Atwood machine problem - symbolically solving for mass?

1. Jan 9, 2010

therest

atwood machine problem -- symbolically solving for mass?

1. The problem statement, all variables and given/known data
Symbolically solve the equation derived for the acceleration of the Atwood's machine in part 1, A-3 for the mass m in terms of M, g, and the acceleration, a. (No numbers.)

This question is part of a lab, so of course some other information is necessary. We were finding theoretical and experimental acceleration for an Atwood's machine when the masses on both sides were the same except for a few pennies balanced on one of the masses. After finding this value, we were using it to fine the theoretical and experimental weights of the pennies.

We were distinguishing m from M: m was the much smaller mass of the pennies, and M was the mass of the weights on both sides.

2. Relevant equations
How I solved for acceleration can be found in attached image #1. The basic equation used is Newton's second law.

3. The attempt at a solution
My weak attempt at solving for mass is in the second attached image. It all seems to cancel out and I'm getting really confused. Can anyone tell me where I made a mistake, or whether I'm just using really faulty, circular logic?

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2. Jan 9, 2010

Staff: Mentor

Re: atwood machine problem -- symbolically solving for mass?

You only solved for the acceleration in terms of the tension. Finish the job. (You should be able to solve those two equations together to get both T and a in terms of the masses.)

Here it seems like you gave up on what you started earlier and tried a different approach. Nothing wrong with the basic idea, but you seemed to have gotten messed up with the 'net mass'.

Stick to those two equations you started with. Solve them together and you'll be fine.

3. Jan 9, 2010

therest

Re: atwood machine problem -- symbolically solving for mass?

Thanks Doc Al! I tried that and pretty much failed. I think my algebra is getting pretty bad of late, so I'm having some trouble with getting the tension and acceleration.

I tried:
a=(T-Mg)/M & a=(M + m)-T
a+T=(M+m)g
T=(M+m)g-a
plugged that in to the first one, and got
a=((M+m)g-a)-Mg)/M
which still has a in it. I kind of puzzled over this for a few minutes, was unable to progress further and tried something else:

a=(T-Mg)/M & -a=(M+m)g-T (the negative is because a is in the opposite direction)
T-a=(M+m)g
T=(M+m)g+a
so T=(M+m)g-(T-Mg)/M
which still has T in it. So how do I get the T out? Is substitution not the right thing to try with this? Or am I just not going far enough with it?

4. Jan 10, 2010

Staff: Mentor

Re: atwood machine problem -- symbolically solving for mass?

Your algebra is getting a bit mixed up. In several cases, you've ended up with things like 'a + T'. That means you're adding an acceleration with a force, which is physically meaningless. (It means you made a mistake somewhere.)

(M + m)g - T = (M + m)a
T - Mg = Ma

How can you combine these two equations to eliminate T? (Don't be in a rush to isolate 'a' in each.)

5. Jan 24, 2010

therest

Re: atwood machine problem -- symbolically solving for mass?

Sorry for the long gap in-between posts. I let this assignment slip away from me for a while.

(M+m)g-T = (M+m)a
T-Mg=ma
(M+m)g-Mg=ma + (M+m)a (can I distribute this (M+m)a?)
Mg+mg-Mg=ma+Ma+ma
mg=2ma + Ma
mg / 2(M+m) = a

Is this correct? Can I use this equation to symbolically solve for mass? (If so, I'm only inputting it into F=ma and simplifying, right?)

6. Jan 25, 2010

Staff: Mentor

Re: atwood machine problem -- symbolically solving for mass?

Careful. That second equation should have M, not m. Redo it.

When you add and then simplify the equations, remember that you are solving for 'm', not 'a'.

7. Jan 26, 2010

therest

Re: atwood machine problem -- symbolically solving for mass?

Sigh. It's a bit embarrassing that I'm having so much difficulty with this. Sorry about that, thanks for your continued help.

(M + m)g - T = (M + m)a
T - Mg = Ma

(M + m)g - Mg = (M + m)a + Ma
Mg + mg - Mg = Ma + ma + Ma
mg = 2Ma + ma
mg - ma = 2Ma
m (g - a) = 2Ma
m = 2Ma / (g - a)

Does that look right?

Oh, also, this would also make a = mg / (2M+m) with some rearrangement, right? I hope this is right, because that would actually make sense to me...

Last edited: Jan 27, 2010
8. Jan 27, 2010

Staff: Mentor

Re: atwood machine problem -- symbolically solving for mass?

Perfect!
Right!

You got it.