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For each block we take +
x
downhill (which is toward the lowerleft in these diagrams)
and +
y
in the direction of the normal force. Applying Newton’s second law to the
x
and
y
directions of both blocks
A
and
B
, we arrive at four equations:
sin
cos
0
sin
cos
0
AA
A
NA
A
BB
B
NB
B
Wf
T
m
a
FW
T
m
a
θ
−−
=
−=
−+
=
which, when combined with Eq. 62 (
Ak
A
N
A
f
F
μ
=
where
k A
= 0.10 and
Bk
B
N
B
f
F
=
f
B
where
k B
= 0.20), fully describe the dynamics of the system so long as the blocks have
the same acceleration and
T
> 0.
(a) From these equations, we find the acceleration to be
2
sin
cos
3.5 m/s .
kA
A
kB
B
AB
WW
ag
μμ
θθ
§·
+
=−
=
¨¸
+
©¹
(b) We solve the above equations for the tension and obtain
()
0.21 N.
T
=
+
Simply returning the value for
a
found in part (a) into one of the above equations is
certainly fine, and probably easier than solving for
T
algebraically as we have done, but
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 Spring '08
 Any
 Physics, Force, Friction

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