## Masses on a Pulley System

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 as shown below. (a) If a1 and a2 are the accelerations of m1 and m2 respectively, what is the relation between these accelerations? Express (b) the tensions in the strings and (c) the accelerations a1 and a2 in terms of their masses m1, and m2, and g.

2. Relevant equations
F=ma

3. The attempt at a solution
a) I believe the accelerations should be the same seeing as they are connected through the strings and are part of the same system.

b) I'm really not sure how to go about with the tensions, I think that the force of gravity will be included in the tension of mass 2, and the force that this block is pulling on mass 1 will have something to do with the tension there.

c) Again, I haven't worked with pulleys in a long time, and can't remember exactly how the accelerations and masses work in relation to one another. I think you have to find the individual force on each object, and then this can help you find the net force, which you divide by the total mass of the system to get the acceleration of the system.

Thanks a bunch
 i'm pretty sure the accelerations are the same as well, but i am having trouble visualizing this pulley system...how is m1 connected to m2 through the first pulley?
 I believe that the string on the top of pulley 1 is just tied somewhere to hold it up in the air, and that the other string coming from it is the one that goes through p2 to p1. I don't really know why p1 needs to be there at all to be honest.

## Masses on a Pulley System

well, if you ignore pulley#1, you can use the formula :

a = 9.81 * m2/(m1+m2)

where a is the common acceleration and m2 is the hanging object.
 Thanks, that equation makes sense now that I see it in front of me. Now I just need to figure out how their tensions are related.
 The accelerations aren't going to be the same. Look at what happens when block 1 moves left 1 unit, Pulley 1 moves left only half a unit as does block 2. So the acceleration of block 2 is half that of block 1.

 Quote by nealh149 The accelerations aren't going to be the same. Look at what happens when block 1 moves left 1 unit, Pulley 1 moves left only half a unit as does block 2. So the acceleration of block 2 is half that of block 1.
Totally correct.This is what i call application. The home work helpers always appreciate this type of thinking.

When the system is of more than one pulley than we really have to think about the diffrent acc taking place.

The tensions are differant in the two strings.Now just as there is a relation in the acc it is so for the strings also. The tension gets divided into t/2 and t/2 at P1 if T is the tension at P2

 Quote by FedEx Totally correct.This is what i call application. The home work helpers always appreciate this type of thinking. When the system is of more than one pulley than we really have to think about the diffrent acc taking place. The tensions are differant in the two strings.Now just as there is a relation in the acc it is so for the strings also. The tension gets divided into t/2 and t/2 at P1 if T is the tension at P2
Thanks, that's actually exactly what I came up with after reading nealh's post.

Now for the accelerations in terms of m1, m2, and g.

Would a2 = (m2*g)/(m1+m2)
and a1 = (m2*g)/(2(m1+m2))

I believe this would still represent the 2:1 ratio explained in part a, as long as the actual equation part was right.

 Quote by Adrianw2 Thanks, that's actually exactly what I came up with after reading nealh's post. Now for the accelerations in terms of m1, m2, and g. Would a2 = (m2*g)/(m1+m2) and a1 = (m2*g)/(2(m1+m2)) I believe this would still represent the 2:1 ratio explained in part a, as long as the actual equation part was right.
No there is a mistake

Remember that the tension in the string for the block m1 is T/2 but the tension in the string for the block m2 is T.
 So that 2 in the denominator would cover that wouldn't it? If F1 = 0.5*F2 then a1 = 0.5*a2. Isn't that where the tension comes into play in that equation?

 Quote by Adrianw2 So that 2 in the denominator would cover that wouldn't it? If F1 = 0.5*F2 then a1 = 0.5*a2. Isn't that where the tension comes into play in that equation?
Right, tension should be grouped with mass times gravity, I believe.