Atwood machine problem - symbolically solving for mass?

[therest]

atwood machine problem -- symbolically solving for mass?

Homework Statement

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.

Homework Equations

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

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?

 

[Doc Al]

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

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.)

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?

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.

 

[therest]

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 into 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?

 

[Doc Al]

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.)

It's easier than you think. Start with the two force equations:

(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.)

 

[therest]

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

I tried again, following your advice:
(M+m)g-T = (M+m)a
T-Mg=ma
added the two, and got
(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?)

 

[Doc Al]

I tried again, following your advice:
(M+m)g-T = (M+m)a
T-Mg=ma

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'.

 

[therest]

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...

 

[Doc Al]

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?

Perfect!

Oh, also, this would also make a = mg / (2M+m) with some rearrangement, right?

Right!

You got it.