Tension in the String of a Pulley System

[JamesEarl]

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

Homework Equations

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.

 

[JamesEarl]

Anyone? This is due tomorrow guys

 

[jamesrc]

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.

 

[JamesEarl]

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.

 

[JamesEarl]

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]

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.

 

[JamesEarl]

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?

 

[Jebus_Chris]

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.