# Homework Help: Pulleys - Relationship between mass and aceleration

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1. Sep 22, 2015

### AlonsoDeMaria

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Last edited: Sep 22, 2015
2. Sep 22, 2015

### DEvens

So, I see you pointing at an exam solution. I don't see your attempt.

The solution you pointed at has a typo. And the explanation for it is wonky. You get the ratio of acceleration of the two masses by conserving string. That is, the string does not change length.

3. Sep 22, 2015

### AlonsoDeMaria

Oh, I am sorry. I didn't want to write all my attempt.

This was my full attempt
I used linear equations to solve this part:
Following the exam solution:
2T-10m=m*ax
T-10m=m*ay (I multiplied this by 2) and got 2T-20m=2m*ay

Then I tried to make a comparison between the accelerations, so I deleted "T" subtracting the first and second equations:
2T-10m - (2T-20m)=m*ax-2m*ay
10m=m(ax-2ay)

Then 10=(ax-2ay). Then, ay=(10-ax)/2
I replaced ay on the second equation:
T-10m=m*(10-ax)/2
Then, 2T-20m=10m - m*ax
2T-30m= - m*ax
-2T+30m= m*ax
Then I took the first equation and replaced m*a
2T-10m = -2T + 30m
4t=40m , T=10m

And then
20m-10m=m*ax, 10m=m*ax, ax=10m/s2

I need to how to get that ax=2ay
And if my answer was correct.

4. Sep 22, 2015

### DEvens

Did you read to the bottom of my post? You get the ratio of accelerations by noting that the length of string does not change. Suppose that the mass X moves up by 1 unit. How far down will Y move?

5. Sep 22, 2015

### AlonsoDeMaria

2 units.

I am checking all again.

6. Sep 22, 2015

### AlonsoDeMaria

So I was reading that ay= - 2 ax.

What did I do wrong on my process?

7. Sep 22, 2015

### AlonsoDeMaria

So the correct was to write: ay=(ax-10)/2
Even so, using that leads to nowhere.

So the only possibility to solve that was just knowing "conserving string" law.

Is there any other way to solve that without that law?
(Or better for me to create another thread?)

Thank you very much for your help.