Work-Energy involving two forces on a pulley.

[Ascendant78]

Homework Statement

Homework Equations

The Attempt at a Solution

Well, I tried working on this problems a few different ways and kept hitting a dead end. The work in "relevant equations" was my last attempt. I have spent well over an hour on this problem and I am ridiculously frustrated.

Our professor never covered this type of problem in our course and I just can't seem to figure it out on my own. I found the force of the rope the man is holding onto to be 935N, which I am almost certain is right. However, I am not sure how to go about solving work-energy for the pulley. I don't know how to incorporate torque forces into the work-energy equation. If someone could help point me in the right direction, I would greatly appreciate it.

 

[Simon Bridge]

The problem statement has no question in it.
Are you supposed to find how hard the brake was applied?

I see a bunch of calculations but it is hard for me to follow your reasoning at each stage.
You appear to be wanting to use conservation of energy arguments - so the initial gravitational and kinetic energy turns into work by friction? But I don't see anything that looks like a calculation of the work done by friction.

The use of free body diagrams suggests that you wanted to use a force argument though - balance of torques, and kinematic equations?

Could it be that thinking about two methods at once has lead to some confusion?

 

[Ascendant78]

Oh, sorry. Yes, the answer in the back of the book is for the force the brake applies to the spinning pulley. Also, yes to the two methods getting me a bit confused.

What I did first was used work-energy to find out how much force the rope attached to the man must have applied in order to stop him at the 75m mark. The value I got for that was 935N. However, when it came to solving work-energy for the pulley, that is where I got lost. I wasn't sure how to analyze the two forces (friction and force of rope attached to man) in the work-energy equation. I figured torque should be considered since the amount of force would be dependent on the distance from the center of the pulley, but I am not sure on that one. I really am completely lost on it.

 

[Simon Bridge]

The tension in the rope is doing work on the man? OK.
If you push the break with normal force $N$, then the force of friction is $f=\mu N$

The friction causes a torque - the work done by a torque is $W_\tau=\tau\theta$ ... there $\theta$ is the total angular displacement of the pulley in radians.

But I'd suggest you do the whole energy from everywhere in one go ... otherwise you'll get some confusion about what to do about that $F_{rope}$.

initially there is energy in:

KE for the man
KE for the pulley
height of the man above final position

finally the energy went into: work from friction.

I'm curious that there are two values for the coefficient of friction.

 

[Ascendant78]

Thanks so much! I got the right answer! Our book didn't even have that formula for work done by torque in it, so where I was going wrong was applying the torque over the distance it was applied (75m) rather than the angle. I even tried a few other things, but of course none of it came out right. Oh, and as far as the two values for the friction, our book shows (static, kinetic) values for all problems, even if you only need one of the values.

One thing I am curious about... the only problem they demonstrated that came close to this involved only variables. However, in that problem they seemed to address the forces against the pulley as regular forces rather than torque forces. Can you tell me how I could go about doing that in this problem? I just want to be able to see this from as many angles as possible for future problems where one method may be easier than the other.

Anyway, if you could fill me in on that, I'd really appreciate it. I also really appreciate you helping me out with this problem. It really was driving me crazy and it is extremely frustrating that it was all because it didn't show me the equation for work done by a torque.

 

[Simon Bridge]

In this problem, you could consider the friction force times a distance instead of the torque times the angular displacement. i.e. $W= fd: d=\theta/R$ and $f$ is the friction force. It's as if the brake-block had slid along a surface distance d.

You'll probably also have been given a formula for turning a 2-radius pulley system into an equivalent 1-radius pulley. That way you could just use the "effective friction" times the distance fallen.

 

[Ascendant78]

In this problem, you could consider the friction force times a distance instead of the torque times the angular displacement. i.e. $W= fd: d=\theta/R$ and $f$ is the friction force. It's as if the brake-block had slid along a surface distance d.

You'll probably also have been given a formula for turning a 2-radius pulley system into an equivalent 1-radius pulley. That way you could just use the "effective friction" times the distance fallen.

Ok, now I see it. Once I took into consideration the rotations and the resulting distance the friction is applied for as a result, I got the same value for work. Great, now I have a couple ways I can look at these types of problems. Thanks again so much for the help, I really appreciate it.