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T
mmg
mm
=
+
F
H
G
I
K
J
−=
21
12
105
(
N
.
μμ
θ
)cos
.
(b) These equations lead to an acceleration equal to
ag
=−
+
+
F
H
G
I
K
J
F
H
G
I
K
J
=
sin
cos
.
.
22
11
362 m/s
2
(c) Reversing the blocks is equivalent to switching the labels. We see from our algebraic
result in part (a) that this gives a negative value for
T
(equal in magnitude to the result we
got before). Thus, the situation is as it was before except that the rod is now in a state of
compression.
68. The freebody diagrams for the two boxes are shown below.
T
is the magnitude of the
force in the rod (when
T
> 0 the rod is said to be in tension and when
T
< 0 the rod is
under compression),
2
N
F
G
is the normal force on box 2 (the uncle box),
1
N
F
G
is the the
normal force on the aunt box (box 1),
G
f
1
is kinetic friction force on the aunt box, and
G
f
2
is kinetic friction force on the uncle box. Also,
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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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