# Simple question about tension in rope

1. Oct 30, 2005

### unnamedplayer

Hi all, I was hoping for a little help with this problem. I already have the solution but I don't understand how they got the answer. The question is:

A sledge loaded with bricks has a total mass of 18.0 kg and is pulled at constant speed by a rope. The rope is inclined at 20.0° above the horizontal, and the sledge moves a distance of 20.0 m on a horizontal surface. The coefficient of kinetic friction between the sledge and the surface is 0.500. (a) What is the tension of the rope?

Now I originally thought that the forces in the x direction were: Tcos0 - f (where f = kinetic friction) Therefore, I thought I need to find f by first finding the normal force and then multiplying by the coeff. of kinetic friction. To do this I thought the forces in the y direction were: n + Tsin0 - mg = 0.

This made it a problem solving for n - the normal force. So I looked at the solution and it show they found the tension by saying:

Tcos(20) = umg (where u = coeff of kinetic friction)

That is what I don't understand. How can Tcos(20) = umg? That would mean the normal force is equal to mg which I don't understand how because shouldn't it be less because the rope is kinda of pulling it in that direction too (ie the Tsin(20) portion of the y-direction forces).

Any way I'm sorry for the long post but any help is greatly appreciated!

2. Oct 30, 2005

### alias25

normal reaction plus the compoment of the tension in the virtical should equal mg
i think this may involve a simultaneous equation

Last edited: Oct 30, 2005
3. Oct 30, 2005

### unnamedplayer

Yes that is what I got. In the vertical direction I came up with n + Tsin20 - mg = 0 so n + Tsin20 = mg or n = mg - Tsin20. But then I can't find the normal force because I don't know T.

But the solution just shows Tcos20 = umg (again u = coeff of kinetic friction)
Which I can't seem to be able to understand why.

4. Oct 30, 2005

### Staff: Mentor

Your equation for vertical forces is correct, but what about horizontal forces? Write the equation for horizontal forces (expressing the friction in terms of normal force) and combine it with this one. You'll have two equations and two unknowns. Solve them simultaneously.
The solution given is wrong; your equations are correct.

5. Oct 30, 2005

### Fermat

It doesn't. Thar's wrong!
That's right.
But they are the only forces and since the sled is moving with constant velocity, then the net force is zero. i.e. Tcos20 - f = 0.
That's right too.
I assume that where you have Tcos0 and Tsin0, they should be Tcos20 and Tsin20.

6. Oct 30, 2005

### unnamedplayer

Hi again all. Thank you for all the help! I'll have to have a talking with my TA. I thought I was going crazy for a little bit there :tongue:

Now forgive me but as you've probably notice I'm not really good at this stuff but this is how I proceeded. I made n = mg - Tsin20. My equation for the x-direction was Tcos20 - un so I plugged in n and got Tcos20 - u(mg-Tsin20) and went farther making it Tcos20 - 0.5mg + 0.5Tsin20 = 0 and solved for T getting T = 79.46. That doesn't match what's given in the solution I have but since the procedure was wrong in that I'm assuming the tension they came up with is also wrong. So I'm wondering if that sounds right or do I have to combine some equations (I'm not really sure which ones I'd have to combine thats why I tried doing it like this first). Any way thanks again for all the help!

7. Oct 30, 2005

### Pyrrhus

Is that the problem statement? i suspect there's more to it than that.

8. Oct 30, 2005

### Staff: Mentor

Looks good to me. (Of course it won't match the incorrect solution given.)