# Double Pulley Problem

1. Sep 22, 2008

### jessedevin

1. The problem statement, all variables and given/known data

An object of mass m1 on a frictionless horizontal table is connected to an object of mass m2 through a very light pulley P1 and a light fixed pulley P2.

(a) If a1 and a2 are the accelerations of m1 and m2, respectively, what is the relation between these accelerations?

(b) Express the tensions in the strings in terms of g and the masses m1 and m2.

(c) Express the accelerations a1 and a2 in terms of g and the masses m1 and m2.

2. Relevant equations

$$\Sigma$$F = ma

3. The attempt at a solution

The way that I did it was I made two force diagrams. The first force diagram of m1 I made just has force going to the right, force T1 (tension of rope). Then I have a second force diagram of m2 with a force T2 going up and a force of mg (or weight of m2) going down. but the thing is that I keep on getting one a (acceleration), but the answer has a1 and a2. Can someone please tell me what I am doing wrong with my force diagrams. I think if i can have the correct force diagram, then I will be able to do the problem (hopefully!).

2. Sep 22, 2008

### LowlyPion

The movable pulley gives you a mechanical advantage of 2, but it means the tension translated to m1 is only 1/2 from T2, and hence so too will the acceleration of m1 be different.

Stated as Tensions that's T2 = 2T1

Last edited: Sep 23, 2008
3. Sep 23, 2008

### jessedevin

Okay, so if T2= 2T1, then the force diagram looks like this (this is what i got):
The force diagram of m1 has T1 going to the right, m1g going down, and the normal vector n going up.
The force diagram of m2 has T2 going up and m2g going down.

So heres the way I did it, but its wrong for some reason, and i dont no why:
$$\Sigma$$Fm1= T1 = m1a1
$$\Sigma$$Fm2= T2 - m2g=m2a2.

So what i did to find the relationship between the accelerations was substitute T2 = 2T1.
2m1a1 - m2g= m2a2
a1= m2(a2 + g)/2m1
But it says that i am wrong, and there is not suppose to be the gravity constant in the relationship between the accelerations. Can someone please tell me what am i doing wrong?

4. Sep 23, 2008

### LowlyPion

You can't avoid g.

Edit: But in part (a) the relationship of the accelerations falls out of the pulley doesn't it?

Last edited: Sep 23, 2008
5. Sep 23, 2008

### jessedevin

I mean for part (a)

6. Sep 23, 2008

### jessedevin

Should i determine part (b) or (c) before I do (a)

7. Sep 23, 2008

### LowlyPion

Sorry, I edited my previous post when I realized you were talking about a).

Here it is again. But in part (a) the relationship of the accelerations falls out of the pulley doesn't it?

8. Sep 23, 2008

### jessedevin

What do you mean? That a1=a2? Im really confused...

9. Sep 23, 2008

### LowlyPion

Visualize how far you have to pull the rope at the m1 side of the pulley to move the weight on the m2 side. Don't you have to move a distance of 2 for every 1 you move on the m2 side? Isn't that where the mechanical advantage comes from? You pull only 1/2 as much, but over twice the distance to move things on the m2 side? For any velocity at m2 then there is 2 times the velocity at m1.

10. Sep 23, 2008

### jessedevin

Okay i get that for any velocity at m2 there is 2 times the velocity at m1, so a1= 2a2. But Im confused on how to set of the force equations.Did i write the force equations right? Can you give me a hint on how to set it up?

Force eqns:
$$\Sigma$$ Fm1= T1 = m1a1
$$\Sigma$$ Fm2= T2 - m2g=m2a2.

Last edited: Sep 23, 2008
11. Sep 23, 2008

### LowlyPion

Ok, one equation is wrong. T2 is minus and M2*g is plus as you need to preserve the same sign convention between the two equations if you want to solve them.

12. Sep 23, 2008

### LowlyPion

If 2x1 = x2
and 2v1 = v2
What is a1 again?

13. Sep 23, 2008

### jessedevin

a1=2a2 ...lol

Last edited: Sep 23, 2008
14. Sep 23, 2008

### jessedevin

how is that eqn wrong. m2g is the force going down (hence, negative) and T2 is going up, isnt the force eqn i wrote right? Also can you give me any more hints on writing each tension T in terms of m1, m2 and gravity?

15. Sep 23, 2008

### LowlyPion

You need to make Tension the same sign in whatever direction you choose.

You chose to the right in equation 1 and it translates to the left in equation 2. As the problem stands it is net down. So T must be < m2g

16. Sep 23, 2008

### jessedevin

Okay i am still drawing a blank on this problem... So lets go back from the beginning. My text book does not explain mechanical advantage, so could you please explain a little more in depth on mechanical advantage on having two pulleys in relation to the two masses?

17. Sep 23, 2008

### LowlyPion

Maybe this will help?
http://en.wikipedia.org/wiki/Pulley#Types_of_systems

18. Sep 23, 2008

### jessedevin

It helped a little, but still kind of foggy. Is there any more hints you could give me to help me solve the problem? I am completely stuck.

19. Sep 24, 2008

### jessedevin

Can anyone give me more hints to answer this problem?

20. Oct 11, 2008

### astronomerc

I think, we set the T's equal , solve for a1 which will give us a2. I don't think the a's fall out when solving for the tension, but I think we can find the a's.... I hope.