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brendan3eb
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Homework Statement
A 10 kg monkey climbs up a massless rope that runs over a frictionless tree limb and back down to a 15 kg package on the ground. (a) What is the magnitude of the least acceleration the monkey must have if it is to lift the package off the ground? If, after the package has been lifted, the monkey stops its climb and holds onto the rope, what are (b) the magnitude and (c) the direction of the monkey's acceleration, and (d) what is the tension in the rope?
Homework Equations
weight=mg
F=ma=T-mg
The Attempt at a Solution
I was able to get part A with the following work:
Sum of forces on monkey = MA=T-Mg
Sum of forces on banana box = ma=T-mg
when the package goes from being on the ground to being lifted off the ground it's acceleration will change from 0 to some number, thus the minimum rope tension required will be just greater than T when a=0
ma=T-mg
a=(T-mg)/m
0=(T-mg)/m
T=mg=(15 kg)(9.8 m/s^2)=147 N
MA=T-Mg
A=(T-Mg)/M
A=[147 N - (10 kg)(9.8 m/s^2)]/10 kg = 4.9 m/s^2
I should also mention that I have the answers from the back of the book:
(a) 4.9 m/s^2
(b) 2.0 m/s^2
(c) upward
(d) 120 N
So I got part A right, but my attempts at parts B-D did not work. Here is what I tried and my results:
Since the monkey is no longer pulling on the rope, the accelerations of the monkey and the box should have the same magnitude but opposite directions.
MA=T-Mg
T=Mg+MA
mA=T-mg
T=mA + mg
set the two Ts to equal each other
Mg+MA = mA + mg
(10 kg)(9.8 m/s^2) + (10 kg)A = (15 kg)A + (15 kg)(9.8 m/s^2)
when you solve, you get A=9.8 m/s^2
I even tried changing the signs around for the two different equations to see if that would give me the right answer, but no luck. Any help would be much appreciated :)