# Friction and normal force on an incline

1. Oct 30, 2012

### sgstudent

I have an incline A that is very steep reaching a vertical height of h and another one B which is less steep with the same vertical height. So using the work energy theorem: in A, KE+work done against friction=mgh so the work done against friction and initial KE is equal to the gain in gravitational potential energy. In B, KE+work done against friction=mgh also. So again the work done against friction and intial KE is equal to the increase in mgh.

So from this we can say that the work done against friction is equal to each other as in both cases the initial KE is the same as each other. (KE+work done=KE+work done). But in A, the distance is lesser than in B. Since work done=forceXdistance so the friction in A is greater than in B?

And since friction is the coefficient of friction multiplied by the upwards force, since the roughness is assumed to be the same for this example, as A is greater than B and the coefficient is the same so won't A have a greater normal upwards force? Thanks for the help

image:http://postimage.org/image/590thw81f/full/

Last edited: Oct 31, 2012
2. Oct 30, 2012

### Nguyen Quang

You are contradicting yourself. As you have assumed, the frictional force is the same in both cases. As a result, the work done in B would be more than the work done in A because the distance covered by a 'body' sliding down slope A would be less than B, hence, less work is done against friction in A as compared to B.

By Principle of Conservation of Energy, the final total energy must be equal to the initial total energy. Given that in both cases, the 'body' possesses the same amount of initial Gravitational Potential Energy, the final total energy must be the same. As a result of larger work done against friction in B, the 'body' in A will possess more Kinetic Energy at the end of the slope.

Now, visiting the question about the Normal Force, you can draw out the free-body diagram to see it for yourself. The resultant force of Friction and Normal force should be opposite and of the same magnitude with Weight. What you get would be a larger Normal force in case B.

Hope that helps :)

3. Oct 31, 2012

### sgstudent

I'm not sure what you mean because in http://www.physicsclassroom.com/mmedia/energy/au.cfm it shows that the work done to bring it up is the same. So KE+work done against =mgh for both cases right?

I didnt say that the friction was the same. I mentioned that the work done against friction is the same. Again you contradict yourself with the normal force. In both examples the normal force is different due to the different angles of inclination as I mentioned in the start. Since friction is μN and N is different the friction cannot be the same?

Is there something wrong with my theory here? Thanks for the help

Hi I also edited the first post to make it clearer.

Last edited: Oct 31, 2012
4. Oct 31, 2012

### Nguyen Quang

I agree with you on the part where KE+work done against =mgh. I did not disagree with it. It adheres to the principle of Conservation of Energy.

I will get this straight. The Normal force and friction force are different in both cases. The work done against friction is also different in both case.

The real problem is that your question does not specify any details on the slope as well as the 'body' in motion. As such, I have to assume one variable to be constant to tell you the exact answer. If you want to consider both variables then there will be no exact answer.

5. Oct 31, 2012

### nasu

If they both start with the same KE, they will not reach the same height, h, due to the friction.
So you have either the same KE and different values of h or same values of h but different initial KEs.
It would be useful for both yourself and people reading your posts, to specify the actual problem, with all conditions. Playing with equations out of context is quite useless and confusing.

6. Oct 31, 2012

### haruspex

It's not clear which way the block moves. If it starts at the top at rest and slides down then yes, final KE+ work against friction = mgh. But it sounds like it starts at the bottom with some initial KE and comes to rest at the top. In this case friction moves to the other side: initial KE = mgh + work against friction.
It depends what's the same and what's different in the two set-ups.
If the coefficient of friction is the same and the height is the same, the work done against friction cannot be the same. The normal force varies as cos(θ) (angle of slope to horizontal), so the dynamic friction does too. The distance the force moves, for fixed h, varies as cosec(θ). So the work done must vary as cot(θ). It follows that the change in KE must be different.