Forces Problem Involving A Tree Branch acting as a monkey's pulley

1. Oct 6, 2007

brendan3eb

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
weight=mg
F=ma=T-mg

3. 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 :)

2. Oct 6, 2007

Staff: Mentor

Careful with signs. If the monkey's acceleration is +A, what must be the box's acceleration?

3. Oct 6, 2007

brendan3eb

Ah! Thank you, I was thinking about that but with regard to mg. I was changing mg to be positive or negative and was getting an acceleration of 2 m/s^2

Once again, thanks! :P

4. Sep 30, 2011

theunloved

Can someone please explain more on the bold part ? I don't understand it

5. Sep 30, 2011

Staff: Mentor

When the monkey hauls on the rope, if tension in the rope exactly equals the package's weight, the package will barely leave the ground. If he exerts a force greater than the weight of the package, the package will move upwards.

6. Sep 30, 2011

PeterO

The monkey stops climbing, but still holds on to the rope, so will be pulling on the rope due to the effects of gravity on the monkey.
For(b) there is the weight force of a 10kg monkey pulling one way, and the weight force of a 15 kg package pulling the other way.
I see that as the net force of a 5kg weight acting on a 25kg system.
A weight force of a 5kg mass, acting on 5 kg gives acceleration g
A weight force of a 5kg mass acting on 25 kg should give acceleration g/5

The tension in the rope will exceed the individual weight force of the 10kg monkey by enough to cause that acceleration.

7. Oct 1, 2011

theunloved

Thanks for explaining this to me.