Tricky Pulley System: How to Solve for Tension and Acceleration?

[Coldie]

Hi guys, this question is driving me up the wall because I always thought pulley system questions were easy and enjoyable to do, but this one refuses to cooperate!

The question:
Someone exerts a force F directly up on the axle of a pulley. Two objects, m1 of mass 1.2kg and m2 of mass 1.9kg, are attached to the opposite ends of the string, which passes over the pulley. The object m2 is in contact with the floor.
a) What is the largest value the force F may have so that m2 will remain at rest on the floor?
b) What is the tension in the string if the upward force F is 110N?
c) With the tension determined in part b, what is the acceleration of m1?

Now, for part a I added the applied force to the gravitational force of m1 since I figured that both forces are pulling up on m2, and I came up with 6.68N.

For part b, I'm confused because I don't know how the upwards force on the pulley is distributed between the left and right weights. I've tried a bunch of different things, but I can't get the tension force to be equal on both weights, telling me there's a problem. Can anyone help?

Thanks.

 

[Doc Al]

Start by analyzing the forces acting on the pulley. I assume that the pulley is massless and the connecting rope is massless and unstretchable. Hint: What's the net force on the pulley?

 

[Chi Meson]

Part a, have the system be both masses plus pulley. With this you can find a term for the net acceleration of the two masses. At the point in question, the acceleration of m2 will be zero.

 

[Coldie]

Start by analyzing the forces acting on the pulley. I assume that the pulley is massless and the connecting rope is massless and unstretchable. Hint: What's the net force on the pulley?

Yes, the pulley and rope are massless and frictionless, sorry for not mentioning that.

The net force on the pulley is 110 - 3.1(9.8) = 79.62N. Thus, the net acceleration of the system will be 25.68m/s^2. However, I'm still left with the problem of how this acceleration is, well, distributed among the two masses. I imagine that pulling the pulley up will accelerate the smaller weight's ascent up to the pulley and the larger weight's descent away from it, but I'm just not sure how to gauge the changes in acceleration of each.

Part a, have the system be both masses plus pulley. With this you can find a term for the net acceleration of the two masses. At the point in question, the acceleration of m2 will be zero.

I found the net acceleration above, but why will m2's acceleration be zero? Wouldn't both masses be lifted into the air?

Sorry to seem to thick, I'm in my second year of post-secondary and haven't taken Physics since High School

 

[Chi Meson]

a) What is the largest value the force F may have so that m2 will remain at rest on the floor?

If m2 is at rest on the floor, then it's acceleration must be zero.

My very first hint turns out to be unnecessary.

I get 37.2N as an answer for a). DocAl, do you agree?

 

[Coldie]

I solved a and got 6.68N, as stated in the first post... is my solution wrong? Right now I'm asking for help on part b.

[edit]
Sorry, you stated right in your post that you were talking about part a, I somehow overlooked that!

 

[Chi Meson]

The net force on the pulley is 110 - 3.1(9.8) = 79.62N. Thus, the net acceleration of the system will be 25.68m/s^2. However, I'm still left with the problem of how this acceleration is, well, distributed among the two masses. I imagine that pulling the pulley up will accelerate the smaller weight's ascent up to the pulley and the larger weight's descent away from it, but I'm just not sure how to gauge the changes in acceleration of each.

The net force on the pulley will be 110N - 2T. T does not have to equal the the combined weight. In fact, it will not.

Edit: OK another useless hint, since the pulley's massless.

With a frictionless and massless pulley, the tension on both sides of the string must be equal. Doesn't that mean that T is half of the applied force?

 

[Coldie]

Wow, thanks, it hadn't crossed my mind to look at it that way. Am I right in subtracting the weight of the masses to find the net force on the pulley? If so, then halfing the weight of both masses combined would give the tension force?

 

[Chi Meson]

Wow, thanks, it hadn't crossed my mind to look at it that way. Am I right in subtracting the weight of the masses to find the net force on the pulley? If so, then halfing the weight of both masses combined would give the tension force?

You don't really care about the pulley. The net force on the system is not all that relevant (I was at first thinking this was a much harder problem where the pulley did have mass). Look at the sum of forces on the individual masses if T is half the pulling force.

 

[Coldie]

I'm aware that the pulley isn't relevant except in the manner that it effects the interaction of the two masses. I only mention it because the net force on the pulley is the net force on the system, right?

I don't understand why the tension force would be half of the pulling force as opposed to the full pulling force plus the weights of the masses- why isn't the tension in the rope the combination of both?

If the tension force is one half of the pulling force, the net force on m1 will be 55 - 1.2(9.8) = 43.24N and the net force on m2 will be 55 - 1.9(9.8) = 36.38N.

This seems to be the right answer. Can you please tell me how you knew that the tension force is one half the pulling force?

Thanks!

 

[Chi Meson]

Because the tension has no choice but to be exactly the same in both sides of the pulley (if the pulley is frictionless and massless). What other two numbers could be exactly the same and still add up to the total pulling force?

And to show why they must add up to the total pulling force, now you can consider the massless pulley. One string pulls up, two strings pull down. Force up must equal force down (since massless, no force is "used" to accelerate the pulley).

 

[Doc Al]

This seems to be the right answer. Can you please tell me how you knew that the tension force is one half the pulling force?

By analyzing forces on the massless pulley, which is why I suggested you start there. The forces must add to zero, which immediately tells you (as Chi Meson has explained) that the tension is half the pulling force. That's the key to quickly solving this, as you've discovered by now.