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:(adsbygoogle = window.adsbygoogle || []).push({});

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!

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# Homework Help: Simple question about tension in rope

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