# Tension in the String of a Pulley System

## Homework Statement

The figure below shows two boxes, m1 and m2. m1 = 10 kg is sitting on a frictionless horizontal surface, while m2 = 7 kg is hanging vertically. The two masses are connected by a massless string that is wrapped around a massless, frictionless pulley.

Figure: http://class.phys.psu.edu/WebAssign_figures/two-box connected.jpg

What is the acceleration of m1?
What is the tension in the string?

m1= 10kg
m2= 7kg

ax=m2g/(m1+m2)

## The Attempt at a Solution

So I got the right answer for acceleration, which was 4.04 m/s2. However, I've already made two attempts at the tension and both were wrong. I tried looking at it from two different standpoints. After drawing a free body diagram, it looks like T=m1ax. When I plug in the numbers, I get 40.4, which is incorrect. I then tried looking at it from the other block. T=m2(9.8-ax). The answer for this was the same, and therefore incorrect. I then tried -40.4, which was also wrong. Anyone have any ideas? The only hunch I have is that the tensions offset, making the answer 0. However, this really doesn't make any sense because there has to be some tension in the rope.

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Anyone? This is due tomorrow guys

jamesrc
Gold Member
I don't see anything wrong with your calculation of tension - the fact that you got the same answer looking at either mass should tell you that you got the right answer (and no, negative or 0 tension does not make sense for this problem).

I take it you are submitting your answers online - perhaps they are looking for you to round the answer to a specific number of digits. If that's not it, your program may just have the wrong answer.

Thanks for the response james, I checked with my TA and apparently I do, in fact, have the WRONG answer. I have no idea what I'm doing wrong, it's really driving me nuts.

Actually, is it possible that the tensions of each component (40.4) need to be added together (80.8) to get the total tension of the string?

jamesrc
Gold Member
I'm sorry JamesEarl, but if 40.4 is not the answer, then I don't know what is.

Free-body diagram of m1 yields the following equation:
m1*a = T

Free-body diagram of m2 yields the following equation:
m2*a = m2*g - T

the string is inextensible and massless; there are no other components in the system with inertia; there is no friction in the system

Since the string doesn't stretch, both masses accelerate with a (as already expressed in the equations above).

Solving the equations:
a = (m2/(m1+m2))*g
T = m1*a = ((m1*m2)/(m1+m2))*g
equivalently,
T = m2*(g-a) = ((m1*m2)/(m1+m2))*g
Plugging in the numbers:
a = 4.04 m/s^2
T = 40.4 N

If that's not the solution, it may be time for me to retire.

That's exactly what I'm thinking, but apparently I'm definitely wrong. I went to my TA and showed him my approach and my answer, and he said there's something I'm not accounting for. I have NO idea what he's talking about. Do I add up the tensions on either side of the pulley?

Tension should be uniform. If you accerlation is right and T=m1a I don't see how the tension couold be anything other than 40.4N.