Unlock document.
This document is partially blurred.
Unlock all pages and 1 million more documents.
Get Access
3–41.
The single elastic cord ABC is used to support the 40-lb
load.Determine the position xand the tension in the cord
that is required for equilibrium. The cord passes through
the smooth ring at Band has an unstretched length of 6 ft
and stiffness of k=50 lb>ft.
A
C
B
x
5 ft
1 ft
SOLUTION
3–42.
SOLUTION
Equations of Equilibrium:
Ans.WB=18.3 lb
+c©Fy=0; 2(10) cos 23.58° - WB = 0
:
+©Fx=0; 10 sin 23.58° - 10 sin 23.58° = 0 (Satisfied!)
A “scale”is constructed with a 4-ft-long cord and the 10-lb
block D.The cord is fixed to a pin at Aand passes over two
small pulleys at Band C.Determine the weight of the
suspended block at B if the system is in equilibrium when
s = 1.5 ft.
s 1.5 ft
C
A
1 ft
3–45.
If t
h
e
b
uc
k
et an
d
i
ts contents
h
ave a tota
l
we
i
g
h
t of 20
lb
,
determine the force in the supporting cables DA,DB,
and DC.
SOLUTION
Ans.
Ans.
Ans.F
D
C
=15.6 lb
F
DB =1.11 lb
F
DA =10.0 lb
©F
z=0; 3
4.5F
DA +3
3.5F
DC -20 =0
©F
y=0; -1.5
4.5F
DA -F
DB +1
3.5F
DC =0
4.5 ft
2.5 ft
3ft
A
C
z
3–46.
SOLUTION
Determine the stretch in each of the two springs required to
hold the 20-kg crate in the equilibrium position shown.
Each spring has an unstretched length of 2 m and a stiffness
of k=360 N>m.
z
O
C
B
A12 m
© 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:
FAB =219 N
FAC =FAD =54.8 N
SOLUTION
3–47.
Determine the force in each cable needed to support the
20-kg owerpot.
D
y
C
A
B
4 m
2 m
2 m
6 m
3 m
z
*3–48.
SOLUTION
FAC =-FAC j
Determine the tension in the cables in order to support the
100-kg crate in the equilibrium position shown.
2.5 m 2 m
2 m
2 m
1 m
A
z
D
y
B
C
(-2 - 0)i+(2 - 0)j+(1- 0)k
2
3–49.
SOLUTION
Equations of Equilibrium: Equilibrium requires
Equating the i, j, and kcomponents yields
(1)
(2)
(3)
W
hen cable AD is subjected to maximum tension,.Thus,by
substituting this value into Eq
s. (1) through (3), we have
Ans.m=102 kg
FAB =FAC =2000 N
FAD =3000 N
1
3 FAD -9.81m=0
-FAC +2
3 FAD =0
FAB -2
3 FAD =0
aFAB -2
3 FAD bi+a-FAC +2
3 FAD bj+a1
3FAD -9.81mbk=0
FAB i+(-FAC j)+a-2
3 FAD i+2
3 FAD j+1
3 FAD kb+[-m(9.81)k]=0
©F=0;
FAB +FAC +FAD +W=0
W=[-m(9.81)k]
FAD =FAD
B
(-2 - 0)i+(2 - 0)j+(1 - 0)k
2(-2 - 0)2+(2 - 0)2+(1 - 0)2
R
=-
2
3FAD i+2
3 FAD j+1
3 FAD
FAC =-FAC j
Determine the maximum mass of the crate so that the tension
developed in any cable does not exceeded 3 kN.
2.5 m 2 m
2 m
2 m
1 m
A
z
D
y
B
C
© 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:
FAC =113 lb
FAB =257 lb
FAD =210 lb
SOLUTION
3–50.
Determine the force in each cable if F = 500 lb. z
A
B
y
D
F
1 ft
3 ft
6 ft
2 ft
1 ft
C
© 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.
*3–52.
SOLUTION
Determine the tension developed in cables AB and AC and
the force developed along strut AD for equilibrium of the
400-lb crate.
z
5.5 ft
2 ft
2 ft
A
B
C
4 ft
© 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.
3–53.
If the tension developed in each of the cables cannot
exceed 300 lb,determine the largest weight of the crate that
can be supported. Also,what is the force developed along
strut AD?
z
5.5 ft
2 ft
2 ft
A
B
C
4 ft
SOLUTION
3–54.
SOLUTION
Equations of Equilibrium: Equilibrium requires
Equating the i,j,and kcomponents yields
(1)
(2)
(3)
Solving Eqs. (1) through (3) yields
Ans.F
AB =79.2 lb
F
AC =119 lb
F
AD =283 lb
2
7 F
AB +3
7 F
AC +4
5 F
AD -300=0
-6
7 F
AB -6
7 F
AC +3
5 F
AD =0
-3
7 F
AB +2
7 F
AC =0
¢
-3
7 F
AB i-6
7 F
AB j+2
7 F
AB k
≤
+
¢
2
7 F
AC i-6
7 F
AC j+3
7 F
AC k
≤
+
¢
3
5 F
AD j+4
5 F
AD k
≤
+(-300k)=0
gF=0;
FAB +FAC +FAD +W=0
W={-300k} lb
FAD =F
AD C(0 -0)i+(3 -0)j+(4 -0)k
2(0 -0)2+(3 -0)2+(4 -0)2S=3
5 F
AD j+4
5 F
AD k
FAC =F
AC C(2 -0)i+(-6-0)j+(3 -0)k
2(2 -0)2+(-6-0)2+(3 -0)2S=2
7 F
AC i -6
7 F
AC j+3
7 F
AC k
2(-3-0)2+(-6-0)2+(2 -0)2S=-
7 F
7 F
7 F
y
A
B
C
D
x
z
6 ft
3 ft
3 ft
2 ft 2 ft
3 ft
4 ft
Determine the tension developed in each cable for
equilibrium of the 300-lb crate.
