# Help with a force/torque equation

1. Jun 23, 2004

### Brianjw

Help with a force/torque problem

I have been sitting on this problem for well over 2 hours now and thought I'd try to maybe get some input from a third party. Below is the question as well as the work I've done and where I think I've made mistakes and can't figure out how to move from there:

(Sorry for the horrible LaTex code, first time posting)

Problem:

A large 16.0-kg roll of paper with radius R = 18.0cm rests against the wall and is held in place by a bracket attached to a rod through the center of the roll . The rod turns without friction in the bracket, and the moment of inertia of the paper and rod about the axis is .260 kg $$m^2$$ . The other end of the bracket is attached by a frictionless hinge to the wall such that the bracket makes an angle of 30.0 degrees with the wall. The weight of the bracket is negligible. The coefficient of kinetic friction between the paper and the wall is $$u_k = .25$$ . A constant vertical force F = 40.0 N is applied to the paper, and the paper unrolls.

Question:
What is the magnitude of the force that the rod exerts on the paper as it unrolls?

For an image of this you can look at:
http://members.cox.net/brianjw/roll.jpg

Here is what I've done so far:
T = unknown tension
$$\sum F_x = N_{wall} - T* \sin 30 = 0$$
$$\sum F_y = N_{rod} - 40N + T* \cos 30 - 16kg * 9.8 \frac{m}{s^2}= 0$$
$$\sum T = 40.0N * .18m -N_{wall}* .25 * .18m = I * \alpha, I = .260 kg * m^2$$

Which gives me(Since the center of mass has no linear acceleration):
$$T = \frac {40}{\sin 30 * .25}$$
which doesn't seem to get the right answer.

I suspect using the 40N is incorrect since the force isn't applied directly on the bar. I'm sure if I can just figure out my mistakes I should be okay. Thanks for the help, and post questions if you need anything else. I think I posted all that I have

Last edited by a moderator: Apr 21, 2017
2. Jun 23, 2004

### jdavel

brianjw,

Your force units problem is coming from that .18m. I don't think you want that in there. The force N has to be balanced by the horizontal component of the compression force in the bracket. You aren't given the force in the bracket, but you know it has to be along the direction of the bracket and you know the vertical component of that force has to be balanced by all the other vertical forces. Then I think the horizontal force component in the bracket should be the vertical component times tan30 (not sin30).

Try messing around with that and see what you get. This sure is an ugly problem! Who gave you this?

3. Jun 23, 2004

### Brianjw

Physics teacher, entry level college physics using calc. Requirement for my CS degree. I like physics, but its problems like these that make me think twice about it.

So are you saying change this equation:

$$\sum F_x = N_{wall} - T* \sin 30 = 0$$

To:
$$\sum F_x = N_{wall} - T* \tan 30 = 0$$

If not, I'm extremely confused at what I'm doing then.

4. Jun 23, 2004

### e(ho0n3

I recommend you make a new post whenever you change formulas or whatnot. That way, we can keep a log of the changes and point out the mistakes.

You are really confusing. You say T is an unknown tension, but of what? And why are you saying that this tension is equal to the torque? Keep track of the symbols you use to avoid confusion. I'm assuming that T is the 'tension' on the bracket. And what are you using as $\alpha$? I recommed you first study this sytem when it is at rest (in that, F = 0).

5. Jun 23, 2004

### jdavel

brian,

The previous poster makes a good point. You're getting your variables mixed up here. Now you have Nwall (which I didn't see in your original post) and then there's fwall. What are these? Also the suggestion to start by finding all the forces without the roll turning is a good one. Try that.

6. Jun 23, 2004

### Brianjw

Sorry about that, I didn't realize I made two T's. Here are my new forumlas:

$$\sum F_x = N_{wall} - T* \sin 30 = 0$$
$$\sum F_y = N_{rod} - 40N + T* \cos 30 - 16kg * 9.8 \frac{m}{s^2}= 0$$
$$\sum \tau = 40.0N * .18m -N_{wall}* .25 * .18m = I * \alpha, I = .260 kg * m^2$$

$$N_{wall}$$ is the Normal of the wall to the roll.
$$f_{wall}$$ is the frictional force to the wall, Since its a kinetic friction doesn't that mean it doesn't have any affect in the x or y axis?

T is the unkown tension of the bracket holding it up. Whats a bit confusing is that they want the force on the bar and not the bracket. So I'm not sure if I'm figuring out the bar with these forumlas or just the bracket? or is it considered the same thing?

$$\alpha$$ is the angular acceleratoin which is another part of the problem I need to solve after this. I thought by changing the angular acceleration to the Acceleration of the center of mass over it's radius I could zero out the equation?

7. Jun 23, 2004

### e(ho0n3

The kinetic friction in this case acts to slow down the rotation of the roll of paper even though the roll isn't going anywhere. In other words, the kinectic friction is countering the force F that is rolling out the paper. At least to my understanding.

Think of the bracket as a rope pulling on the bar. The tension of the 'rope' is acting on the bar.

I'm finding this confusing too since $\tau = I\alpha = Ia/R$ and since $a$ is 0, the torque is 0. Hmm...I need to think about this.

8. Jun 23, 2004

### jdavel

"Since its a kinetic friction doesn't that mean it doesn't have any affect in the x or y axis?"

No, it has an effect. Suppose the coeeficient of friction were infinite. Then no matter how hard you pulled on the right side, the roll wouldn't turn. So as you pulled on the right, there would have to be an equal and opposite torque on the left keeping the roll from turning. So the torques cancel. But the FORCES that cause those equal and opposite torques are in the SAME direction, namely down! Now since the coefficient of friction is finite here, you don't get an equal force down on the right (or and equal an opposite torque, which is why the roll turns), but it's still there, and you have to add it to the total downward force.

I think you're right about the tension in the bracket being the same thing as the force the rod exerts on the roll (Newton's 3rd law).

9. Jun 23, 2004

### Brianjw

I think my most confusing aspect right now is the 40 N being applied to cause it to roll. How would I find the force it applies to the bar since its not being applied directly to the bar?

Other then that, do my formulas look correct at least?

Thanks alot for the help, this problem is just driving me nuts and I still have more to do.

10. Jun 23, 2004

### Brianjw

What I meant by this is that we're only given the kinetic friction and not the static friction. Which would be the friction that would have an affect in the positive Y-axis. Would the static friction affect the $$\sum F_y$$ since its kinetic?

11. Jun 23, 2004

### Staff: Mentor

translational equilibrium

There seems to be a lot of extraneous information given. Is this just one part of the problem?

The forces on the roll are: Normal force of wall, friction, tension in the rod, the weight, and the applied 40N force. The roll is in translational equilibrium (the cm is at rest) so just apply the equilibrium condition for horizontal and vertical components. You should be able to easily solve for T.

12. Jun 23, 2004

### Brianjw

When you say the forces are in equilibrium that is saying the all forces in the system zero out. That would imply for me to calculate the forces on the roll minus the kinetic friction and without he 40N force since thats the force that is causing it to rotate.

So if it was not rotating, the force on the bar would be the tention from the bracket, Normal force of the wall, and the weight? Or would that be incorrect?

13. Jun 23, 2004

### Staff: Mentor

Right.
Why would you ignore the 40N force? It is certainly acting on the roll! Include all the forces acting on the roll.
Unless I'm missing something (that can happen ) the rotation is irrelevant. The tension in the rod is the force that the rod exerts on the roll. The normal force and the weight are forces on the roll, not the rod. (By "bar" you mean "rod", right?)

Do this: Take each of the forces acting on the roll (I listed them in my last post: there are 5) and find the x and y components. Add up the x-components and set equal to zero; do the same for the y-components. See what happens.

14. Jun 23, 2004

### Brianjw

Thanks will do as soon as I get off from work. I think that'll help alot!

15. Jun 23, 2004

### Staff: Mentor

Good luck with it. Let us know how you make out.

16. Jun 24, 2004

### Brianjw

Well it worked, I got the Force on the rod. But I don't get what I'm doing wrong on the second part. It seems really straight forward to me and I'm not getting the right answer.

I solved for the tension in the problem and got 226N. Now it would like me to find the angular acceleration in the problem. I know $$\sum \tau = I\alpha$$

In this problem we have 2 forces acting as torques. We have the friction and the 40N force. Therefore(Treating clockwise as the positive direction):

$$\sum\tau = 40.0N * .18m - .18m * 226N * Sin30* .25 = I\alpha$$ where we know I = .260 kg m^2 Should be simple to solve. I got the 226 * sin30 from the previous question. Since the tension is 226N I need just the X component from it.

Using that I should get:
$$\alpha = \frac{40.0N * .18m - .18m * 226N * Sin30* .25 }{.260 kg*m^2}$$

when I submit that answer it comes back as wrong and don't know why.

Last edited: Jun 24, 2004
17. Jun 24, 2004

### Staff: Mentor

recheck the tension

Your method for finding $\alpha$ looks good to me. But recheck that value for tension; I got 266N, not 226N. (But I did it pretty quick and may have fat fingered something.) Perhaps you just copied it wrong to part 2?

18. Jun 24, 2004

### Brianjw

Bah, you were right, damn me and my copying abilities heh. Well that problem is finished finally. Only got 3 more as part of the assignment that I can't figure out yet, gonna give those another go now.

Thanks!