# Homework Help: Force problem

1. Oct 10, 2005

### xidis

encountered this problem on a quiz in my physics class. The teacher has one way of doing it, which I believe is wrong.

A box sits on an incline plane set at 35 degrees to the horizontal. Find the coefficient of friction if the box is being pulled up the ramp at a constant velocity (net force is zero). The box is 10 Kg.

The force equations are:
Fw = mg
Fn = Fw * cos(angle)
Fa = Fw * sin(angle)
Ff = Fn * coefficient of friction

My teacher believes the answer to the problem can be found by taking the tangent of the angle. He gets this by saying the force applied of the box is equal to the force friction Fa = Ff. From this equality, you can get tangent of the angle equals coeffient of friction.

I and several of my fellow students believe he is wrong. We believe he is completely disregarding the force due to gravity which wants the box to move down the hill. So, in order for the box to move up the hill, the force moving the box up the hill must overcome the force that wants to move the box down the hill and the force that wants to keep the box in place (friction). Thus, Fup = Fdown + Ffriction. This problem does not give enough information to solve it that way. I don't see how my teacher is correct because his idea doesn't make sense in the real world or even with math. Any thoughts?

2. Oct 10, 2005

### Tide

Listen to your teacher! If the box is moving at constant speed then the net force acting on the box is 0. What can you infer from that? :)

3. Oct 10, 2005

### xidis

Im not disagreeing with that. I say that the force pulling the box up the ramp is equal to the force of friction and the force pulling the box down the ramp, due to gravity. My teacher says the force going up the ramp is equal to force friction.

I say:
Fup = Fdown + Ffriction

My teacher:
Fdown = Ff = Fup

4. Oct 10, 2005

### arildno

Since for any coefficient of friction there exist a force so that you may pull the box up with constant velocity, the information given is insufficient to specify a single solution.

5. Oct 10, 2005

### Tide

If the force upward along the incline is not zero then the box must accelerate up or down the incline.

6. Oct 10, 2005

### xidis

Tide, the net force of the box is zero. Fup - (Fdown + Ffriction) = 0.

7. Oct 10, 2005

### Tide

Oh, wait! I see what you're saying. You have, in fact, already set the total force to zero. Nevermind! :)

8. Oct 10, 2005

### arildno

Time for a session, Tide..