Moving Pulley Mechanics Question

Boorglar

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

A string is attached on a pulley, with Mass 1 (2 kg) hanging from the left and Mass 2 (5 kg) hanging from the right. The pulley itself is being pulled upwards with a force of 100N.

Find the tension T in the string, and the accelerations of both masses.

We consider the pulley and string to be massless, and no friction. Gravitational acceleration is g=9.8 m/s^2

F = 100N
m1 = 2kg
m2 = 5kg
a1 = acceleration of Mass 1
a2 = acceleration of Mass 2
a = acceleration of the whole system
T = tension in the string

2. Relevant equations

m1 a1 = T - m1 g
m2 a2 = T - m2 g

3. The attempt at a solution

First I found the acceleration of the system:

the total force on the system is F - (m1+m2)g (because gravity pulls it downwards). So

a = F/(m1+m2) - g.


Since the string is not extendable, the relative accelerations of the masses with respect to the pulley are equal in magnitude and with opposite signs:

a1 - a = -(a2 - a)

a2 = 2a - a1 = 2F/(m1+m2) - 2g - a1

Replacing a2 in the Relevant equations:

m1 a1 = T - m1 g
m2 [2F/(m1+m2) - 2g - a1] = T - m2 g

After solving this system for T (by multiplying the second equation by m1/m2 and adding both equations to eliminate the a1 term), I get:

T = 2m1m2F/(m1+m2)^2 = 2000/49 = 40.82 N

But the answer is supposed to be 50N. And the accelerations I get are also not the same as in the answers.

Where did I go wrong?

Doc Al

Careful with this. Here you're finding the acceleration of the center of mass of the system, which is not what you want.

Hint: If you analyze forces on the pulley, you'll immediately find the tension in the string.

Boorglar

Well, I wanted first to find the acceleration of the center of mass because I had to find the relative accelerations of m1 and m2 with respect to the moving center of mass (because they have to be opposite signs).

Isn't the formula for relative acceleration with respect to a moving system :

acceleration of the object - acceleration of the system ?

Doc Al

The acceleration of m1 and m2 with respect to the pulley are equal and opposite. But the acceleration of the pulley is not the acceleration of the center of mass.

Boorglar

Ok, well, in this case I have no idea how to solve it...

Can you explain how I would find the acceleration of the pulley?

Doc Al

Why not do as I suggested? Analyze the forces on the pulley.
Why would you want that? (You certainly could find it, in a manner similar to what you tried. But why do you want that?)

Boorglar

Well, Itried to analyze the forces on the pulley but I couldn't go much further... I guess I don't see what you mean. Maybe I am completely misunderstanding something?

The forces on the pulley are the upwards force F = 100 N and the downwards weight of the string with masses, right?

So the net force on pulley = F - (m1+m2)g?

Or is it F - 2T? (because the two strings pull the pulley downwards with the same tension T?

I tried to use both and I still didn't get the right answer...

Doc Al

No. The weight of the masses does not act on the pulley. The tension from the string does.

That's it. (Why do you say you didn't get the right answer?)

Boorglar

Ah wait, are you saying that T = F/2 ???

This does give the right answer! Thanks!


I just don't understand something:

The net force on the pulley is F - 2T. Since the pulley is massless, if the force were anything other than 0, the pulley would have infinite acceleration, which is impossible.
So the net force has to be zero and so F-2T = 0 so T = F/2 = 50N

But this means that the pulley can never accelerate? I don't quite understand that because if I try to find the acceleration of the pulley I would get 0/0... And this is the first time I ever see a problem where the way to find a force is to realize that if it is not 0, then you get infinities...

Doc Al

Right. As simple as that.

So far, so good.

Not at all. Of course it can accelerate.
That just means that you cannot use that method to find the pulley's acceleration.

It comes up whenever you treat something as massless (which is just a very useful approximation, of course). For example, consider a massless rope. Since the net force must be zero, the tension force on each end of it must be the same. That's why we say that the tension is the same throughout a segment of massless rope.