBLEM 2.40
For the post of Prob. 2.36, determine (a) the required
tension in rope AC if the resultant of the three forces
exerted at point C is to be horizontal, (b) the corresponding
magnitude of the resultant.
SOLUTION
Rx = ΣFx = −
Rx = −
48
TAC + 640 N
73
Ry = ΣFy = −
Ry = −
(a)
960
24
4
(500 N) + (200 N)
TAC +
1460
25
5
1100
7
3
TAC + (500 N) − (200 N)
1460
25
5
55
TAC + 20 N
73
(2)
For R to be horizontal, we must have Ry = 0.
Set Ry = 0 in Eq. (2):
−
55
TAC + 20 N = 0
73
TAC = 26.545 N
(b)
(1)
TAC = 26.5 N
Substituting for TAC into Eq. (1) gives
48
(26.545 N) + 640 N
73
Rx = 622.55 N
Rx = −
R = Rx = 623 N
R = 623 N
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PRO
OBLEM 2.4
41
Deteermine (a) thee required tenssion in cable AC
A , knowing that
t
the resulttant
of thhe three forcess exerted at Point
P
C of booom BC must be
b directed aloong
BC, (b) the correspponding magnnitude of the resultant.
TION
SOLUT
Using thhe x and y axess shown:
Rx = ΣFx = TAC sin10° + (50 lb) cos355° + (75 lb) coos 60°
= TAC sin10° + 78.458 lb
(1)
Ry = ΣFy = (50 lb)sin 35° + (75 lb)sinn 60° − TAC cos10°
Ry = 93.631 lb − TAC cos10°
(a )
(2)
Seet Ry = 0 in Eq.
E (2):
93.631 lb − TAC cos10° = 0
TAC = 95.075 lb
(b )
TAC = 95.1 lb
Suubstituting forr TAC in Eq. (11):
Rx = (95.075 lb) sin10° + 78.4558 lb
= 94.968 lb
R = Rx
R = 95.0 lb
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PRO
OBLEM 2.42
2
For thhe block of Problems
P
2.377 and 2.38, deetermine (a) the
required value of α if the resultannt of the threee forces shownn is
to be parallel
p
to the incline, (b) thhe correspondiing magnitudee of
the ressultant.
SOLUT
TION
Select thhe x axis to bee along a a′.
Then
Rx = ΣFx = (60 lb)
l + (80 lb)coos α + (120 lb) sin α
(1)
Ry = ΣFy = (80 lb)sin
l
α − (1200 lb)cos α
(2)
and
(a )
Seet Ry = 0 in Eq.
E (2).
(80 lbb)sin α − (120 lb) cos α = 0
D
Dividing
each term
t
by cos α gives:
(800 lb) tan α = 1220 lb
1220 lb
8 lb
80
α = 566.310°
tanα =
(b )
α = 56.3°
Suubstituting forr α in Eq. (1)) gives:
Rx = 60 lb + (800 lb)cos56.31°° + (120lb)sinn 56.31° = 204..22 lb
Rx = 204 lb
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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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PROBLEM 2.44
Two cables are tied together at C and are loaded as shown. Knowing
that α = 30°, determine the tension (a) in cable AC, (b) in cable BC.
SOLUTION
Free-Body Diagram
Force Triangle
Law of sines:
TAC
T
6 kN
= BC =
sin 60° sin 35° sin 85°
(a)
TAC =
6 kN
(sin 60°)
sin 85°
TAC = 5.22 kN
(b)
TBC =
6 kN
(sin 35°)
sin 85°
TBC = 3.45 kN
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PROBLEM 2.45
Two cables are tied together at C and loaded as shown.
Determine the tension (a) in cable AC, (b) in cable BC.
SOLUTION
Free-Body Diagram
1.4
4.8
α = 16.2602°
1.6
tan β =
3
β = 28.073°
tan α =
Force Triangle
Law of sines:
TAC
TBC
1.98 kN
=
=
sin 61.927° sin 73.740° sin 44.333°
(a)
TAC =
1.98 kN
sin 61.927°
sin 44.333°
TAC = 2.50 kN
(b)
TBC =
1.98 kN
sin 73.740°
sin 44.333°
TBC = 2.72 kN
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PROBLEM 2.46
Two cabbles are tied together at C and are looaded as show
wn.
Knowingg that P = 5000 N and α = 60°, determinne the tensionn in
(a) in caable AC, (b) inn cable BC.
SOLUT
TION
Free-Bod
dy Diagram
Law of sines:
s
Forrce Triangle
TAC
T
500 N
= BC =
sin 35° ssin 75° sin 700°
(a )
TAC =
5500 N
sin 35°
sin 70°
TAC = 305 N
(b )
TBC =
5500 N
sin 75°
siin 70°
TBC = 514 N
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PRO
OBLEM 2.4
47
Two cables are tieed together at C and are loaaded as shownn. Determine the
tensiion (a) in cable AC, (b) in caable BC.
TION
SOLUT
Free-Bod
dy Diagram
F
Force
Trianggle
W = mg
= (20
00 kg)(9.81 m//s 2 )
= 196
62 N
s
Law of sines:
TAC
TBC
1962 N
A
=
=
sin 15° sin 105° sin 60°
(a )
TAC =
(1962 N
N) sin 15°
sinn 60°
TAC = 586 N
(b )
TBC =
(1962 N)sin
N
105°
sinn 60°
TBC = 2190 N
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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
s
Law of sines:
Force Trianggle
TAC
T
1200 lb
= BC =
sinn 110° sin 5° sin 65°
(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 30D + TCB sin 30D − P coss 45° − 200N = 0
For P = 200N we have,
−0.5TCAA + 0.5TCB + 2112.13 − 200 = 0 (1)
ΣFy = 0
TCA coos30° − TCB cos30D − P sin 45D = 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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PROBLEM 2.50
0
Two cables are tied togethher at C annd are loadded as show
wn.
Deterrmine the rannge of valuees of P for which both cables remaain
taut.
SOLUT
TION
Free-B
Body Diagram
m
ΣFx = 0
− TCA sin 30D + TCB sin 30D − P coss 45° − 200N = 0
For TCA = 0 we have,,
0.5TCB + 0.70711P − 200
2 = 0 (1)
ΣFy = 0
D
TCA cos30
c
° − TCB cos30
c
− P sin 45D = 0 ; agaain setting TCAA = 0 yields,
0.866603TCB − 0.707711P = 0 (2))
Adding equations (1) and (2) givess,
1.36603TCB
410N and P = 179.315N
ce TCB = 146.4
C = 200 henc
Substituuting for TCB = 0 into the eqquilibrium equuations and sollving simultanneously gives,
−0.5TCA + 0.70711P − 200 = 0
0.86603TCA − 0.707111P = 0
0N , P = 6699.20N Thus for
fo both cabless to remain taaut, load P muust be within the
Andd TCA = 546.40
range off 179.315 N an
nd 669.20 N.
179.3 N <P< 669 N
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P
PROBLEM
2
2.51
Tw
wo forces P and
a Q are appplied as show
wn to an aircrraft
coonnection. Knnowing that the connection is
i in equilibriuum
annd that P = 500 lb and Q = 650 lb, determine the
m
magnitudes
of the
t forces exerrted on the rodds A and B.
TION
SOLUT
Freee-Body Diagrram
Resolvinng the forces into
i
x- and y-ddirections:
R = P + Q + FA + FB = 0
Substituuting componeents:
R = −(500 lb) j + [(650 lb) cos 50
5 °]i
− [(650 lb)siin 50°]j
c 50°)i + ( FA sin 50°) j = 0
+ FB i − ( FA cos
In the y-direction
(onee unknown forrce):
−500 lb − (650 lb)sin 50° + FA sin 50° = 0
Thus,
FA =
500 lb + (650 lb)sin 500°
sin 50°
= 1302.70 lb
In the x-direction:
Thus,
FA = 1303 lb
(6550 lb)cos50° + FB − FA cos550° = 0
FB = FA cos500° − (650 lb) cos50
c
°
= (1302.700 lb) cos50° − (650 lb) cos500°
= 419.55 lbb
FB = 420 lb
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P
PROBLEM
M 2.52
Two forces P and Q are appplied as show
T
wn to an aircrraft
c
connection.
Knowing thhat the connnection is in
e
equilibrium
annd that the maagnitudes of thhe forces exerrted
o rods A and
on
a
B are FA = 750 lb annd FB = 400 lb,
d
determine
the magnitudes of
o P and Q.
TION
SOLUT
Free-Boody Diagram
Resolvinng the forces into
i
x- and y-ddirections:
R = P + Q + FA + FB = 0
Substituuting componeents:
R = − Pj + Q cos 500°i − Q sin 50°j
°
− [(750 lb) cos 50°]i
+ [(750 lb)sin 50°]j + (400 lbb)i
In the x-direction
(onee unknown forrce):
Q cos 50
5 ° − [(750 lb) cos 50°] + 4000 lb = 0
Q=
(750 lbb) cos 50° − 4000 lb
cos 50°
= 127.7100 lb
In the y-direction:
− P − Q sin 50° + (750 lb)sin 50° = 0
P = −Q sin 50° + (750 lb) sin 50
5 °
= −(127.710 lbb) sin 50° + (7550 lb) sin 50°
= 476.70 lb
P = 477 lb; Q = 127.7 lb
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PR
ROBLEM 2.53
A welded
w
connecction is in equuilibrium undder the action of
the four forces shown. Knnowing that FA = 8 kN and
a
FB = 16 kN, deteermine the magnitudes
m
off the other two
t
forcces.
SOLUT
TION
F
Free-Body
Diiagram of Connection
ΣFx = 0:
3
3
FB − FC − FA = 0
5
5
FA = 8 kN
With
FB = 16 kN
4
4
FC = (16 kN)) − (8 kN)
5
5
FC = 6.40 kN
3
3
Σ Fy = 0: − FD + FB − FA = 0
5
5
With FA and FB as abo
ove:
3
3
FD = (16 kN)) − (8 kN)
5
5
FD = 4.80 kN
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PROBLEM 2.54
4
A wellded connectioon is in equiliibrium under the
t action of the
four forces
fo
shown. Knowing thaat FA = 5 kN and FD = 6 kN,
k
determ
mine the magnnitudes of the other
o
two forces.
SOLUT
TION
F
Free-Body
Diiagram of Connection
3
3
ΣFy = 0: − FD − FA + FB = 0
5
5
or
3
FB = FD + FA
5
With
FA = 5 kN, FD = 8 kN
5⎡
3
⎤
FB = ⎢6 kN + (5 kN) ⎥
3⎣
5
⎦
ΣFx = 0: − FC +
FB = 15.00 kN
4
4
FB − FA = 0
5
5
4
( FB − FA )
5
4
= (15 kN − 5 kN)
5
FC =
FC = 8.00 kN
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P
PROBLEM
2
2.55
A sailor is beeing rescued using a boaatswain’s chhair
thhat is suspennded from a pulley that can roll freeely
onn the support cable ACB
A
and iss pulled at a
coonstant speed by cable CD.
C Knowinng that α = 30°
3
annd β = 10° and that thee combined weight of the
t
booatswain’s chair and the sailorr is 200 lb,
deetermine thee tension (a) in the suppoort cable AC
CB,
(bb) in the tracction cable CD.
C
TION
SOLUT
Free-B
Body Diagram
m
ΣFx = 0:: TACB cos 100° − TACB cos 30
3 ° − TCD cos 30
3 °=0
TCD = 0.1377158TACB
(1)
ΣFy = 0:: TACB sin 10° + TACB sin 300° + TCD sin 300° − 200 = 0
0.67365TACB + 0.5TCD = 200
(a )
Suubstitute (1) in
nto (2):
0.667365TACB + 0.5(0.137158T
0
TACB ) = 200
TACB = 269.46 lbb
(b )
Frrom (1):
(2)
TACB = 269 lb
TCD = 0.1371588(269.46 lb)
TCD = 37.0 lb
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P
PROBLEM
2.56
A sailor is beinng rescued usiing a boatswaain’s chair thatt is
suuspended from
m a pulley thatt can roll freelly on the suppport
caable ACB andd is pulled at a constant speeed by cable CD
C .
K
Knowing
that α = 25° and β = 15° and thhat the tensionn in
caable CD is 200 lb, determinne (a) the com
mbined weightt of
thhe boatswain’ss chair and thhe sailor, (b) thhe tension in the
suupport cable ACB
A .
SOLUT
TION
Free-B
Body Diagram
m
ΣFx = 0: TACB
cos 15° − TACB cos 25° − (20 lb)cos 255° = 0
A
TACB
= 304.04 lb
l
A
ΣFy = 0: (3004.04 lb)sin 15° + (304.04 lb)sin
l
25°
+ (20 lb)sin 255° − W = 0
W = 215.664 lb
(a )
W = 216 lb
(b) TACB = 304 lb
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PROBLEM 2.57
For the cables of prob. 2.44, find the value of α for which the tension
is as small as possible (a) in cable bc, (b) in both cables
simultaneously. In each case determine the tension in each cable.
SOLUTION
Free-Body Diagram
Force Triangle
(a) For a minimum tension in cable BC, set angle between cables to 90 degrees.
α = 35.0D
By inspection,
TAC = (6 kN) cos35D
TAC = 4.91 kN
TBC = (6 kN)sin 35D
TBC = 3.44 kN
(b) For equal tension in both cables, the force triangle will be an isosceles.
α = 55.0D
Therefore, by inspection,
TAC = TBC = (1 / 2)
6 kN
cos35°
TAC = TBC = 3.66 kN
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PROBLEM 2.58
For the cables of Prooblem 2.46, itt is known thaat the maximuum
allowablle tension is 600
6 N in cable AC and 7500 N in cable BC
B .
Determine (a) the maaximum forcee P that can be
b applied at C,
(b) the corresponding
c
value of α.
TION
SOLUT
Free-Body
F
Diagram
(a )
L of cosines
Law
Forrce Triangle
P 2 = (600)2 + (7750)2 − 2(600)(750) cos (25°° + 45°)
P = 784.02 N
(b )
L of sines
Law
P = 784 N
sin β
sin (25° + 45
4 °)
=
600 N
784.02 N
β = 46.0°
∴ α = 46.0° + 25°
α = 71.0°
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P
PROBLEM
2.59
For the situatioon described in Figure P2.448, determine (a)
thhe value of α for
f which the tension in roppe BC is as sm
mall
ass possible, (b) the corresponnding value off the tension.
TION
SOLUT
Free-Body
F
Diagram
Forrce Triangle
To be sm
mallest, TBC must
m be perpenndicular to thee direction of TAC.
(a )
(b )
T
Thus,
α = 5.00°
TBC = (1200 lb)sinn 5°
α = 5.00°
TBC = 104.6 lb
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PROBLE
EM 2.60
Two cabless tied togetherr at C are loaaded as shownn. Determine the
range of vaalues of Q for which the tennsion will nott exceed 60 lbb in
either cablee.
TION
SOLUT
Free-Boody Diagram
ΣFx = 0: −TBC − Q coos60° + 75 lb = 0
TBC = 75 lbb − Q cos60°
(1)
ΣFy = 0: TAC − Q sin 60° = 0
TAC = Q sinn 60°
Requiremeent:
From Eq. (2):
(
(2)
TAC = 60 lbb:
Q sin
s 60° = 60 lb
Q = 69.3 lb
Requiremeent:
From Eq. (1):
(
TBC = 60 lbb:
75 lb − Q cos 60° = 60 lb
Q = 300.0 lb 30.0 lb ≤ Q ≤ 69.3 lb
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PROBLEM 2.61
A movable bin and its contents have a combined weight of 2.8 kN. Determine
the shortest chain sling ACB that can be used to lift the loaded bin if the
tension in the chain is not to exceed 5 kN.
SOLUTION
Free-Body Diagram
tan α =
h
0.6 m
(1)
Isosceles Force Triangle
Law of sines:
sin α =
1
2
(2.8 kN)
TAC
TAC = 5 kN
sin α =
1
2
(2.8 kN)
5 kN
α = 16.2602°
From Eq. (1): tan16.2602° =
h
0.6 m
∴ h = 0.175000 m
Half-length of chain = AC = (0.6 m) 2 + (0.175 m) 2
= 0.625 m
Total length:
= 2 × 0.625 m
1.250 m
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PROBLEM 2.62
For W = 800 N, P = 200 N, and d = 600 mm,
determine the value of h consistent with
equilibrium.
SOLUTION
Free-Body Diagram
TAC = TBC = 800 N
(h
AC = BC =
ΣFy = 0: 2(800 N)
800 =
Data:
P
⎛d ⎞
1+ ⎜ ⎟
2
⎝h⎠
2
+ d2
h
2
h + d2
)
−P=0
2
P = 200 N, d = 600 mm and solving for h
800 N =
200 N
⎛ 600 mm ⎞
1+ ⎜
⎟
2
h
⎝
⎠
2
h = 75.6 mm
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PROBLE
EM 2.63
Collar A is connected
c
as shown
s
to a 500-lb load and can
c
slide on a frictionless horizontal
h
rodd. Determine the
magnitude of the force P required to maintain the
equilibrium of the collaar when (a) x = 4.5 in., (b)
x = 15 in.
TION
SOLUT
(a )
F
Free
Body: Co
ollar A
Forcce Triangle
P 50 lb
=
4.55 20.5
(b )
F
Free
Body: Co
ollar A
P = 10.98 lb
Forcce Triangle
P 50 lb
=
25
15
P = 30.0 lb
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PROBLEM
M 2.64
Collar A is connected
c
as shown
s
to a 500-lb load and can
c
slide on a frictionless
f
hoorizontal rod.. Determine the
distance x foor which the collar is in eqquilibrium whhen
P = 48 lb.
SOLUT
TION
Free Bo
ody: Collar A
Force Trianglee
1
N 2 = (500)2 − (48)2 = 196
N = 14.00 lb
Similar Triangles
x
488 lb
=
20 in. 144 lb
x = 68.6 in.
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PROBLEM 2.65
Three forces are applied to a bracket as shown. The directions of the two 150-N
forces may vary, but the angle between these forces is always 50°. Determine
the range of values of α for which the magnitude of the resultant of the forces
acting at A is less than 600 N.
SOLUTION
Combine the two 150-N forces into a resultant force Q:
Q = 2(150 N) cos 25°
= 271.89 N
Equivalent loading at A:
Using the law of cosines:
(600 N) 2 = (500 N) 2 + (271.89 N)2 + 2(500 N)(271.89 N) cos(55° + α )
cos(55° + α ) = 0.132685
Two values for α :
55° + α = 82.375
α = 27.4°
or
55° + α = −82.375°
55° + α = 360° − 82.375°
α = 222.6°
For R < 600 lb:
27.4° < α < 222.6D
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PROBL
LEM 2.66
A 200-kgg crate is to bee supported by
b the rope-annd-pulley arrangement show
wn.
Determinee the magnituude and directtion of the forrce P that mu
ust be exerted on
the free ennd of the rope to maintain equilibrium. (H
Hint: The tensiion in the ropee is
the same on
o each side of
o a simple pu
ulley. This cann be proved byy the methodss of
Ch. 4.)
SOLUT
TION
Free-Boody Diagram:: Pulley A
⎛ 5 ⎞
ΣFx = 0: − 2 P ⎜
c α =0
⎟ + P cos
⎝ 281 ⎠
cos α = 0.59655
α = ±53.377°
For α = +53.377°:
⎛ 166 ⎞
ΣFy = 0: 2 P ⎜
⎟ + P sin 533.377° − 1962 N = 0
⎝ 281 ⎠
P = 7244 N
53.4°
For α = −53.377°:
⎛ 166 ⎞
ΣFy = 0: 2 P ⎜
5
°) − 19662 N = 0
⎟ + P sin( −53.377
⎝ 281 ⎠
P = 17773
53.4°
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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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PR
ROBLEM 2..68
Solvve Parts b andd d of Problem
m 2.67, assumiing
thatt the free end of the rope iss attached to the
cratte.
PROBLEM 2.677 A 600-lb crrate is supporrted
as
by several ropee-and-pulley arrangements
a
the
shown. Determinne for each arrangement
a
tenssion in the roppe. (See the hint
h for Problem
2.666.)
TION
SOLUT
Free-Boody Diagram of Pulley and
d Crate
(b)
ΣFy = 0: 3T − (600 lbb) = 0
1
T = (600 lb)
3
T = 200 lb
(d )
ΣFy = 0: 4T − (600 lbb) = 0
T=
1
(600 lb))
4
T = 150.0 lb
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PRO
OBLEM 2.6
69
A load Q is appliied to the pullley C, whichh can roll on the
cablee ACB. The pulley
p
is heldd in the position shown byy a
seconnd cable CAD
D, which paasses over thee pulley A and
a
50 N, determ
suppoorts a load P. Knowing that P = 75
mine
(a) thhe tension in cable
c
ACB, (b)) the magnitudde of load Q.
SOLUT
TION
Free-Boody Diagram:: Pulley C
ΣFx = 0: TACBB (cos 25° − coss55°) − (750 N)cos55°
N
=0
(a)
Hencce:
TACB = 12992.88 N
TACB
= 1293 N
A
ΣFy = 0: TACB (sin
( 25° + sin 55°) + (750 N) sin 55° − Q = 0
(b)
(1292.88 N)(sin 25° + sin 55°) + (750 N) sin 55° − Q = 0
or
Q = 2219.8 N
Q = 2220 N
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PRO
OBLEM 2.7
70
An 1800-N load Q is applied too the pulley C,
C which can roll
r
on thhe cable ACB. The pulley is held in the poosition shown by
a seccond cable CAD,
CA which passes
p
over thhe pulley A and
a
suppoorts a load P. Determine (a)
( the tensionn in cable AC
CB,
(b) thhe magnitude of
o load P.
SOLUT
TION
Free-Boody Diagram:: Pulley C
ΣFx = 0:
0 TACB (cos 25
2 ° − cos55°) − P cos55° = 0
P = 0.58010
0
TACB (1)
or
ΣFy = 0: TACB (sin 25
2 ° + sin 55°) + P sin 55° − 1800 N = 0
1.24177T
TACB + 0.819155P = 1800 N (2)
or
(a)
( into Equattion (2):
Substittute Equation (1)
1.244177TACB + 0.81915(0.580110TACB ) = 18000 N
Hence:
TACB = 10488.37 N
TACB
= 1048 N
A
(b)
(
Using (1),
P = 0.58010(1048.337 N) = 608.166 N
P = 608 N
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PROBLEM 2.71
Determine (a) the x, y, and z components of the 600-N force,
(b) the angles θx, θy, and θz that the force forms with the
coordinate axes.
SOLUTION
(a)
Fx = (600 N)sin 25° cos30D
Fx = 219.60 N
Fx = 220 N
Fy = (600 N) cos 25°
Fy = 544 N
Fy = 543.78 N
Fz = (380.36 N)sin 25° sin 30D
Fz = 126.785 N
(b)
cos θ x =
cos θ y =
cos θ z =
Fx 219.60 N
=
F
600 N
Fy
Fz = 126.8 N
θ x = 68.5°
543.78 N
600 N
θ y = 25.0°
Fz 126.785 N
=
600 N
F
θ z = 77.8°
F
=
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PROBLEM 2.72
Determine (a) the x, y, and z components of the 450-N force,
(b) the angles θx, θy, and θz that the force forms with the
coordinate axes.
SOLUTION
(a)
Fx = −(450 N)cos 35° sin 40D
Fx = −236.94 N
Fx = −237 N
Fy = (450 N)sin 35°
Fy = 258 N
Fy = 258.11 N
Fz = (450 N) cos35° cos 40D
Fz = 282.38 N
(b)
cos θ x =
cos θ y =
cos θ z =
Fx -236.94 N
=
F
450 N
Fy
Fz = 282 N
θx = 121.8°
258.11 N
450 N
θ y = 55.0°
Fz 282.38 N
=
450 N
F
θ z = 51.1°
F
=
Note: From the given data, we could have computed directly
θ y = 90D − 35D = 55D , which checks with the answer obtained.
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PROBLEM 2.73
A gun is aimed at a point A located 35° east of north. Knowing that the barrel of the gun forms an angle
of 40° with the horizontal and that the maximum recoil force is 400 N, determine (a) the x, y, and z
components of that force, (b) the values of the angles θx, θy, and θz defining the direction of the recoil
force. (Assume that the x, y, and z axes are directed, respectively, east, up, and south.)
SOLUTION
Recoil force
F = 400 N
∴ FH = (400 N)cos 40°
= 306.42 N
(a)
Fx = − FH sin 35°
= −(306.42 N)sin 35°
= −175.755 N
Fx = −175.8 N
Fy = − F sin 40°
= −(400 N)sin 40°
= −257.12 N
Fy = −257 N
Fz = + FH cos 35°
= +(306.42 N) cos 35°
= +251.00 N
(b)
cos θ x =
cos θ y =
cos θ z =
Fx −175.755 N
=
F
400 N
Fy
F
=
−257.12 N
400 N
Fz 251.00 N
=
400 N
F
Fz = +251 N
θ x = 116.1°
θ y = 130.0°
θ z = 51.1°
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PROBLEM 2.74
Solve Problem 2.73, assuming that point A is located 15° north of west and that the barrel of the gun
forms an angle of 25° with the horizontal.
PROBLEM 2.73 A gun is aimed at a point A located 35° east of north. Knowing that the barrel of the
gun forms an angle of 40° with the horizontal and that the maximum recoil force is 400 N, determine
(a) the x, y, and z components of that force, (b) the values of the angles θx, θy, and θz defining the direction
of the recoil force. (Assume that the x, y, and z axes are directed, respectively, east, up, and south.)
SOLUTION
Recoil force
F = 400 N
∴ FH = (400 N)cos 25°
= 362.52 N
(a)
Fx = + FH cos15°
= +(362.52 N)cos15°
= +350.17 N
Fx = +350 N
Fy = − F sin 25°
= −(400 N)sin 25°
= −169.047 N
Fy = −169.0 N
Fz = + FH sin15°
= +(362.52 N)sin15°
= +93.827 N
(b)
cos θ x =
cosθ y =
cos θ z =
Fx +350.17 N
=
400 N
F
Fy
F
=
−169.047 N
400 N
Fz +93.827 N
=
400 N
F
Fz = +93.8 N
θ x = 28.9°
θ y = 115.0°
θ z = 76.4°
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PROBLEM 2.75
The angle between spring AB and the post DA is 30°. Knowing
that the tension in the spring is 50 lb, determine (a) the x, y,
and z components of the force exerted on the circular plate at
B, (b) the angles θx, θy, and θz defining the direction of the
force at B.
SOLUTION
Fh = F cos 60°
= (50 lb) cos 60°
Fh = 25.0 lb
Fx = − Fh cos 35°
Fy = F sin 60D
Fz = − Fh sin 35D
Fx = (−25.0 lb)cos35D
Fy = (50.0 lb)sin 60D
Fz = (−25.0 lb)sin 35D
Fx = −20.479 lb
Fy = 43.301 lb
Fz = −14.3394 lb
(a)
Fx = −20.5 lb

Fy = 43.3 lb
Fz = −14.33 lb
(b)
cos θ x =
cos θ y =
cos θ z =
Fx −20.479 lb
=
F
50 lb
Fy
F
=
43.301 lb
50 lb
Fz -14.3394 lb
=
50 lb
F
θ x = 114.2°
θ y = 30.0°
θ z = 106.7°
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PROBLEM 2.76
The angle between spring AC and the post DA is
30°. Knowing that the tension in the spring is 40 lb,
determine (a) the x, y, and z components of the force
exerted on the circular plate at C, (b) the angles θx,
θy, and θz defining the direction of the force at C.
SOLUTION
Fh = F cos 60°
= (40 lb) cos 60°
Fh = 20.0 lb
(a)
Fx = Fh cos 35°
= (20.0 lb) cos 35°
Fx = 16.3830 lb
Fy = F sin 60°
= (40 lb)sin 60°
Fy = 34.641 lb
Fz = −Fh sin 35°
= −(20.0 lb)sin 35°
Fz = −11.4715 lb

Fx = 16.38 lb
Fy = 34.6 lb
Fz = −11.47 lb
(b)
cos θ x =
cos θ y =
cos θ z =
Fx 16.3830 lb
=
F
40 lb
Fy
F
=
34.641 lb
40 lb
Fz -11.4715 lb
=
40 lb
F
θ x = 65.8°
θ y = 30.0°
θ z = 106.7°
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P
PROBLEM
2.77
Cable AB is 65 ft long, and
C
a
the tensioon in that caable is 3900 lb.
D
Determine
(a) the x, y, andd z componentts of the forcee exerted by the
c
cable
on the anchor
a
B, (b) the angles θ x , θ y , and θ z defining the
d
direction
of thaat force.
TION
SOLUT
56 ft
65 ft
= 0.86154
θ y = 30.51°
cos θ y =
From trianngle AOB:
Fx = − F sin θ y cos 20°
(a)
= −(3900 lbb)sin 30.51° coos 20°
Fx = −1861 lb
Fy = + F cosθ y = (3900 lb)(00.86154)
Fy = +3360 lb
Fz = +(3900 lbb)sin 30.51° siin 20°
(b)
cos θ x =
Froom above:
Fx
18861 lb
= − 0.47771
=−
F
39900 lb
θ y = 30.551°
cos θ z =
Fz
677 lb
=+
= + 0.1736
3900 lb
F
Fz = +677 lb
θ x = 118.5°
θ y = 30.5°
θ z = 80.0°
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PROBLEM 2.78
Cable AC is 70 ft long, and the tension in that cable is 5250 lb.
Determine (a) the x, y, and z components of the force exerted by the
cable on the anchor C, (b) the angles θx, θy, and θz defining the
direction of that force.
SOLUTION
AC = 70 ft
OA = 56 ft
F = 5250 lb
In triangle AOB:
56 ft
70 ft
θ y = 36.870°
cos θ y =
FH = F sin θ y
= (5250 lb)sin 36.870°
= 3150.0 lb
(a)
Fx = −FH sin 50° = −(3150.0 lb)sin 50° = −2413.0 lb
Fx = −2410 lb
Fy = + F cosθ y = +(5250 lb) cos36.870° = +4200.0 lb
Fy = +4200 lb
Fz = −FH cos50° = −3150cos50° = −2024.8 lb
(b)
cos θ x =
From above:
Fz = −2025 lb
Fx −2413.0 lb
=
F
5250 lb
θ y = 36.870°
θz =
Fz −2024.8 lb
=
5250 lb
F
θ x = 117.4°
θ y = 36.9°
θ z = 112.7°
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PROBLEM 2.79
Determine the magnitude and direction of the force F = (240 N)i – (270 N)j + (680 N)k.
SOLUTION
F = Fx2 + Fy2 + Fz2
F = (240 N) 2 + (−270 N) 2 + (−680 N) 2
cos θ x =
cos θ y =
cos θ y =
Fx 240 N
=
F 770 N
Fy
F
=
−270 N
770 N
Fz 680 N
=
F 770 N
F = 770 N
θ x = 71.8°
θ y = 110.5°
θ z = 28.0°
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PROBLEM 2.80
Determine the magnitude and direction of the force F = (320 N)i + (400 N)j − (250 N)k.
SOLUTION
F = Fx2 + Fy2 + Fz2
F = (320 N) 2 + (400 N) 2 + (−250 N) 2
cos θ x =
cos θ y =
cos θ y =
Fx 320 N
=
F 570 N
Fy
F
=
400 N
570 N
Fz −250 N
=
570 N
F
F = 570 N
θ x = 55.8°
θ y = 45.4°
θ z = 116.0°
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PROBLEM 2.81
A force acts at the origin of a coordinate system in a direction defined by the angles θx = 69.3° and θz
= 57.9°. Knowing that the y component of the force is –174.0 lb, determine (a) the angle θy, (b) the other
components and the magnitude of the force.
SOLUTION
cos 2 θ x + cos 2 θ y + cos 2 θ z = 1
cos 2 (69.3°) + cos 2 θ y + cos 2 (57.9°) = 1
cos θ y = ±0.7699
(a)
(b)
Since Fy < 0, we choose cosθ y = −0.7699
∴ θ y = 140.3°
Fy = F cos θ y
−174.0 lb = F (−0.7699)
F = 226.0 lb
F = 226 lb
Fx = F cosθ x = (226.0 lb)cos69.3°
Fx = 79.9 lb
Fz = F cosθ z = (226.0 lb)cos57.9°
Fz = 120.1lb
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PROBLEM 2.82
A force acts at the origin of a coordinate system in a direction defined by the angles θx = 70.9° and
θy = 144.9°. Knowing that the z component of the force is –52.0 lb, determine (a) the angle θz, (b) the
other components and the magnitude of the force.
SOLUTION
cos 2 θ x + cos 2 θ y + cos 2 θ z = 1
cos 2 70.9D + cos 2 144.9° + cos 2 θ z ° = 1
cos θ z = ±0.47282
(a)
(b)
Since Fz < 0, we choose cosθ z = −0.47282
∴ θ z = 118.2°
Fz = F cos θ z
−52.0 lb = F (−0.47282)
F = 110.0 lb
F = 110.0 lb
Fx = F cosθ x = (110.0 lb)cos70.9°
Fx = 36.0 lb
Fy = F cosθ y = (110.0 lb)cos144.9°
Fy = −90.0 lb
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PROBLEM 2.83
A force F of magnitude 210 N acts at the origin of a coordinate system. Knowing that Fx = 80 N,
θz = 151.2°, and Fy < 0, determine (a) the components Fy and Fz, (b) the angles θx and θy.
SOLUTION
Fz = F cosθ z = (210 N)cos151.2°
(a)
= −184.024 N
Then:
So:
Hence:
Fz = −184.0 N
F 2 = Fx2 + Fy2 + Fz2
(210 N) 2 = (80 N) 2 + ( Fy )2 + (184.024 N)2
Fy = − (210 N) 2 − (80 N) 2 − (184.024 N) 2
= −61.929 N
(b)
cos θ x =
cos θ y =
Fx
80 N
=
= 0.38095
F 210 N
Fy
F
=
Fy = −62.0 lb
θ x = 67.6°
61.929 N
= −0.29490
210 N
θ y = 107.2°
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PROBLEM 2.84
A force F of magnitude 1200 N acts at the origin of a coordinate system. Knowing that
θx = 65°, θy = 40°, and Fz > 0, determine (a) the components of the force, (b) the angle θz.
SOLUTION
cos 2 θ x + cos 2 θ y + cos 2 θ z = 1
cos 2 65D + cos 2 40° + cos 2 θ z ° = 1
cos θ z = ±0.48432
(b)
(a)
Since Fz > 0, we choose cosθ z = 0.48432, or θ z = 61.032D
∴ θ z = 61.0°
F = 1200 N
Fx = F cosθ x = (1200 N) cos65D
Fx = 507 N
Fy = F cosθ y = (1200 N) cos 40°
Fy = 919 N
Fz = F cos θ z = (1200 N) cos 61.032°
Fz = 582 N
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PROBLEM 2.85
A frame ABC is supported in part by cable DBE that
passes through a frictionless ring at B. Knowing that the
tension in the cable is 385 N, determine the components
of the force exerted by the cable on the support at D.
SOLUTION
JJJG
DB = (480 mm)i − (510 mm) j + (320 mm)k
DB = (480 mm)2 + (510 mm2 ) + (320 mm)2
= 770 mm
F = F λ DB
JJJG
DB
=F
DB
385 N
=
[(480 mm)i − (510 mm)j + (320 mm)k ]
770 mm
= (240 N)i − (255 N) j + (160 N)k
Fx = +240 N, Fy = −255 N, Fz = +160.0 N
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PROBLEM 2.86
For the frame and cable of Problem 2.85, determine the
components of the force exerted by the cable on the
support at E.
PROBLEM 2.85 A frame ABC is supported in part by
cable DBE that passes through a frictionless ring at B.
Knowing that the tension in the cable is 385 N, determine
the components of the force exerted by the cable on the
support at D.
SOLUTION
JJJG
EB = (270 mm)i − (400 mm) j + (600 mm)k
EB = (270 mm)2 + (400 mm)2 + (600 mm)2
= 770 mm
F = F λ EB
JJJG
EB
=F
EB
385 N
=
[(270 mm)i − (400 mm)j + (600 mm)k ]
770 mm
F = (135 N)i − (200 N) j + (300 N)k
Fx = +135.0 N, Fy = −200 N, Fz = +300 N
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PROBLEM 2.87
In order to move a wrecked truck, two cables are attached
at A and pulled by winches B and C as shown. Knowing
that the tension in cable AB is 2 kips, determine the
components of the force exerted at A by the cable.
SOLUTION
Cable AB:
λAB
TAB
JJJG
AB (−46.765 ft)i + (45 ft) j + (36 ft)k
=
=
AB
74.216 ft
−46.765i + 45 j + 36k
= TAB λAB =
74.216
(TAB ) x = −1.260 kips
(TAB ) y = +1.213 kips
(TAB ) z = +0.970 kips
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PROBLEM 2.88
In order to move a wrecked truck, two cables are attached
at A and pulled by winches B and C as shown. Knowing
that the tension in cable AC is 1.5 kips, determine the
components of the force exerted at A by the cable.
SOLUTION
Cable AB:
λAC
TAC
JJJG
AC ( −46.765 ft)i + (55.8 ft) j + (−45 ft)k
=
=
AC
85.590 ft
−46.765i + 55.8 j − 45k
= TAC λAC = (1.5 kips)
85.590
(TAC ) x = −0.820 kips
(TAC ) y = +0.978 kips
(TAC ) z = −0.789 kips
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PROBLEM 2.89
A rectangular plate is supported by three cables as
shown. Knowing that the tension in cable AB is 408 N,
determine the components of the force exerted on the
plate at B.
SOLUTION
We have:
JJJG
BA = +(320 mm)i + (480 mm)j - (360 mm)k
Thus:
FB = TBA λBA = TBA
BA = 680 mm
JJJG
BA
12
9 ⎞
⎛ 8
= TBA ⎜ i + j - k ⎟
BA
⎝ 17 17 17 ⎠
⎛8
⎞ ⎛ 12
⎞ ⎛ 9
⎞
⎜ 17 TBA ⎟ i + ⎜ 17 TBA ⎟ j − ⎜ 17 TBA ⎟ k = 0
⎝
⎠ ⎝
⎠ ⎝
⎠
Setting TBA = 408 N yields,
Fx = +192.0 N, Fy = +288 N, Fz = −216 N
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PROBLEM 2.90
A rectangular plate is supported by three cables as
shown. Knowing that the tension in cable AD is 429 N,
determine the components of the force exerted on the
plate at D.
SOLUTION
We have:
JJJG
DA = −(250 mm)i + (480 mm)j + (360 mm)k
Thus:
FD = TDA λDA = TDA
DA = 650 mm
JJJG
DA
48
36 ⎞
⎛ 5
j+ k⎟
= TDA ⎜ − i +
DA
65 ⎠
⎝ 13 65
⎛5
⎞ ⎛ 48
⎞ ⎛ 36
⎞
− ⎜ TDA ⎟ i + ⎜ TDA ⎟ j + ⎜ TDA ⎟ k = 0
⎝ 13
⎠ ⎝ 65
⎠ ⎝ 65
⎠
Setting TDA = 429 N yields,
Fx = −165.0 N, Fy = +317 N, Fz = +238 N
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PROBLEM 2.91
Find the magnitude and direction of the resultant of the two forces
shown knowing that P = 300 N and Q = 400 N.
SOLUTION
P = (300 N)[− cos 30° sin15°i + sin 30° j + cos 30° cos15°k ]
= − (67.243 N)i + (150 N) j + (250.95 N)k
Q = (400 N)[cos 50° cos 20°i + sin 50° j − cos 50° sin 20°k ]
= (400 N)[0.60402i + 0.76604 j − 0.21985]
= (241.61 N)i + (306.42 N) j − (87.939 N)k
R =P+Q
= (174.367 N)i + (456.42 N) j + (163.011 N)k
R = (174.367 N)2 + (456.42 N) 2 + (163.011 N) 2
= 515.07 N
cos θ x =
cos θ y =
cos θ z =
Rx 174.367 N
=
= 0.33853
515.07 N
R
Ry
R
=
456.42 N
= 0.88613
515.07 N
Rz 163.011 N
=
= 0.31648
515.07 N
R
R = 515 N
θ x = 70.2°
θ y = 27.6°
θ z = 71.5°
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PROBLEM 2.92
Find the magnitude and direction of the resultant of the two forces
shown knowing that P = 400 N and Q = 300 N.
SOLUTION
P = (400 N)[− cos 30° sin15°i + sin 30° j + cos 30° cos15°k ]
= − (89.678 N)i + (200 N) j + (334.61 N)k
Q = (300 N)[cos50° cos 20°i + sin 50° j − cos 50° sin 20°k ]
= (181.21 N)i + (229.81 N)j − (65.954 N)k
R =P+Q
= (91.532 N)i + (429.81 N) j + (268.66 N)k
R = (91.532 N)2 + (429.81 N)2 + (268.66 N) 2
= 515.07 N
cos θ x =
cos θ y =
cos θ z =
Rx 91.532 N
=
= 0.177708
R 515.07 N
Ry
R = 515 N
θ x = 79.8°
429.81 N
= 0.83447
515.07 N
θ y = 33.4°
Rz 268.66 N
=
= 0.52160
R 515.07 N
θ z = 58.6°
R
=
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PROBLEM 2.93
Knowing that the tension is 425 lb in cable AB and 510 lb
in cable AC, determine the magnitude and direction of the
resultant of the forces exerted at A by the two cables.
SOLUTION
JJJG
AB = (40 in.)i − (45 in.) j + (60 in.)k
AB = (40 in.) 2 + (45 in.) 2 + (60 in.) 2 = 85 in.
JJJG
AC = (100 in.)i − (45 in.) j + (60 in.)k
AC = (100 in.)2 + (45 in.) 2 + (60 in.)2 = 125 in.
JJJG
⎡ (40 in.)i − (45 in.) j + (60 in.)k ⎤
AB
TAB = TAB λAB = TAB
= (425 lb) ⎢
⎥
AB
85 in.
⎣
⎦
TAB = (200 lb)i − (225 lb) j + (300 lb)k
JJJG
⎡ (100 in.)i − (45 in.) j + (60 in.)k ⎤
AC
TAC = TAC λAC = TAC
= (510 lb) ⎢
⎥
125 in.
AC
⎣
⎦
TAC = (408 lb)i − (183.6 lb) j + (244.8 lb)k
R = TAB + TAC = (608)i − (408.6 lb) j + (544.8 lb)k
Then:
and
R = 912.92 lb
R = 913 lb
cos θ x =
608 lb
= 0.66599
912.92 lb
cos θ y =
408.6 lb
= −0.44757
912.92 lb
θ y = 116.6°
cos θ z =
544.8 lb
= 0.59677
912.92 lb
θ z = 53.4°
θ x = 48.2°
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PROBLEM 2.94
Knowing that the tension is 510 lb in cable AB and 425 lb
in cable AC, determine the magnitude and direction of the
resultant of the forces exerted at A by the two cables.
SOLUTION
JJJG
AB = (40 in.)i − (45 in.) j + (60 in.)k
AB = (40 in.) 2 + (45 in.) 2 + (60 in.) 2 = 85 in.
JJJG
AC = (100 in.)i − (45 in.) j + (60 in.)k
AC = (100 in.)2 + (45 in.) 2 + (60 in.)2 = 125 in.
JJJG
⎡ (40 in.)i − (45 in.) j + (60 in.)k ⎤
AB
TAB = TAB λAB = TAB
= (510 lb) ⎢
⎥
AB
85 in.
⎣
⎦
TAB = (240 lb)i − (270 lb) j + (360 lb)k
JJJG
⎡ (100 in.)i − (45 in.) j + (60 in.)k ⎤
AC
TAC = TAC λAC = TAC
= (425 lb) ⎢
⎥
125 in.
AC
⎣
⎦
TAC = (340 lb)i − (153 lb) j + (204 lb)k
R = TAB + TAC = (580 lb)i − (423 lb) j + (564 lb)k
Then:
and
R = 912.92 lb
R = 913 lb
cos θ x =
580 lb
= 0.63532
912.92 lb
θ x = 50.6°
cos θ y =
−423 lb
= −0.46335
912.92 lb
θ y = 117.6°
cos θ z =
564 lb
= 0.61780
912.92 lb
θ z = 51.8°
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PROBLEM 2.95
For the frame of Problem 2.85, determine the magnitude and
direction of the resultant of the forces exerted by the cable at B
knowing that the tension in the cable is 385 N.
PROBLEM 2.85 A frame ABC is supported in part by cable
DBE that passes through a frictionless ring at B. Knowing that
the tension in the cable is 385 N, determine the components of
the force exerted by the cable on the support at D.
SOLUTION
JJJG
BD = − (480 mm)i + (510 mm) j − (320 mm)k
BD = (480 mm) 2 + (510 mm) 2 + (320 mm) 2 = 770 mm
JJJG
BD
FBD = TBD λ BD = TBD
BD
(385 N)
[−(480 mm)i + (510 mm) j − (320 mm)k ]
=
(770 mm)
= −(240 N)i + (255 N) j − (160 N)k
JJJG
BE = −(270 mm)i + (400 mm) j − (600 mm)k
BE = (270 mm) 2 + (400 mm) 2 + (600 mm) 2 = 770 mm
JJJG
BE
FBE = TBE λ BE = TBE
BE
(385 N)
[−(270 mm)i + (400 mm) j − (600 mm)k ]
=
(770 mm)
= −(135 N)i + (200 N) j − (300 N)k
R = FBD + FBE = −(375 N)i + (455 N) j − (460 N)k
R = (375 N)2 + (455 N)2 + (460 N)2 = 747.83 N
R = 748 N
cos θ x =
−375 N
747.83 N
θ x = 120.1°
cos θ y =
455 N
747.83 N
θ y = 52.5°
cos θ z =
−460 N
747.83 N
θ z = 128.0°
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PROBLEM 2.96
For the plate of Prob. 2.89, determine the tensions in cables AB
and AD knowing that the tension in cable AC is 54 N and that
the resultant of the forces exerted by the three cables at A must
be vertical.
SOLUTION
We have:
JJJG
AB = −(320 mm)i − (480 mm)j + (360 mm)k AB = 680 mm
JJJG
AC = (450 mm)i − (480 mm) j + (360 mm)k AC = 750 mm
JJJG
AD = (250 mm)i − (480 mm) j − ( 360 mm ) k AD = 650 mm
Thus:
TAB
TAC
TAD
JJJG
AB TAB
= TAB λ AB = TAB
=
( −320i − 480 j + 360k )
AB 680
JJJG
AC 54
= TAC λAC = TAC
=
( 450i − 480 j + 360k )
AC 750
JJJG
AD TAD
= TAD λAD = TAD
=
( 250i − 480 j − 360k )
AD 650
Substituting into the Eq. R = ΣF and factoring i, j, k :
250
⎛ 320
⎞
TAB + 32.40 +
TAD ⎟ i
R = ⎜−
650
⎝ 680
⎠
480
⎛ 480
⎞
+ ⎜−
TAB − 34.560 −
TAD ⎟ j
650
⎝ 680
⎠
360
⎛ 360
⎞
+⎜
TAB + 25.920 −
TAD ⎟ k
650
⎝ 680
⎠
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PROBLEM 2.96 (Continued)
Since R is vertical, the coefficients of i and k are zero:
i:
−
320
250
TAB + 32.40 +
TAD = 0
680
650
(1)
360
360
TAB + 25.920 −
TAD = 0
680
650
(2)
k:
Multiply (1) by 3.6 and (2) by 2.5 then add:
−
252
TAB + 181.440 = 0
680
TAB = 489.60 N
TAB = 490 N
Substitute into (2) and solve for TAD :
360
360
(489.60 N) + 25.920 −
TAD = 0
680
650
TAD = 514.80 N
TAD = 515 N
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PROBLEM 2.97
The boom OA carries a load P and is supported by
two cables as shown. Knowing that the tension in
cable AB is 183 lb and that the resultant of the load P
and of the forces exerted at A by the two cables must
be directed along OA, determine the tension in cable
AC.
SOLUTION
Cable AB:
TAB = 183 lb
TAB
TAB
Cable AC:
TAC
TAC
Load P:
JJJG
AB
(−48 in.)i + (29 in.) j + (24 in.)k
= TAB λ AB = TAB
= (183 lb)
AB
61 in.
= −(144 lb)i + (87 lb) j + (72 lb)k
JJJG
AC
(−48 in.)i + (25 in.) j + (−36 in.)k
= TAC λ AC = TAC
= TAC
AC
65 in.
48
25
36
= − TAC i + TAC j − TAC k
65
65
65
P = Pj
For resultant to be directed along OA, i.e., x-axis
Rz = 0: ΣFz = (72 lb) −
36
′ =0
TAC
65
TAC = 130.0 lb
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PROBLEM 2.98
For the boom and loading of Problem. 2.97, determine
the magnitude of the load P.
PROBLEM 2.97 The boom OA carries a load P and is
supported by two cables as shown. Knowing that the
tension in cable AB is 183 lb and that the resultant of
the load P and of the forces exerted at A by the two
cables must be directed along OA, determine the
tension in cable AC.
SOLUTION
See Problem 2.97. Since resultant must be directed along OA, i.e., the x-axis, we write
Ry = 0: ΣFy = (87 lb) +
25
TAC − P = 0
65
TAC = 130.0 lb from Problem 2.97.
Then
(87 lb) +
25
(130.0 lb) − P = 0
65
P = 137.0 lb
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PROBLEM 2.99
A container is supported by three cables that are attached to
a ceiling as shown. Determine the weight W of the container,
knowing that the tension in cable AB is 6 kN.
SOLUTION
Free-Body Diagram at A:
The forces applied at A are:
TAB , TAC , TAD , and W
where W = W j. To express the other forces in terms of the unit vectors i, j, k, we write
JJJG
AB = − (450 mm)i + (600 mm) j
AB = 750 mm
JJJG
AC = + (600 mm) j − (320 mm)k
AC = 680 mm
JJJG
AD = + (500 mm)i + (600 mm) j + (360 mm)k AD = 860 mm
JJJG
AB
(−450 mm)i + (600 mm) j
TAB = λ ABTAB = TAB
= TAB
and
AB
750 mm
⎛ 45 60 ⎞
= ⎜ − i + j ⎟ TAB
⎝ 75 75 ⎠
TAC = λ AC TAC = TAC
TAD = λ ADTAD = TAD
JJJG
AC
(600 mm)i − (320 mm) j
= TAC
AC
680 mm
32 ⎞
⎛ 60
= ⎜ j − k ⎟ TAC
68 ⎠
⎝ 68
JJJG
AD
(500 mm)i + (600 mm) j + (360 mm)k
= TAD
AD
860 mm
60
36 ⎞
⎛ 50
=⎜ i+
j + k ⎟ TAD
86
86
86
⎝
⎠
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PROBLEM 2.99 (Continued)
Equilibrium condition:
ΣF = 0: ∴ TAB + TAC + TAD + W = 0
Substituting the expressions obtained for TAB , TAC , and TAD ; factoring i, j, and k; and equating each of
the coefficients to zero gives the following equations:
From i:
From j:
From k:
45
50
TAB + TAD = 0
75
86
(1)
60
60
60
TAB + TAC + TAD − W = 0
75
68
86
(2)
32
36
TAC + TAD = 0
68
86
(3)
−
−
Setting TAB = 6 kN in (1) and (2), and solving the resulting set of equations gives
TAC = 6.1920 kN
TAC = 5.5080 kN
W = 13.98 kN
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PROBLEM 2.100
A container is supported by three cables that are attached to
a ceiling as shown. Determine the weight W of the container,
knowing that the tension in cable AD is 4.3 kN.
SOLUTION
See Problem 2.99 for the figure and analysis leading to the following set of linear algebraic equations:
45
50
TAB + TAD = 0
75
86
(1)
60
60
60
TAB + TAC + TAD − W = 0
75
68
86
(2)
32
36
TAC + TAD = 0
68
86
(3)
−
−
Setting TAD = 4.3 kN into the above equations gives
TAB = 4.1667 kN
TAC = 3.8250 kN
W = 9.71 kN
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PROB
BLEM 2.101
1
Three cables
c
are useed to tether a balloon as shhown. Determine
the verttical force P exerted
e
by the balloon at A knowing
k
that the
tension in cable AD is 481 N.
TION
SOLUT
FREE-BOD
DY DIAGRA
AM AT A
The forcces applied at A are:
TAB , TAC , TAD , andd P
where P = Pj. To exp
press the otherr forces in term
ms of the unit vectors i, j, k, we write
JJJG
AB = 7.00 m
AB = − (4.20 m)i − (5.60 m) j
JJJG
AC = (2.40 m)i − (5.60
(
m) j + (4.20 m)k AC = 7.40 m
JJJG
AD = − (5.60 m) j − (3.30 m)k
D = 6.50 m
AD
JJJG
AB
and
TAB = TAB λ AB = TABB
= (− 0.6i − 0.8 j)TAB
AB
JJJG
AC
TAC = TAC λ AC = TAC
= (0.322432i − 0.756776 j + 0.56757k )TAC
A
AC
JJJG
AD
TAD = TAD λ AD = TAD
= (− 0.886154 j − 0.507769k )TAD
A
AD
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P
PROBLEM
2
2.101
(Con
ntinued)
Equilibrrium condition
n:
ΣF = 0: TABB + TAC + TAD + Pj = 0
Substituuting the expreessions obtaineed for TAB , TAC
nd factoring i,, j, and k:
A , and TAD an
(−
− 0.6TAB + 0.32432TAC )i + (−0.8TAB − 0.75676TAC − 0.886154TAD + P)j
)
+ (0.567577TAC − 0.507699TAD )k = 0
Equatingg to zero the coefficients
c
off i, j, k:
− 0.6TAB
A + 0.32432TAC = 0
(1)
− 0.8TAB − 0.75676TAC − 0.86154TAD + P = 0
(2)
0.56757T
TAC − 0.50769TAD = 0
(3)
Setting TAD = 481 N in (2) and (3),, and solving the
t resulting seet of equations gives
TAC = 4330.26 N
TAD = 2332.57 N
P = 926 N
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PROBLEM 2.102
Three cables are used to tether a balloon as shown. Knowing
that the balloon exerts an 800-N vertical force at A, determine
the tension in each cable.
SOLUTION
See Problem 2.101 for the figure and analysis leading to the linear algebraic Equations (1), (2), and (3).
− 0.6TAB + 0.32432TAC = 0
(1)
− 0.8TAB − 0.75676TAC − 0.86154TAD + P = 0
(2)
0.56757TAC − 0.50769TAD = 0
(3)
From Eq. (1):
TAB = 0.54053TAC
From Eq. (3):
TAD = 1.11795TAC
Substituting for TAB and TAD in terms of TAC into Eq. (2) gives
− 0.8(0.54053TAC ) − 0.75676TAC − 0.86154(1.11795TAC ) + P = 0
2.1523TAC = P ; P = 800 N
800 N
2.1523
= 371.69 N
TAC =
Substituting into expressions for TAB and TAD gives
TAB = 0.54053(371.69 N)
TAD = 1.11795(371.69 N)
TAB = 201 N, TAC = 372 N, TAD = 416 N
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PROBLEM 2.103
A 36-lb triangular plate is supported by three wires as shown. Determine
the tension in each wire, knowing that a = 6 in.
SOLUTION
By Symmetry TDB = TDC
The forces applied at D are:
Free-Body Diagram of Point D:
TDB , TDC , TDA , and P
where P = Pj = (36 lb)j. To express the other forces in terms of the unit vectors i, j, k, we write
JJJG
DA = (16 in.)i − (24 in.) j
DA = 28.844 in.
JJJG
DB = −(8 in.)i − (24 in.) j + (6 in.)k DB = 26.0 in.
JJJG
DC = −(8 in.)i − (24 in.) j − (6 in.)k DC = 26.0 in.
JJJG
DA
and
TDA = TDA λDA = TDA
= (0.55471i − 0.83206 j)TDA
DA
JJJG
DB
TDB = TDB λ DB = TDB
= (−0.30769i − 0.92308 j + 0.23077k )TDB
DB
JJJG
DC
TDC = TDC λ DC = TDC
= (−0.30769i − 0.92308 j − 0.23077k )TDC
DC
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PROBLEM 2.103 (Continued)
Equilibrium condition:
ΣF = 0: TDA + TDB + TDC + (36 lb)j = 0
Substituting the expressions obtained for TDA , TDB , and TDC and factoring i, j, and k:
(0.55471TDA − 0.30769TDB − 0.30769TDC )i + (−0.83206TDA − 0.92308TDB − 0.92308TDC + 36 lb) j
+ (0.23077TDB − 0.23077TDC )k = 0
Equating to zero the coefficients of i, j, k:
0.55471TDA − 0.30769TDB − 0.30769TDC = 0
(1)
− 0.83206TDA − 0.92308TDB − 0.92308TDC + 36 lb = 0
(2)
0.23077TDB − 0.23077TDC = 0
(3)
Equation (3) confirms that TDB = TDC . Solving simultaneously gives,
TDA = 14.42 lb;
TDB = TDC = 13.00 lb
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PROBLEM 2.104
Solve Prob. 2.103, assuming that a = 8 in.
PROBLEM 2.103 A 36-lb triangular plate is supported by three wires
as shown. Determine the tension in each wire, knowing that a = 6 in.
SOLUTION
By Symmetry TDB = TDC
The forces applied at D are:
Free-Body Diagram of Point D:
TDB , TDC , TDA , and P
where P = Pj = (36 lb)j. To express the other forces in terms of the unit vectors i, j, k, we write
JJJG
DA = (16 in.)i − (24 in.) j
DA = 28.844 in.
JJJG
DB = −(8 in.)i − (24 in.) j + (8 in.)k DB = 26.533 in.
JJJG
DC = −(8 in.)i − (24 in.) j − (8 in.)k DC = 26.533 in.
JJJG
DA
and
TDA = TDA λDA = TDA
= (0.55471i − 0.83206 j)TDA
DA
JJJG
DB
TDB = TDB λDB = TDB
= (−0.30151i − 0.90453j + 0.30151k )TDB
DB
JJJG
DC
TDC = TDC λDC = TDC
= (−0.30151i − 0.90453j − 0.30151k )TDC
DC
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PROBLEM 2.104 (Continued)
Equilibrium condition:
ΣF = 0: TDA + TDB + TDC + (36 lb)j = 0
Substituting the expressions obtained for TDA , TDB , and TDC and factoring i, j, and k:
(0.55471TDA − 0.30151TDB − 0.30151TDC )i + (−0.83206TDA − 0.90453TDB − 0.90453TDC + 36 lb) j
+ (0.30151TDB − 0.30151TDC )k = 0
Equating to zero the coefficients of i, j, k:
0.55471TDA − 0.30151TDB − 0.30151TDC = 0
(1)
− 0.83206TDA − 0.90453TDB − 0.90453TDC + 36 lb = 0
(2)
0.30151TDB − 0.30151TDC = 0
(3)
Equation (3) confirms that TDB = TDC . Solving simultaneously gives,
TDA = 14.42 lb;
TDB = TDC = 13.27 lb
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PR
ROBLEM 2.105
2
A crate is suppoorted by threee cables as shhown. Determine
thee weight of thee crate knowinng that the tennsion in cable AC
A
is 544
5 lb.
Solutionn The forces ap
pplied at A aree:
TAB , TAC , TADD and W
where P = Pj. To exp
press the otherr forces in term
ms of the unit vectors i, j, k, we write
JJJG
AB = − (36 in.)i + (60 in.) j − (27 in.)k
AB = 75 in.
JJJG
AC = (60 in.) j + (332 in.)k
AC = 68 in.
JJJG
AD = (40 in.)i + (660 in.) j − (27 inn.)k
AD = 77 in.
JJJG
AB
and
TAB = TAB λAB = TAB
AB
= (− 0.48i + 0.88 j − 0.36k )TABB
JJJG
AC
TAC = TAC λAC = TAC
A
AC
= (0.88235 j + 0.47059k )TACC
JJJG
AD
TAD = TAD λ AD = TAD
A
AD
= (0.51948i + 0.77922 j − 0.335065k )TAD
Equilibrrium Condition
n with
W = − Wj
ΣF = 0: TAB + TAC
A + TAD − Wj = 0
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PROBLEM 2.105 (Continued)
Substituting the expressions obtained for TAB , TAC , and TAD and factoring i, j, and k:
(−0.48TAB + 0.51948TAD )i + (0.8TAB + 0.88235TAC + 0.77922TAD − W ) j
+ (−0.36TAB + 0.47059TAC − 0.35065TAD )k = 0
Equating to zero the coefficients of i, j, k:
− 0.48TAB + 0.51948TAD = 0
(1)
0.8TAB + 0.88235TAC + 0.77922TAD − W = 0
(2)
− 0.36TAB + 0.47059TAC − 0.35065TAD = 0
(3)
Substituting TAC = 544 lb in Equations (1), (2), and (3) above, and solving the resulting set of equations
using conventional algorithms, gives:
TAB = 374.27 lb
TAD = 345.82 lb
W = 1049 lb
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P
PROBLEM
2
2.106
A 1600-lb cratte is supporteed by three cables as show
wn.
Determine the tension
t
in eachh cable.
TION
SOLUT
The forcces applied at A are:
TAB , TAC , TADD and W
where P = Pj. To exp
press the otherr forces in term
ms of the unit vectors i, j, k, we write
JJJG
AB = − (36 in.)i + (60 in.) j − (27 in.)k
AB = 75 in.
JJJG
AC = (60 in.) j + (332 in.)k
AC = 68 in.
JJJG
AD = (40 in.)i + (660 in.) j − (27 inn.)k
AD = 77 in.
JJJG
AB
and
TAB = TAB λAB = TAB
AB
= (− 0.48i + 0.88 j − 0.36k )TABB
JJJG
AC
TAC = TAC λAC = TAC
A
AC
= (0.88235 j + 0.47059k )TACC
JJJG
AD
TAD = TAD λ AD = TAD
A
AD
= (0.51948i + 0.77922 j − 0.335065k )TAD
Equilibrrium Condition
n with
W = − Wj
ΣF = 0: TAB + TAC
A + TAD − Wj = 0
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PROBLEM 2.106 (Continued)
Substituting the expressions obtained for TAB , TAC , and TAD and factoring i, j, and k:
(−0.48TAB + 0.51948TAD )i + (0.8TAB + 0.88235TAC + 0.77922TAD − W ) j
+ (−0.36TAB + 0.47059TAC − 0.35065TAD )k = 0
Equating to zero the coefficients of i, j, k:
−0.48TAB + 0.51948TAD = 0
(1)
0.8TAB + 0.88235TAC + 0.77922TAD − W = 0
(2)
−0.36TAB + 0.47059TAC − 0.35065TAD = 0
(3)
Substituting W = 1600 lb in Equations (1), (2), and (3) above, and solving the resulting set of equations
using conventional algorithms gives,
TAB = 571 lb
TAC = 830 lb
TAD = 528 lb
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PROBLEM 2.107
Three cables are connected at A, where the forces P and Q are
applied as shown. Knowing that Q = 0, find the value of P for
which the tension in cable AD is 305 N.
SOLUTION
ΣFA = 0: TAB + TAC + TAD + P = 0 where P = Pi
JJJG
AB = −(960 mm)i − (240 mm)j + (380 mm)k
AB = 1060 mm
JJJG
AC = −(960 mm)i − (240 mm) j − (320 mm)k AC = 1040 mm
JJJG
AD = −(960 mm)i + (720 mm) j − (220 mm)k AD = 1220 mm
JJJG
AB
19 ⎞
⎛ 48 12
TAB = TAB λAB = TAB
= TAB ⎜ − i − j + k ⎟
AB
53
53 ⎠
⎝ 53
JJJG
AC
3
4 ⎞
⎛ 12
TAC = TAC λAC = TAC
= TAC ⎜ − i − j − k ⎟
AC
⎝ 13 13 13 ⎠
305 N
TAD = TAD λAD =
[(−960 mm)i + (720 mm) j − (220 mm)k ]
1220 mm
= −(240 N)i + (180 N) j − (55 N)k
Substituting into ΣFA = 0, factoring i, j, k , and setting each coefficient equal to φ gives:
i: P =
48
12
TAB + TAC + 240 N
53
13
(1)
j:
12
3
TAB + TAC = 180 N
53
13
(2)
k:
19
4
TAB − TAC = 55 N
53
13
(3)
Solving the system of linear equations using conventional algorithms gives:
TAB = 446.71 N
TAC = 341.71 N
P = 960 N
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PROBLEM 2.108
Three cables are connected at A, where the forces P and Q
are applied as shown. Knowing that P = 1200 N, determine
the values of Q for which cable AD is taut.
SOLUTION
We assume that TAD = 0 and write
ΣFA = 0: TAB + TAC + Qj + (1200 N)i = 0
JJJG
AB = −(960 mm)i − (240 mm)j + (380 mm)k AB = 1060 mm
JJJG
AC = −(960 mm)i − (240 mm) j − (320 mm)k AC = 1040 mm
JJJG
AB ⎛ 48 12
19 ⎞
TAB = TAB λAB = TAB
= ⎜ − i − j + k ⎟ TAB
53 ⎠
AB ⎝ 53 53
JJJG
AC ⎛ 12
3
4 ⎞
TAC = TAC λAC = TAC
= ⎜ − i − j − k ⎟ TAC
AC ⎝ 13 13 13 ⎠
Substituting into ΣFA = 0, factoring i, j, k , and setting each coefficient equal to φ gives:
i: −
48
12
TAB − TAC + 1200 N = 0
53
13
(1)
j: −
12
3
TAB − TAC + Q = 0
53
13
(2)
k:
19
4
TAB − TAC = 0
53
13
(3)
Solving the resulting system of linear equations using conventional algorithms gives:
TAB = 605.71 N
TAC = 705.71 N
Q = 300.00 N
0 ≤ Q < 300 N
Note: This solution assumes that Q is directed upward as shown (Q ≥ 0), if negative values of Q
are considered, cable AD remains taut, but AC becomes slack for Q = −460 N.
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PROB
BLEM 2.109
A rectaangular plate is supportedd by three caables as show
wn.
Knowinng that the teension in cablle AC is 60 N,
N determine the
weight of the plate.
SOLUT
TION
We notee that the weight of the pllate is equal in magnitude to the force P exerted byy the support on
Point A.
Freee Body A :
ΣF = 0: TAB + TAC + TAD + Pj = 0
We havee:
JJJG
AB = −(320 mm)i − (480 mm)j + (360
(
mm)k AB
A = 680 mm
m
JJJG
AC = (4
450 mm)i − (4480 mm) j + (3360 mm)k A
AC = 750 mm
m
JJJG
AD = (2
250 mm)i − (4480 mm) j − ( 3360 mm ) k AD
A = 650 mm
m
Thus:
TAB
TAC
TAD
JJJG
AB ⎛ 8
12
9 ⎞
= ⎜ − i − j + k ⎟ TAB
= TAB λ AB = TAB
A
AB ⎝ 17 17 17 ⎠
JJJG
AC
= ( 0.6i − 0.64 j + 0.488k ) TAC
= TAC λAC = TAC
A
AC
JJJG
9.6
7.2 ⎞
AD ⎛ 5
=⎜ i−
= TAD λAD = TAD
j−
k TAD
A
13
13 ⎟⎠
AD ⎝ 13
Dim
mensions in mm
m
Substituuting into the Eq.
E ΣF = 0 annd factoring i, j, k :
5
⎞
⎛ 8
A ⎟i
⎜ − 17 TAB + 0.6TAC + 13 TAD
⎠
⎝
12
9.6
⎞
⎛
TAD + P ⎟ j
+ ⎜ − TABB − 0.64TAC −
17
13
⎠
⎝
7..2
⎞
⎛ 9
TAD ⎟ k = 0
+ ⎜ TAB + 0.48TAC −
133
⎝ 17
⎠
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PROBLEM 2.109 (Continued)
Setting the coefficient of i, j, k equal to zero:
i:
−
8
5
TAB + 0.6TAC + TAD = 0
17
13
(1)
j:
−
12
9.6
TAB − 0.64TAC −
TAD + P = 0
7
13
(2)
9
7.2
TAB + 0.48TAC −
TAD = 0
17
13
(3)
8
5
TAB + 36 N + TAD = 0
17
13
(1′)
9
7.2
TAB + 28.8 N −
TAD = 0
17
13
(3′)
k:
Making TAC = 60 N in (1) and (3):
−
Multiply (1′) by 9, (3′) by 8, and add:
554.4 N −
12.6
TAD = 0 TAD = 572.0 N
13
Substitute into (1′) and solve for TAB :
TAB =
17 ⎛
5
⎞
⎜ 36 + × 572 ⎟
8 ⎝
13
⎠
TAB = 544.0 N
Substitute for the tensions in Eq. (2) and solve for P:
12
9.6
(544 N) + 0.64(60 N) +
(572 N)
17
13
= 844.8 N
P=
Weight of plate = P = 845 N
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PROBLEM 2.110
A rectangular plate is supported by three cables as shown.
Knowing that the tension in cable AD is 520 N, determine the
weight of the plate.
SOLUTION
See Problem 2.109 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and
(3) below:
8
5
TAB + 0.6TAC + TAD = 0
17
13
(1)
12
9.6
TAB + 0.64 TAC −
TAD + P = 0
17
13
(2)
9
7.2
TAB + 0.48TAC −
TAD = 0
17
13
(3)
−
−
Making TAD = 520 N in Eqs. (1) and (3):
8
TAB + 0.6TAC + 200 N = 0
17
(1′)
9
TAB + 0.48TAC − 288 N = 0
17
(3′)
−
Multiply (1′) by 9, (3′) by 8, and add:
9.24TAC − 504 N = 0 TAC = 54.5455 N
Substitute into (1′) and solve for TAB :
TAB =
17
(0.6 × 54.5455 + 200) TAB = 494.545 N
8
Substitute for the tensions in Eq. (2) and solve for P:
12
9.6
(494.545 N) + 0.64(54.5455 N) +
(520 N)
17
13
= 768.00 N
P=
Weight of plate = P = 768 N
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PROBLEM 2.111
A transmission tower is held by three guy wires attached
to a pin at A and anchored by bolts at B, C, and D. If the
tension in wire AB is 840 lb, determine the vertical force
P exerted by the tower on the pin at A.
SOLUTION
ΣF = 0: TAB + TAC + TAD + Pj = 0
We write
Free-Body Diagram at A:
JJJG
AB = −20i − 100 j + 25k AB = 105 ft
JJJG
AC = 60i − 100 j + 18k AC = 118 ft
JJJG
AD = −20i − 100 j − 74k AD = 126 ft
JJJG
AB
TAB = TAB λ AB = TAB
AB
20
5 ⎞
⎛ 4
j + k ⎟ TAB
= ⎜− i −
21 ⎠
⎝ 21 21
JJJG
AC
TAC = TAC λAC = TAC
AC
50
9 ⎞
⎛ 30
j + k ⎟ TAC
=⎜ i−
59
59 ⎠
⎝ 59
JJJG
AD
TAD = TAD λAD = TAD
AD
50
37 ⎞
⎛ 10
j − k ⎟ TAD
= ⎜− i −
63
63
63 ⎠
⎝
Substituting into the Eq. ΣF = 0 and factoring i, j, k :
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PROBLEM 2.111 (Continued)
30
10
⎛ 4
⎞
⎜ − 21 TAB + 59 TAC − 63 TAD ⎟ i
⎝
⎠
50
50
⎛ 20
⎞
+ ⎜ − TAB − TAC − TAD + P ⎟ j
59
63
⎝ 21
⎠
9
37
⎛ 5
⎞
+ ⎜ TAB + TAC − TAD ⎟ k = 0
21
59
63
⎝
⎠
Setting the coefficients of i, j, k , equal to zero:
i:
−
4
30
10
TAB + TAC − TAD = 0
21
59
63
(1)
j:
−
20
50
50
TAB − TAC − TAD + P = 0
21
59
63
(2)
5
9
37
TAB + TAC − TAD = 0
21
59
63
(3)
k:
Set TAB = 840 lb in Eqs. (1) – (3):
30
10
TAC − TAD = 0
59
63
(1′)
50
50
TAC − TAD + P = 0
59
63
(2′)
9
37
TAC − TAD = 0
59
63
(3′)
−160 lb +
−800 lb −
200 lb +
Solving,
TAC = 458.12 lb TAD = 459.53 lb
P = 1552.94 lb
P = 1553 lb
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PROBLEM 2.112
A transmission tower is held by three guy wires attached
to a pin at A and anchored by bolts at B, C, and D. If the
tension in wire AC is 590 lb, determine the vertical force
P exerted by the tower on the pin at A.
SOLUTION
ΣF = 0: TAB + TAC + TAD + Pj = 0
We write
Free-Body Diagram at A:
JJJG
AB = −20i − 100 j + 25k AB = 105 ft
JJJG
AC = 60i − 100 j + 18k AC = 118 ft
JJJG
AD = −20i − 100 j − 74k AD = 126 ft
JJJG
AB
TAB = TAB λ AB = TAB
AB
20
5 ⎞
⎛ 4
= ⎜− i −
j + k ⎟ TAB
21 ⎠
⎝ 21 21
JJJG
AC
TAC = TAC λAC = TAC
AC
30
50
9 ⎞
⎛
=⎜ i−
j + k ⎟ TAC
59
59 ⎠
⎝ 59
JJJG
AD
TAD = TAD λAD = TAD
AD
50
37 ⎞
⎛ 10
= ⎜− i −
j − k ⎟ TAD
63 ⎠
⎝ 63 63
Substituting into the Eq. ΣF = 0 and factoring i, j, k :
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PROBLEM 2.112 (Continued)
30
10
⎛ 4
⎞
⎜ − 21 TAB + 59 TAC − 63 TAD ⎟ i
⎝
⎠
50
50
⎛ 20
⎞
+ ⎜ − TAB − TAC − TAD + P ⎟ j
21
59
63
⎝
⎠
9
37
⎛ 5
⎞
+ ⎜ TAB + TAC − TAD ⎟ k = 0
59
63
⎝ 21
⎠
Setting the coefficients of i, j, k , equal to zero:
i:
−
4
30
10
TAB + TAC − TAD = 0
21
59
63
(1)
j:
−
20
50
50
TAB − TAC − TAD + P = 0
21
59
63
(2)
5
9
37
TAB + TAC − TAD = 0
21
59
63
(3)
k:
Set TAC = 590 lb in Eqs. (1) – (3):
4
10
TAB + 300 lb − TAD = 0
21
63
(1′)
20
50
TAB − 500 lb − TAD + P = 0
21
63
(2′)
−
−
5
37
TAB + 90 lb − TAD = 0
21
63
Solving,
TAB = 1081.82 lb TAD = 591.82 lb
(3′)
P = 2000 lb
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PROBLEM 2.113
In trying to move across a slippery icy surface, a 175-lb man uses
two ropes AB and AC. Knowing that the force exerted on the man
by the icy surface is perpendicular to that surface, determine the
tension in each rope.
SOLUTION
Free-Body Diagram at A
30 ⎞
⎛ 16
N= N⎜ i+
j⎟
34 ⎠
⎝ 34
and W = W j = −(175 lb) j
TAC = TAC λ AC = TAC
JJJG
AC
(−30 ft)i + (20 ft) j − (12 ft)k
= TAC
AC
38 ft
6 ⎞
⎛ 15 10
= TAC ⎜ − i + j − k ⎟
⎝ 19 19 19 ⎠
TAB = TAB λ AB = TAB
JJJG
AB
(−30 ft)i + (24 ft) j + (32 ft)k
= TAB
AB
50 ft
12
16 ⎞
⎛ 15
= TAB ⎜ − i +
j+ k⎟
25
25 ⎠
⎝ 25
Equilibrium condition: ΣF = 0
TAB + TAC + N + W = 0
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PROBLEM 2.113 (Continued)
Substituting the expressions obtained for TAB , TAC , N, and W; factoring i, j, and k; and equating each of
the coefficients to zero gives the following equations:
15
15
16
TAB − TAC +
N =0
25
19
34
(1)
From j:
12
10
30
TAB + TAC +
N − (175 lb) = 0
25
19
34
(2)
From k:
16
6
TAB − TAC = 0
25
19
(3)
From i:
−
Solving the resulting set of equations gives:
TAB = 30.8 lb; TAC = 62.5 lb
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PROBLEM 2.114
Solve Problem 2.113, assuming that a friend is helping the man
at A by pulling on him with a force P = −(45 lb)k.
PROBLEM 2.113 In trying to move across a slippery icy
surface, a 175-lb man uses two ropes AB and AC. Knowing that
the force exerted on the man by the icy surface is perpendicular
to that surface, determine the tension in each rope.
SOLUTION
Refer to Problem 2.113 for the figure and analysis leading to the following set of equations, Equation (3)
being modified to include the additional force P = (−45 lb)k.
15
15
16
TAB − TAC +
N =0
25
19
34
(1)
12
10
30
TAB + TAC +
N − (175 lb) = 0
25
19
34
(2)
16
6
TAB − TAC − (45 lb) = 0
25
19
(3)
−
Solving the resulting set of equations simultaneously gives:
TAB = 81.3 lb
TAC = 22.2 lb
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PROBLEM 2.115
For the rectangular plate of Problems 2.109 and 2.110,
determine the tension in each of the three cables knowing that
the weight of the plate is 792 N.
SOLUTION
See Problem 2.109 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and
(3) below. Setting P = 792 N gives:
8
5
TAB + 0.6TAC + TAD = 0
17
13
(1)
12
9.6
TAB − 0.64TAC −
TAD + 792 N = 0
17
13
(2)
9
7.2
TAB + 0.48TAC −
TAD = 0
17
13
(3)
−
−
Solving Equations (1), (2), and (3) by conventional algorithms gives
TAB = 510.00 N
TAB = 510 N
TAC = 56.250 N
TAC = 56.2 N
TAD = 536.25 N
TAD = 536 N
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PROBLEM 2.116
For the cable system of Problems 2.107 and 2.108,
determine the tension in each cable knowing that
P = 2880 N and Q = 0.
SOLUTION
ΣFA = 0: TAB + TAC + TAD + P + Q = 0
Where
P = Pi and Q = Qj
JJJG
AB = −(960 mm)i − (240 mm) j + (380 mm)k AB = 1060 mm
JJJG
AC = −(960 mm)i − (240 mm) j − (320 mm)k AC = 1040 mm
JJJG
AD = −(960 mm)i + (720 mm) j − (220 mm)k AD = 1220 mm
JJJG
AB
19 ⎞
⎛ 48 12
TAB = TAB λAB = TAB
= TAB ⎜ − i − j + k ⎟
AB
53 ⎠
⎝ 53 53
JJJG
3
4 ⎞
AC
⎛ 12
TAC = TAC λAC = TAC
= TAC ⎜ − i − j − k ⎟
AC
⎝ 13 13 13 ⎠
JJJG
AD
36
11 ⎞
⎛ 48
TAD = TAD λAD = TAD
j− k⎟
= TAD ⎜ − i +
AD
61
61 ⎠
⎝ 61
Substituting into ΣFA = 0, setting P = (2880 N)i and Q = 0, and setting the coefficients of i, j, k equal to
0, we obtain the following three equilibrium equations:
i: −
48
12
48
TAB − TAC − TAD + 2880 N = 0
53
13
61
(1)
j: −
12
3
36
TAB − TAC + TAD = 0
53
13
61
(2)
k:
19
4
11
TAB − TAC − TAD = 0
53
13
61
(3)
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PROBLEM 2.116 (Continued)
Solving the system of linear equations using conventional algorithms gives:
TAB = 1340.14 N
TAC = 1025.12 N
TAD = 915.03 N
TAB = 1340 N
TAC = 1025 N
TAD = 915 N
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PROBLEM 2.117
For the cable system of Problems 2.107 and 2.108,
determine the tension in each cable knowing that
P = 2880 N and Q = 576 N.
SOLUTION
See Problem 2.116 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below:
−
48
12
48
TAB − TAC − TAD + P = 0
53
13
61
(1)
−
12
3
36
TAB − TAC + TAD + Q = 0
53
13
61
(2)
19
4
11
TAB − TAC − TAD = 0
53
13
61
(3)
Setting P = 2880 N and Q = 576 N gives:
−
48
12
48
TAB − TAC − TAD + 2880 N = 0
53
13
61
(1′)
12
3
36
TAB − TAC + TAD + 576 N = 0
53
13
61
(2′)
19
4
11
TAB − TAC − TAD = 0
53
13
61
(3′)
−
Solving the resulting set of equations using conventional algorithms gives:
TAB = 1431.00 N
TAC = 1560.00 N
TAD = 183.010 N
TAB = 1431 N
TAC = 1560 N
TAD = 183.0 N
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PROBLEM 2.118
For the cable system of Problems 2.107 and 2.108,
determine the tension in each cable knowing that
P = 2880 N and Q = −576 N. (Q is directed downward).
SOLUTION
See Problem 2.116 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below:
−
48
12
48
TAB − TAC − TAD + P = 0
53
13
61
(1)
−
12
3
36
TAB − TAC + TAD + Q = 0
53
13
61
(2)
19
4
11
TAB − TAC − TAD = 0
53
13
61
(3)
Setting P = 2880 N and Q = −576 N gives:
−
48
12
48
TAB − TAC − TAD + 2880 N = 0
53
13
61
(1′)
12
3
36
TAB − TAC + TAD − 576 N = 0
53
13
61
(2′)
19
4
11
TAB − TAC − TAD = 0
53
13
61
(3′)
−
Solving the resulting set of equations using conventional algorithms gives:
TAB = 1249.29 N
TAC = 490.31 N
TAD = 1646.97 N
TAB = 1249 N
TAC = 490 N
TAD = 1647 N
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PROBLEM 2.119
For the transmission tower of Probs. 2.111 and 2.112,
determine the tension in each guy wire knowing that the
tower exerts on the pin at A an upward vertical force of
1800 lb.
PROBLEM 2.111 A transmission tower is held by three
guy wires attached to a pin at A and anchored by bolts at
B, C, and D. If the tension in wire AB is 840 lb, determine
the vertical force P exerted by the tower on the pin at A.
SOLUTION
See Problem 2.111 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and
(3) below:
i:
−
4
30
10
TAB + TAC − TAD = 0
21
59
63
(1)
j:
−
20
50
50
TAB − TAC − TAD + P = 0
21
59
63
(2)
5
9
37
TAB + TAC − TAD = 0
21
59
63
(3)
k:
Substituting for P = 1800 lb in Equations (1), (2), and (3) above and solving the resulting set of equations
using conventional algorithms gives:
4
30
10
TAB + TAC − TAD = 0
21
59
63
(1′)
20
50
50
TAB − TAC − TAD + 1800 lb = 0
21
59
63
(2′)
5
9
37
TAB + TAC − TAD = 0
21
59
63
(3′)
−
−
TAB = 973.64 lb
TAC = 531.00 lb
TAD = 532.64 lb
TAB = 974 lb
TAC = 531 lb
TAD = 533 lb
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PROBLEM 2.120
Three wires are connected at point D, which is
located 18 in. below the T-shaped pipe support
ABC. Determine the tension in each wire when
a 180-lb cylinder is suspended from point D as
shown.
SOLUTION
Free-Body Diagram of Point D:
The forces applied at D are:
TDA , TDB , TDC and W
where W = −180.0 lbj. To express the other forces in terms of the unit vectors i, j, k, we write
JJJG
DA = (18 in.) j + (22 in.)k
DA = 28.425 in.
JJJG
DB = −(24 in.)i + (18 in.) j − (16 in.)k
DB = 34.0 in.
JJJG
DC = (24 in.)i + (18 in.) j − (16 in.)k
DC = 34.0 in.
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PROBLEM 2.120 (Continued)
and
JJJG
DA
DA
= (0.63324 j + 0.77397k )TDA
JJJG
DB
= TDB λDB = TDB
DB
= (−0.70588i + 0.52941j − 0.47059k )TDB
JJJG
DC
= TDC λDC = TDC
DC
= (0.70588i + 0.52941j − 0.47059k )TDC
TDA = TDa λDA = TDa
TDB
TDC
Equilibrium Condition with
W = − Wj
ΣF = 0: TDA + TDB + TDC − Wj = 0
Substituting the expressions obtained for TDA , TDB , and TDC and factoring i, j, and k:
(−0.70588TDB + 0.70588TDC )i
(0.63324TDA + 0.52941TDB + 0.52941TDC − W ) j
(0.77397TDA − 0.47059TDB − 0.47059TDC )k
Equating to zero the coefficients of i, j, k:
−0.70588TDB + 0.70588TDC = 0
(1)
0.63324TDA + 0.52941TDB + 0.52941TDC − W = 0
(2)
0.77397TDA − 0.47059TDB − 0.47059TDC = 0
(3)
Substituting W = 180 lb in Equations (1), (2), and (3) above, and solving the resulting set of equations using
conventional algorithms gives,
TDA = 119.7 lb
TDB = 98.4 lb
TDC = 98.4 lb
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PROBLEM 2.121
A container of weight W is suspended from ring
A, to which cables AC and AE are attached. A
force P is applied to the end F of a third cable
that passes over a pulley at B and through ring A
and that is attached to a support at D. Knowing
that W = 1000 N, determine the magnitude of P.
(Hint: The tension is the same in all portions of
cable FBAD.)
SOLUTION
The (vector) force in each cable can be written as the product of the (scalar) force and the unit vector
along the cable. That is, with
JJJG
AB = −(0.78 m)i + (1.6 m) j + (0 m)k
AB = (−0.78 m) 2 + (1.6 m) 2 + (0) 2
= 1.78 m
TAB = T λ AB = TAB
JJJG
AB
AB
TAB
[−(0.78 m)i + (1.6 m) j + (0 m)k ]
1.78 m
= TAB (−0.4382i + 0.8989 j + 0k )
=
and
TAB
JJJG
AC = (0)i + (1.6 m) j + (1.2 m)k
and
AC = (0 m)2 + (1.6 m) 2 + (1.2 m) 2 = 2 m
JJJG
AC TAC
TAC = T λAC = TAC
=
[(0)i + (1.6 m) j + (1.2 m)k ]
AC 2 m
TAC = TAC (0.8 j + 0.6k )
JJJG
AD = (1.3 m)i + (1.6 m) j + (0.4 m)k
AD = (1.3 m) 2 + (1.6 m)2 + (0.4 m)2 = 2.1 m
JJJG
T
AD
= AD [(1.3 m)i + (1.6 m) j + (0.4 m)k ]
TAD = T λAD = TAD
AD 2.1 m
TAD = TAD (0.6190i + 0.7619 j + 0.1905k )
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PROBLEM 2.121 (Continued)
JJJG
AE = −(0.4 m)i + (1.6 m) j − (0.86 m)k
Finally,
AE = (−0.4 m) 2 + (1.6 m) 2 + (−0.86 m) 2 = 1.86 m
JJJG
AE
TAE = T λAE = TAE
AE
TAE
[−(0.4 m)i + (1.6 m) j − (0.86 m)k ]
=
1.86 m
TAE = TAE (−0.2151i + 0.8602 j − 0.4624k )
With the weight of the container
W = −W j, at A we have:
ΣF = 0: TAB + TAC + TAD − Wj = 0
Equating the factors of i, j, and k to zero, we obtain the following linear algebraic equations:
−0.4382TAB + 0.6190TAD − 0.2151TAE = 0
(1)
0.8989TAB + 0.8TAC + 0.7619TAD + 0.8602TAE − W = 0
(2)
0.6TAC + 0.1905TAD − 0.4624TAE = 0
(3)
Knowing that W = 1000 N and that because of the pulley system at B TAB = TAD = P, where P is the
externally applied (unknown) force, we can solve the system of linear Equations (1), (2) and (3) uniquely
for P.
P = 378 N
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PROBLEM 2.122
Knowing that the tension in cable AC of the
system described in Problem 2.121 is 150 N,
determine (a) the magnitude of the force P, (b)
the weight W of the container.
PROBLEM 2.121 A container of weight W is
suspended from ring A, to which cables AC and
AE are attached. A force P is applied to the end F
of a third cable that passes over a pulley at B and
through ring A and that is attached to a support at
D. Knowing that W = 1000 N, determine the
magnitude of P. (Hint: The tension is the same in
all portions of cable FBAD.)
SOLUTION
Here, as in Problem 2.121, the support of the container consists of the four cables AE, AC, AD, and AB,
with the condition that the force in cables AB and AD is equal to the externally applied force P. Thus,
with the condition
TAB = TAD = P
and using the linear algebraic equations of Problem 2.131 with TAC = 150 N, we obtain
(a)
P = 454 N
(b)
W = 1202 N
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PROBLEM 2.123
Cable BAC passes through a frictionless ring A and is
attached to fixed supports at B and C, while cables AD and
AE are both tied to the ring and are attached, respectively, to
supports at D and E. Knowing that a 200-lb vertical load P is
applied to ring A, determine the tension in each of the three
cables.
SOLUTION
Free Body Diagram at A:
Since TBAC = tension in cable BAC, it follows that
TAB = TAC = TBAC
TAB = TBAC λ AB = TBAC
(−17.5 in.)i + (60 in.) j
60 ⎞
⎛ −17.5
i+
j
= TBAC ⎜
62.5 in.
62.5 ⎟⎠
⎝ 62.5
(60 in.)i + (25 in.)k
25 ⎞
⎛ 60
= TBAC ⎜ j + k ⎟
65 in.
65
65
⎝
⎠
(80 in.)i + (60 in.) j
3 ⎞
⎛4
= TAD ⎜ i + j ⎟
100 in.
5
5
⎝
⎠
TAC = TBAC λ AC = TBAC
TAD = TAD λ AD = TAD
TAE = TAE λ AE = TAE
(60 in.) j − (45 in.)k
3 ⎞
⎛4
= TAE ⎜ j − k ⎟
75 in.
5 ⎠
⎝5
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PROBLEM 2.123 (Continued)
Substituting into ΣFA = 0, setting P = (−200 lb) j, and setting the coefficients of i, j, k equal to φ , we
obtain the following three equilibrium equations:
17.5
4
TBAC + TAD = 0
62.5
5
From
i: −
From
3
4
⎛ 60 60 ⎞
j: ⎜
+ ⎟ TBAC + TAD + TAE − 200 lb = 0
5
5
⎝ 62.5 65 ⎠
From
k:
(1)
25
3
TBAC − TAE = 0
65
5
(2)
(3)
Solving the system of linear equations using conventional algorithms gives:
TBAC = 76.7 lb; TAD = 26.9 lb; TAE = 49.2 lb
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PROBLEM 2.124
Knowing that the tension in cable AE of Prob. 2.123 is 75 lb,
determine (a) the magnitude of the load P, (b) the tension in
cables BAC and AD.
PROBLEM 2.123 Cable BAC passes through a frictionless
ring A and is attached to fixed supports at B and C, while
cables AD and AE are both tied to the ring and are attached,
respectively, to supports at D and E. Knowing that a 200-lb
vertical load P is applied to ring A, determine the tension in
each of the three cables.
SOLUTION
Refer to the solution to Problem 2.123 for the figure and analysis leading to the following set of
equilibrium equations, Equation (2) being modified to include Pj as an unknown quantity:
−
17.5
4
TBAC + TAD = 0
62.5
5
(1)
3
4
⎛ 60 60 ⎞
+ ⎟ TBAC + TAD + TAE − P = 0
⎜
5
5
⎝ 62.5 65 ⎠
25
3
TBAC − TAE = 0
65
5
(2)
(3)
Substituting for TAE = 75 lb and solving simultaneously gives:
(a)
P = 305 lb
(b)
TBAC = 117.0 lb; TAD = 40.9 lb
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P
PROBLEM
2
2.125
Collars A and B are connecteed by a 525-m
mm-long wire and
a
caan slide freeely on fricctionless rodds. If a force
P = (341 N)j is
i applied to collar A, deetermine (a) the
teension in the wire
w when y = 155 mm, (bb) the magnituude
off the force Q required to maintain
m
the eqquilibrium of the
syystem.
TION
SOLUT
For bothh Problems 2.125 and 2.1266:
Freee-Body Diagrrams of Collars:
( AB)2 = x2 + y 2 + z 2
Here
(00.525 m)2 = (00.20 m)2 + y 2 + z 2
y 2 + z 2 = 0..23563 m2
or
Thus, when
w
y given, z is determinedd,
Now
λAB
JJJG
AB
=
AB
1
(0.20i − yj + zk )m
0.525 m
= 0.38095i − 1.90476 yj + 1.90476 zk
=
Where y and z are in units
u
of meterrs, m.
From thhe F.B. Diagraam of collar A::
ΣF = 0:: N x i + N z k + Pj + TAB λ AB = 0
Setting the
t j coefficieent to zero givees
P − (1.990476 y )TAB = 0
P = 341 N
With
TAB =
341 N
1.90476 y
Now, frrom the free bo
ody diagram of
o collar B:
ΣF = 0: N x i + N y j + Qk − TAB λAB = 0
Setting the
t k coefficieent to zero givves
Q − TAB (1.90476 z ) = 0
And usiing the above result
r
for TAB , we have
Q = TAB z =
341 N
(3441 N)( z )
(1..90476 z ) =
y
(
(1.90476)
y
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PROBLEM 2.125 (Continued)
Then from the specifications of the problem, y = 155 mm = 0.155 m
z 2 = 0.23563 m 2 − (0.155 m)2
z = 0.46 m
and
341 N
0.155(1.90476)
= 1155.00 N
TAB =
(a)
TAB = 1155 N
or
and
Q=
(b)
341 N(0.46 m)(0.866)
(0.155 m)
= (1012.00 N)
Q = 1012 N
or
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PROBLEM 2.126
Solve Problem 2.125 assuming that y = 275 mm.
PROBLEM 2.125 Collars A and B are connected by a
525-mm-long wire and can slide freely on frictionless
rods. If a force P = (341 N)j is applied to collar A,
determine (a) the tension in the wire when y = 155 mm,
(b) the magnitude of the force Q required to maintain the
equilibrium of the system.
SOLUTION
From the analysis of Problem 2.125, particularly the results:
y 2 + z 2 = 0.23563 m 2
341 N
1.90476 y
341 N
Q=
z
y
TAB =
With y = 275 mm = 0.275 m, we obtain:
z 2 = 0.23563 m 2 − (0.275 m)2
z = 0.40 m
and
TAB =
(a)
341 N
= 651.00
(1.90476)(0.275 m)
TAB = 651 N
or
and
Q=
(b)
341 N(0.40 m)
(0.275 m)
Q = 496 N
or
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PROB
BLEM 2.12
27
Two sttructural mem
mbers A and B are bolted to a bracket as
shown.. Knowing thhat both mem
mbers are in compression
c
a
and
that thee force is 15 kN in membeer A and 10 kN
k in memberr B,
determ
mine by trigonometry the magnitude
m
and direction of the
resultannt of the forcees applied to the
t bracket byy members A and
a
B.
TION
SOLUT
Using thhe force triang
gle and the law
ws of cosines and
a sines,
we havee
γ = 180° − (400° + 20°)
= 120°
Then
R 2 = (15 kN)2 + (10 kN) 2
− 2(15 kN
N)(10 kN) cos120°
= 475 kN 2
R = 21.794 kN
N
and
Hence:
10 kN
N 21.794 kN
N
=
sin α
sin120°
N ⎞
⎛ 10 kN
sin120°
sin α = ⎜
21.794
k ⎟⎠
kN
⎝
= 0.39737
α = 23.414
φ = α + 50° = 73.414
R = 21.88 kN
73.4°
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PROBLEM 2.128
Determine the x and y components of each of the forces shown.
SOLUTION
Compute the following distances:
OA = (24 in.)2 + (45 in.) 2
= 51.0 in.
OB = (28 in.)2 + (45 in.)2
= 53.0 in.
OC = (40 in.) 2 + (30 in.) 2
= 50.0 in.
102-lb Force:
106-lb Force:
200-lb Force:
Fx = −102 lb
24 in.
51.0 in.
Fx = −48.0 lb
Fy = +102 lb
45 in.
51.0 in.
Fy = +90.0 lb
Fx = +106 lb
28 in.
53.0 in.
Fx = +56.0 lb
Fy = +106 lb
45 in.
53.0 in.
Fy = +90.0 lb
Fx = −200 lb
40 in.
50.0 in.
Fx = −160.0 lb
Fy = −200 lb
30 in.
50.0 in.
Fy = −120.0 lb
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PROBLEM
M 2.129
A hoist trollley is subjecteed to the threee forces show
wn. Knowing that
t
α = 40°, determine (a) thhe required maagnitude of thhe force P if the
resultant of the three forcces is to be vertical,
v
(b) thhe correspondiing
magnitude of the resultantt.
SOLUT
TION
Rx =
ΣFx = P + (200 lb)sin 40° − (400 lb) cos 40°
4
Rx = P − 177.860
1
lb
Ry =
(1)
ΣFy = (200 lb) cos
c 40° + (400 lb)sin 40°
Ry = 410.32 lb
(a)
Foor R to be verrtical, we mustt have Rx = 0..
Rx = 0 in Eq.
E (1)
Set
0 = P − 1777.860 lb
P = 177.8660 lb
(b)
(2)
P = 177.9 lb
Siince R is to bee vertical:
R = Ry = 410 lb
R = 410 lb
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PRO
OBLEM 2.130
Know
wing that α = 55° and that boom AC exerts on pin C a force directed
alongg line AC, deteermine (a) thee magnitude of
o that force, (b)
( the tensionn in
cable BC.
TION
SOLUT
Freee-Body Diagraam
s
Law of sines:
Force Trianggle
FAC
T
300 lb
= BC =
sinn 35° sin 500° sin 95°
(a)
FAC =
300 lbb
sin 35°
sin 955°
FAC = 172.7 lb
(b)
TBC =
300 lbb
sin 50°
sin 955°
TBC = 231 lb
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PR
ROBLEM 2.131
Twoo cables are tied together at C and looaded as show
wn.
Knoowing that P = 360 N, determ
mine the tenssion (a) in caable
AC, (b) in cable BC.
B
SOLUT
TION
Free Body: C
(a)
ΣFx = 0: −
(b)
ΣFy = 0:
12
4
TAC + (3360 N) = 0
13
5
TAC = 312 N
5
3
N − 480 N = 0
(312 N) + TBC + (360 N)
113
5
TBC = 480 N − 120 N − 216 N
TBC = 144.0 N
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PROBLEM 2.132
2
Two cabbles tied togethher at C are looaded as show
wn. Knowing that
t
the maxximum allow
wable tensionn in each caable is 800 N,
determinne (a) the maagnitude of thhe largest forcce P that can be
applied at
a C, (b) the coorresponding value of α.
TION
SOLUT
Free-Body Diagram:
D
C
Force Trian
ngle
Force triiangle is isoscceles with
2β = 1880° − 85°
β = 47.5°
P = 2(800
2
N)cos 477.5° = 1081 N
(a)
he solution is correct.
c
Since P > 0, th
(b)
P = 1081 N
α = 180° − 50° − 477.5° = 82.5°
α = 82.5°
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PROBLEM 2.133
The end of the coaxial cable AE is attached to the pole AB,
which is strengthened by the guy wires AC and AD. Knowing
that the tension in wire AC is 120 lb, determine (a) the
components of the force exerted by this wire on the pole, (b)
the angles θx, θy, and θz that the force forms with the
coordinate axes.
SOLUTION
(a)
Fx = (120 lb) cos 60° cos 20°
Fx = 56.382 lb
Fx = +56.4 lb
Fy = −(120 lb) sin 60°
Fy = −103.923 lb
Fy = −103.9 lb
Fz = −(120 lb) cos 60° sin 20°
Fz = −20.521 lb
(b)
cos θ x =
cos θ y =
cos θ z =
Fx 56.382 lb
=
F
120 lb
Fy
F
=
−103.923 lb
120 lb
Fz −20.52 lb
=
F
120 lb
Fz = −20.5 lb
θ x = 62.0°
θ y = 150.0°
θ z = 99.8°
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PROBLEM 2.134
Knowing that the tension in cable AC is 2130 N, determine
the components of the force exerted on the plate at C.
SOLUTION
JJJG
CA = −(900 mm)i + (600 mm) j − (920 mm)k
CA = (900 mm)2 + (600 mm)2 + (920 mm)2
TCA
TCA
= 1420 mm
= TCA λ CA
JJJG
CA
= TCA
CA
2130 N
[−(900 mm)i + (600 mm) j − (920 mm)k ]
=
1420 mm
= −(1350 N)i + (900 N) j − (1380 N)k
(TCA ) x = −1350 N, (TCA ) y = 900 N, (TCA ) z = −1380 N
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PROBLEM 2.135
Find the magnitude and direction of the resultant of the two forces
shown knowing that P = 600 N and Q = 450 N.
SOLUTION
P = (600 N)[sin 40° sin 25°i + cos 40° j + sin 40° cos 25°k ]
= (162.992 N)i + (459.63 N) j + (349.54 N)k
Q = (450 N)[cos 55° cos 30°i + sin 55° j − cos 55° sin 30°k ]
= (223.53 N)i + (368.62 N) j − (129.055 N)k
R =P+Q
= (386.52 N)i + (828.25 N) j + (220.49 N)k
R = (386.52 N) 2 + (828.25 N) 2 + (220.49 N) 2
= 940.22 N
cos θ x =
cosθ y =
cos θ z =
Rx 386.52 N
=
R 940.22 N
Ry
R = 940 N
θ x = 65.7°
828.25 N
940.22 N
θ y = 28.2°
Rz 220.49 N
=
R 940.22 N
θ z = 76.4°
R
=
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PROBLE
EM 2.136
A containeer of weight W is suspendded from ring A.
Cable BAC
C passes throuugh the ring annd is attachedd to
fixed suppports at B andd C. Two forrces P = Pi and
a
Q = Qk are
a applied to
t the ring to
t maintain the
container in
i the positioon shown. Knowing
K
that W
= 376 N, determine P and Q. (Hintt: The tensionn is
the same inn both portionss of cable BAC
C.)
TION
SOLUT
TAB = T λ AB
JJJK
AB
=T
AB
(−130 mm)i + (4000 mm) j + (1600 mm)k
=T
4500 mm
40
16 ⎞
⎛ 133
j+ k⎟
=T ⎜− i +
45
45 ⎠
⎝ 455
Free-Bod
dy A:
TAC = T λ AC
JJJK
AC
=T
AC
(−150 mm)i + (4000 mm) j + (−2440 mm)k
=T
4990 mm
40
24 ⎞
⎛ 155
j− k⎟
=T ⎜− i +
49
49 ⎠
⎝ 499
ΣF = 0: TABB + TAC + Q + P + W = 0
t zero:
Setting coefficients off i, j, k equal to
i: −
13
15
T − T +P=0
45
49
0.59501T = P
(1)
j: +
40
40
T + T −W = 0
45
49
1.70521T = W
(2)
k: +
16
24
T − T +Q =0
45
49
0.1342440T = Q
(3)
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PROBLEM 2.136 (Continued)
Data:
W = 376 N 1.70521T = 376 N T = 220.50 N
0.59501(220.50 N) = P
P = 131.2 N
0.134240(220.50 N) = Q
Q = 29.6 N
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PROBLEM 2.137
7
Collars A and B are coonnected by a 25-in.-long wire
w and can slide
freely onn frictionless rods. If a 60-llb force Q is applied to colllar
B as shhown, determ
mine (a) the tension in the wire whhen
x = 9 in.., (b) the coorresponding magnitude of
o the force P
requiredd to maintain thhe equilibrium
m of the system
m.
SOLUT
TION
Free-Body Diagrams
D
of Collars:
C
A:
B:
λAB
JJJG
AB − xi − (20 in.) j + zk
=
=
25 in.
AB
ΣF = 0: Pi + N y j + N z k + TAB λ AB = 0
A
Collar A:
Substituute for λAB an
nd set coefficieent of i equal to
t zero:
P−
TAB x
=0
25 in.
(1)
ΣF = 0: (660 lb)k + N x′ i + N y′ j − TAB λ ABB = 0
B
Collar B:
Substituute for λ AB and
d set coefficiennt of k equal too zero:
60 lb −
x = 9 in.
(a)
TAB z
=0
25 in.
(2)
(9 in.) 2 + (20 in.) 2 + z 2 = (25 in.) 2
z = 12 in.
Frrom Eq. (2):
(b)
F
From
Eq. (1):
60 lb − TAAB (12 in.)
25 in.
P=
TAB = 125.0 lb
(1255.0 lb)(9 in.)
25 in.
P = 45.0 lb
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PROBLEM 2.138
Collars A and B are connected by a 25-in.-long wire and can slide
freely on frictionless rods. Determine the distances x and z for
which the equilibrium of the system is maintained when
P = 120 lb and Q = 60 lb.
SOLUTION
See Problem 2.137 for the diagrams and analysis leading to Equations (1) and (2) below:
P=
TAB x
=0
25 in.
(1)
60 lb −
TAB z
=0
25 in.
(2)
For P = 120 lb, Eq. (1) yields
TAB x = (25 in.)(20 lb)
(1′)
From Eq. (2):
TAB z = (25 in.)(60 lb)
(2′)
x
=2
z
Dividing Eq. (1′) by (2′),
Now write
(3)
x 2 + z 2 + (20 in.)2 = (25 in.)2
(4)
Solving (3) and (4) simultaneously,
4 z 2 + z 2 + 400 = 625
z 2 = 45
z = 6.7082 in.
From Eq. (3):
x = 2 z = 2(6.7082 in.)
= 13.4164 in.
x = 13.42 in., z = 6.71 in.
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PROBLEM 2F1
Two cables are tied together at C and loaded as shown.
Draw the free-body diagram needed to determine the
tension in AC and BC.
SOLUTION
Free-Body Diagram of Point C:
W = (1600 kg)(9.81 m/s 2 )
W = 15.6960(103 ) N
W =15.696 kN
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PROBLEM 2.F2
Two forces of magnitude TA = 8 kips and TB = 15 kips are
applied as shown to a welded connection. Knowing that the
connection is in equilibrium, draw the free-body diagram
needed to determine the magnitudes of the forces TC
and TD.
SOLUTION
Free-Body Diagram of Point E:
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PROBLEM 2.F3
The 60-lb collar A can slide on a frictionless vertical rod and is connected
as shown to a 65-lb counterweight C. Draw the free-body diagram needed
to determine the value of h for which the system is in equilibrium.
SOLUTION
Free-Body Diagram of Point A:
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P
PROBLEM
2
2.F4
A chairlift has been stoppedd in the positiion
shhown. Knowinng that each chair
c
weighs 250
2
N and that the skier
s
in chair E weighs 765 N,
drraw the freee-body diagraams needed to
deetermine the weight
w
of the skier
s
in chair F.
F
SOLUT
TION
Free-Boody Diagram of Point B:
WE = 250 N + 765 N = 10115 N
88.25
= 30.510°
14
1
10
= tan −1
= 22.620°
2
24
θ AB = tan −1
θ BC
Use this free body to determ
mine TAB and TBC.
Free-Boody Diagram of Point C:
θCD = tan −1
11.1
= 10.3889°
6
Use this free body to determ
mine TCD and WF.
Then weight of skier WS is found by
WS = WF − 250 N
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PROBLEM 2.F5
Three cables are used to tether a balloon as shown. Knowing that
the tension in cable AC is 444 N, draw the free-body diagram
needed to determine the vertical force P exerted by the balloon at
A.
SOLUTION
Free-Body Diagram of Point A:
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PROBLEM 2.F6
A container of mass m = 120 kg is supported by three cables
as shown. Draw the free-body diagram needed to determine
the tension in each cable
SOLUTION
Free-Body Diagram of Point A:
W = (120 kg)(9.81 m/s 2 )
= 1177.2 N
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PROBLEM 2.F7
A 150-lb cylinder is supported by two cables AC and BC that are
attached to the top of vertical posts. A horizontal force P, which
is perpendicular to the plane containing the posts, holds the
cylinder in the position shown. Draw the free-body diagram
needed to determine the magnitude of P and the force in each
cable.
SOLUTION
Free-Body Diagram of Point C:
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PROBLEM 2.F8
A transmission tower is held by three guy wires attached
to a pin at A and anchored by bolts at B, C, and D.
Knowing that the tension in wire AB is 630 lb, draw the
free-body diagram needed to determine the vertical force
P exerted by the tower on the pin at A.
SOLUTION
Free-Body Diagram of point A:
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