978-0073398242 Chapter 12 Solution Manual Part 6

subject Type Homework Help
subject Pages 9
subject Words 1271
subject Authors Brian Self, David Mazurek, E. Johnston, Ferdinand Beer, Phillip Cornwell

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page-pf1
PROBLEM 12.31
A 10-lb block B rests as shown on a 20-lb bracket A. The
coefficients of friction are 0.30
s
and 0.25
k
between
block B and bracket A, and there is no friction in the pulley or
between the bracket and the horizontal surface. (a) Determine
the maximum weight of block C if block B is not to slide on
bracket A. (b) If the weight of block C is 10% larger than the
answer found in a determine the accelerations of A, B and C.
page-pf2
PROBLEM 12.31 (Continued)
page-pf3
PROBLEM 12.31 (Continued)
page-pf4
PROBLEM 12.32
Knowing that = 0.30, determine the acceleration of each block when
m
A
= m
B
= m
C
.
page-pf5
Problem 12.32 (Continued)
page-pf6
PROBLEM 12.33
Knowing that = 0.30, determine the acceleration of each block
when m
A
= 5 kg, m
B
= 30 kg, and m
C
= 15 kg.
page-pf7
Problem 12.33 (Continued)
page-pf8
PROBLEM 12.34
A 25-kg block A rests on an inclined surface, and a 15-kg counterweight
B is attached to a cable as shown. Neglecting friction, determine the
acceleration of A and the tension in the cable immediately after the
system is released from rest.
Block A:
:sin20
xAAA AB AA
Fmamg N Tma 
sin 20
AA AB A
ma N T mg
/
25 83.880
BA AB
aNT
(4)
Eliminate
/
BA
a
using Eq. (1), then add Eq. (4) to Eq. (2) and
subtract Eq. (3).
22
55 4.068 or 0.0740 m/s , 0.0740 m/s
AA A
aa  a
20
From Eq. (1),
2
/
0.0740 m/s
BA
a
From Eq. (3),
137.2 NT
137.2 NT
page-pf9
PROBLEM 12.35
Block B of mass 10-kg rests as shown on the upper surface of a 22-kg
wedge A. Knowing that the system is released from rest and neglecting
friction, determine (a) the acceleration of B, (b) the velocity of B relative to
A at 0.5 s.t
: cos 20 sin 50


yByABB BA
Fma NW ma
or
10 ( cos 20 sin 50 )
AB A
Ng a
Equating the two expressions for
AB
N
1
22 210( cos 20 sin 50 )
cos 40
A
A
ag
ga




or
2
(9.81)(1.1 cos 20 cos 40 ) 6.4061 m/s
2.2 cos 40 sin 50
A
a


/
:sin20 cos50


xBx B BBABA
F ma W ma ma
or
/
2
2
sin 20 cos50
(9.81sin 20 6.4061cos 50 ) m/s
7.4730 m/s
BA A
ag a

page-pfa
PROBLEM 12.35 (Continued)

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