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Since the weight of block
A
is 44 N, the least weight for
C
is (110 – 44) N = 66 N.
(b) The second law equations become
T
–
f
= (
W
A
/
g
)
a
F
N
–
W
A
= 0
W
B
–
T
= (
W
B
/
g
)
a
.
In addition,
f
=
μ
k
F
N
. The second equation gives
F
N
=
W
A
, so
f
=
k
W
A
. The third gives
T
=
W
B
– (
W
B
/
g
)
a
. Substituting these two expressions into the first equation, we obtain
W
B
– (
W
B
/
g
)
a
–
k
W
A
= (
W
A
/
g
)
a
.
Therefore,
()
(
)
2
2
(9.8 m/s ) 22 N
0.15 44 N
2.3 m/s .
44 N + 22 N
Bk
A
AB
gW
W
a
WW
−
−
==
=
+
31. (a) Freebody diagrams for the blocks
A
and
C
, considered as a single object, and for
the block
B
are shown below.
T
is the magnitude of the tension force of the rope,
F
N
is
the magnitude of the normal force of the table on block
A
,
f
is the magnitude of the force
of friction,
W
AC
is the combined weight of blocks
A
and
C
(the magnitude of force
G
F
gAC
shown in the figure), and
W
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Force

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