Interacting Systems - Acceleration and Pulleys

[mantillab]

Homework Statement

In the figure (attached), find an expression for the acceleration of m_1 (assume that the table is frictionless).

Homework Equations

Fnet = (m)(a)

The Attempt at a Solution

So I am assuming that the Fnet for the box m1 is T, which would be related to the weight of the pulley/box m2. Determining the value of T for box m1 is where I am having difficulty.

At first I assumed that Fnet box 1 = (m2)(g), but since there are two ropes on the pulley m1, I thought that Fnet box 1 = (m2)(g)/2. Both of these are incorrect when plugged back to find a: a = Fnet/m1.

Also, should the mass for finding the acceleration be just m1, or should I be adding m2 to that as well? Thanks!

 

[mantillab]

Still stuck on this. Any ideas?

 

[Anadyne]

You sound like you're just guessing how they are related. If you draw free body diagrams for M1 and M2, including the tensions, force of gravity, etc, then you won't have to "assume."

 

[mantillab]

I did draw individual force diagrams for each pulley. But I don't understand how the forces interact with each other. For example, for the second pulley between the m1 and m2 pulleys, I understand that tension is pulling in the negative x direction and that weight is pulling in the negative y direction. My problem is that I don't understand how the tension T force related to the weight of the blocks.

 

[Anadyne]

Let's call the tension associated with m1 and the left pulley T1.
Between the two pulleys T2,
and between M2 and the ceiling? T3.

The left pulley is not accelerating. Let's say that the clockwise direction (from T1's side to T2's side) is positive. This means Fnet = -T1 + T2 = 0, which means T1=T2. Does that make sense?

 

[mantillab]

The left pulley is not accelerating. Let's say that the clockwise direction (from T1's side to T2's side) is positive. This means Fnet = -T1 + T2 = 0, which means T1=T2. Does that make sense?

If the pulley on the left is not accelerating, how can you calculate the acceleration of the m1 block?

 

[Anadyne]

The left pulley isn't accelerating because it's attached to the edge there. This is NOT the same as saying that the strings aren't moving. The pulley on M2 IS accelerating along with M2. All they do is tell the rope which way to go. They are assumed to be massless and whatnot.

I think the acceleration of m1 is related to the acceleration of m2.. I know that the acceleration of m2 is less than that of m1 (if you compare to the situation where there is no T3 and it's just hanging at the end). I don't know if it's right to say that m2's acceleration is 1/2 a...

 

[mantillab]

I don't know if it's right to say that m2's acceleration is 1/2 a...

I'm think it is 1/2 because there are two strings acting on that pulley. How to factor that into solving for the acceleration, however, is completely beyond me.

 

[Anadyne]

I think it's 1/2 because think about how much the rope moves. M1 moves right by a distance, but M2 moves down by only half that distance.

Can you write down the equations that you now know?

 

[learningphysics]

Yes acceleration of m2 is 1/2 acceleration of m1. And since, nothing is mentioned about the pulleys... I think we can assume they are massless...

that means we can assume all the tensions are equal.

So just use one variable... T...

 

[physnoob]

I hope it's okay to bump this 2 yrs old thread up, because i have the same problem :P
Could someone please explain to me why the acceleration of m2 is half of m1?

 

[peterbishop]

I have this problem now as well and I would be able to do it fine but I'm not sure how to handle the 2 tensions on m2 to draw the free body diagram and Newton's 2nd law component equations. Are they equal? I assume they have to be equal as it is an ideal massless string, etc. Do we treat it as one force then, 2T? I tried this and it didn't work out right