978-0133915426 Chapter 3 Part 2

subject Type Homework Help
subject Pages 14
subject Words 1267
subject Authors Russell C. Hibbeler

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page-pf1
3–21.
Determine the unstretched length of spring AC if a force
causes the angle for equilibrium. Cord
AB is 2 ft long.Take k=50 lb>ft.
u=60°P=80 lb
SOLUTION
Solving for ,
Ans.l=212 -40
50 =2.66 ft40 =50(212 -l¿)
F
s=kx
F
s=40 lb
Fs
:
+©Fx=0; -Tcos 60° +Fscos 30° =0
+c©Fy=0; Tsin 60° +Fssin 30° -80 =0
f=sin-1
¢
2sin 60°
212
=30°
212
sin 60° =2
sin f
l=212
l=242+22-2(2)(4) cos 60°
2 ft
k
2 ft
A
BC
u
page-pf2
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3–22.
SOLUTION
Ans.F=158 N
T=ks =500(232+(1.5)2-3) =177.05 N
+©F
The springs BA and BC each have a stiffness of 500 N
>
m and an
unstretched length of 3 m. Determine the horizontal force F
applied to the cord which is attached to the small ring B so
that the displacement of the ring from the wall is d = 1.5 m.
F
B
A
k 500 N/m
k 500 N/m
6 m
page-pf3
3–23.
The springs
BA and BC each have a stiffness of 500 N
>
m and an
unstretched length of 3 m. Determine the displacement
d of the
cord from the wall when a force F = 175 N is applied to the cord.
SOLUTION
By trial and error
:
Ans.d=1.56 m
da1-3
29+d2b=0.175
T=ks =500(232+d2-3)
TCd
232+d2S=87.5
:
+©F
x=0; 175 =2Tsin u
F
B
A
k 500 N/m
6 m
page-pf4
SOLUTION
x
*3–24.
Determine the distances x and y for equilibrium if F1 = 800 N
and F2 = 1000 N.
B
CDF1
F2
2 m
y
page-pf5
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Ans:
y=2 m
F1=833 N
SOLUTION
3–25.
Determine the magnitude of F1 and the distance y if x = 1.5 m
and F2 = 1000 N.
B
CDF1
F2
2 m
y
page-pf6
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3–26.
SOLUTION
The 30-kg pipe is supported at A by a system of fivecords.
Determine the force in each cord for equilibrium.
A
H
E
B
C
D
60°
3
4
5
page-pf7
187
3–27.
Each cord can sustain a maximum tension of 500 N.
Determine the largest mass of pipe that can be supported.
SOLUTION
At A:
At B:
By comparison, cord BC carries the largest load. Thus
Ans.m=261.69
9.81 =26.7 kg
W=261.69 N
500 =1.9107 W
FBC =1.9107 W
:
+©Fx=0; -FBC +1.667 Wa4
5b+1.1547 sin 30° =0
FBD =1.667 W
+c©Fy=0; FBD a3
5b-(1.1547 cos 30°)W=0
FAE =0.5774 W
:
+©Fx=0; FAE -(1.1547 W) cos 60° =0
FAB =1.1547 W
+c©Fy=0; FAB sin 60° -W=0
A
E
B
C
D
60°
3
4
5
page-pf8
*3–28.
The street-lights at A and B are suspended from the two
poles as shown. If each light has a weight of 50 lb, determine
the tension in each of the three supporting cables and the
required height h of the pole DE so that cable AB is
horizontal.
SOLUTION
D
A
h
B
C
page-pf9
3–29.
SOLUTION
Ans.
Ans.
Using the result FCD = 339.83 N and applying the equations of equilibrium along the
xand yaxes to the free-body diagram of joint Dshown in Fig. b, we have
(1)
(2)
Solving Eqs. (1) and (2), yields
Ans.F
CA =243 NF
CB =275 N
F
CA a4
5b-F
CB sin 45° =0+c©F
y=0;
339.83 -F
CA a3
5b-F
CD cos 45° =0©F
x=0;
:
+
F
CD =339.83 N =340 N392.4 cos 30° -F
CD =0+c©F
y=0;
F
DE =392.4 N =392 NF
DE sin 30°-20(9.81) =0©F
x=0;
:
Determine the tension developed in each cord required for
equilibrium of the 20-kg lamp.
A
B
D
E
C
45°
30°
3
45
page-pfa
3–30.
SOLUTION
Using the result FCD = 16.99mand applying the equations of equilibrium along the
xand yaxes to the free-body diagram of joint Dshown in Fig. b, we have
(1)
(2)
Solving Eqs. (1) and (2), yields
Notice that cord DE is subjected to the greatest tensile force, and so it will achieve
the maximum allowable tensile force first. Thus
Ans.
m=20.4 kg
F
DE =400 =19.62m
F
CA =12.14mF
CB =13.73m
F
CA a4
5b-F
CB sin 45° =0+c©F
y=0;
16.99m-F
CA a3
5b-F
CD cos 45° =0©F
x=0;
:
+
F
CD =16.99m19.62m cos 30° -F
CD =0©F
x=0;
:
+
Determine the maximum mass of the lamp that the cord
system can support so that no single cord develops a tension
exceeding 400 N.
A
B
D
E
C
45°
30°
3
45
page-pfb
page-pfc
page-pfd
page-pfe
page-pff
195
3–35.
Thus,
Solving by trial and error,
Ans.
Ans.
Also,
Ans.
Ans.l1.40 ft
2
sin 66.75° =l
sin 40°
FAB =82.4 lb
u=66.75°
l=2.34 ft
2
sin 33.25° =l
sin 40°
FAB =175 lb
u=33.25°
sin u+0.8391 cos u=1.25
+c©Fy=0; 160 sin u+FAB sin 40° -200 =0
:
+©Fx=0; FAB cos 40° -160 cos u=0
The ring of negligible size is subjected to a vertical force of
200 lb. Determine the required length lof cord AC such that
the tension acting in AC is 160 lb. Also, what is the force in
cord AB?Hint: Use the equilibrium condition to determine
the required angle for attachment, then determine lusing
trigonometry applied to triangle ABC.
u
40°
θ
BC
A
l2ft
200 lb
SOLUTION
page-pf10
*3–36.
SOLUTION
Geometry: Referring to Fig. b, we can write
Also,
Ans.
Substituting into Eq. (1),yields
Ans.T=687 N
f=45.57°
x=1.38 m
x tan 45.57° +0.75 =(3.5 -x) tan 45.57°
f=cos-1 a3.5
5b=45.57°
3.5 -x
cos f+x
cos f=5
Cable ABC has a length of 5 m. Determine the position x
and the tension developed in ABC required for equilibrium
of the 100-kg sack.Neglect the size of the pulley at B.
A
B
C
x
3.5 m
0.75 m
page-pf11
3–37.
SOLUTION
Geometry
:The angle which the surface make with the horizontal is to be
determined first.
F
ree Body Diagram:The tension in the cord is the same throughout the cord and is
equal to the weight of block
B,.
Equations of Equilibrium
:
[1]
[2]
Solving Eqs
. [1] and [2] yields
Ans.mB=3.58 kg
N=19.7 N
8.4957mB+0.4472N=39.24
+c©F
y=0; mB(9.81) sin 60° +Ncos 63.43° -39.24 =0
N=5.4840mB
:
+©Fx=0; mB(9.81) cos 60° -Nsin 63.43° =0
W
B=mB(9.81)
u=63.43°
tan u
`
x=0.4 m
=dy
dx
`
x=0.4m
=5.0x
`
x=0.4 m
=2.00
u
A4-kg sphere rests on the smooth parabolic surface.
Determine the normal force it exerts on the surface and the
mass of block Bneeded to hold it in the equilibrium
position shown.
mB
B
A
y
x
0.4m
0.4m
60fi
y2.5x
2
page-pf12
3–38.
Determine the forces in cables AC and AB needed to hold
the 20-kg ball Din equilibrium.Take F= 300 N and d= 1 m.
SOLUTION
(1)
(2)
Solving Eqs. (1) and (2) yields
Ans.F
AB =98.6 N F
AC =267 N
0.7809F
AB +0.4472F
AC =196.2
+c©F
y=0; F
AB a5
241 b+F
AC a1
25b-196.2 =0
06247F
AB +0.8944F
AC =300
:
+©F
x=0; 300 -F
AB a4
241 b-F
AC a2
25b=0
A
C
B
D
1.5 m
d
page-pf13
3–39.
The ball Dhas a mass of 20 kg.If a force of F= 100 N is
applied horizontally to the ring at A,determine the largest
dimension dso that the force in cable AC is zero.
SOLUTION
(1)
(2)
Solving Eqs. (1) and (2) yields
From the geometry,
Ans.d=2.42 m
d+1.5 =2 tan 62.99°
u=62.99° F
AB =220.21 N
+c©F
y=0; F
AB sin u-196.2 =0 F
AB sin u=196.2
:
+©F
x=0; 100 -F
AB cos u=0 F
AB cos u=100
A
C
B
D
1.5 m
d
page-pf14
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*3–40.
T
h
e 200-
lb
un
i
form tan
k
i
s suspen
d
e
d
b
y means of a
6
-ft-
long cable, which is attached to the sides of the tank and
passes over the small pulley located at O.If the cable can be
attached at either points Aand B,or Cand D,determine
which attachment produces the least amount of tension in
the cable. What is this tension?
SOLUTION
Free-Body Diagram: By observation, the force F has to support the entire weight
of the tank. Thus, F = 200 lb. The tension in cable AOB or COD is the same
throughout the cable.
Equations of Equilibrium:
(1)
F
rom the function obtained above, one realizes that in order to produce the least
amount of tension in the cable, hence must be as great as possible. Since the
attachment of the cable to point Cand Dproduces a greater
as compared to the attachment of the cable to points Aand ,
the attachment of the cable to point C and D will produce the least amount
of tension in the cable. Ans.
Thus,
Ans.T=100
sin 70.53° =106 lb
B
A
u=cos-12
3=48.19°
B
u
A
u=cos-11
3=70.53°
B
usin u
+c©F
y=0; 200 -2Tsin u=0T=100
sin u
:
+©Fx=0; Tcos u-Tcos u=0(Satisfied!)
A
O
C
1 ft
B
2 ft
F
D
2 ft
2 ft

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