# Help on friction

1. Jan 31, 2005

### tobyguy

A man is pushing a box at a constant velocity. The weight of the box is 30 N and the force is 15 N. How do you arrive at the force of friction and the coefficient of sliding friction.

2. Jan 31, 2005

### Sirus

How much net work is being done on the box? What does that tell you about the frictional force, and consequently the coefficient of kinetic friction?

3. Jan 31, 2005

### tobyguy

Totally lost here but thanks for trying to help.

4. Jan 31, 2005

### gnome

"constant velocity"

What does that tell you about the forces acting on the box?

5. Jan 31, 2005

### tobyguy

I really don't know. Sorry. Thats why I am asking here. Thanks though for trying to help. I appreciate it.

6. Jan 31, 2005

### schattenjaeger

Constant velocity means there is no acceleration, hence a=0, so the force of friction is cancelling out that applied force...or something like that, I think

7. Jan 31, 2005

### gnome

Read the chapter again. Sorry if that sounds harsh, but it's really what you need to do. Otherwise you'll be just as lost with every question.

8. Jan 31, 2005

### tobyguy

See thats the problem there is only a very small section on this and nothing that helped but again thanks for your suggestion.

9. Jan 31, 2005

### Sirus

Consider the following formulas.

$$F_{net}=ma$$

$$W=Fd\cos{\theta}$$

W is work done by a force F over a distance d, in this case. Now consider the other posts in this thread. Does it make any sense?

10. Jan 31, 2005

### tobyguy

I dont know. I have been researching this on the Net and as close as I have come is possibly the coefficient would be 0.5 and force would be 15 =30 x .5 so force would be 15 but I am not sure. Anyway thanks for the help. Bedtime for me. :)

11. Jan 31, 2005

### Sirus

The main concept to understand here is that, because of the first equation in my last post, no net force acts on the box since its velocity is not changing. No work is being done on the box since the kinetic energy is constant, so the frictional forces must be equal in magnitude and opposite in direction to the applied force. Your answers above are correct, therefore.

12. Jan 31, 2005

### gnome

tobyguy: you're right, but you obviously have to work on it some more to convince yourself of that. Is there anything in particular that's puzzling you?

Sirus: his question and today's date both suggest that he is one or two days into a first course in physics, so any talk about work and energy is probably a few weeks premature.

13. Feb 1, 2005

### tobyguy

Correct gnome :) Thanks for your help all !