then
F
N
= 0 and
a
p
= 0. According to the second law equation for the package, this means
F
=
m
p
g
. Substituting
m
p
g
for
F
in the equation for the monkey, we solve for
a
m
:
()
2
2
15 kg 10 kg 9.8 m/s
4.9 m/s .
10 kg
pm
m
m
mm
g
Fm
g
a
−
−
−
==
=
=
(b) As discussed, Newton’s second law leads to
pp
p
gm
a
−=
for the package and
m
Fmgm
a
for the monkey. If the acceleration of the package is downward, then
the acceleration of the monkey is upward, so
a
m
= –
a
p
. Solving the first equation for
F
ga
m
=+
=−
di
bg
and substituting this result into the second equation, we solve for
a
m
:
2
2
15 kg
10 kg 9.8 m/s
2.0 m/s .
15 kg
10 kg
m
g
a
−
−
=
++
(c) The result is positive, indicating that the acceleration of the monkey is upward.
(d) Solving the second law equation for the package, we obtain
(
)
22
15 kg 9.8 m/s
2.0 m/s
120N.
Fmga
=
−
=
57. We take +
y
to be up for both the monkey and the package.
(a) The force the monkey pulls downward on the rope has magnitude
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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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