# Friction questions

1. Oct 22, 2006

### rofljohn

I'm having trouble with friction so bear with me, i might have a lot of questions...

If i have an object moving at a constant speed over a level surface, how do i find the friction force? i'm only given the force being put on the object (470 N) and the mass of the object (23 kgs)?

2. Oct 22, 2006

### Dorothy Weglend

Dorothy

3. Oct 22, 2006

### rofljohn

No not really :| I can't find a formula that gives me kinetic friction

Also on another problem i think i have the answer but I'm not sure. A mass of 37 kg slides down a 17 degree incline at a constant speed. Is the mu equal to ~.31?

4. Oct 22, 2006

### Dorothy Weglend

Yes, that's the right answer. If you can solve this one, you can solve your first one. And you must know the equation F_friction = mu N, or you couldn't have solved this one

Dorothy

5. Oct 22, 2006

### Dorothy Weglend

Umm, actually, you don't even need F = muN to solve the first problem. Sorry about that.

6. Oct 22, 2006

### rofljohn

I still have no clue where to begin on that problem.

I did another one. A mass of 45 kg is sliding down a 34 degree incline with a mu of .32

7. Oct 22, 2006

### Dorothy Weglend

Yes. That is correct also. These problems are harder than the one you can't solve.

You have a force pushing on an object. You have a frictional force opposing that motion.

Consider what would happen without the friction. Since the force is 470 N, with no friction, the object would accelerate at 470 N/23 kg = 20.4 m/s^2.

But what is it's acceleration now? If you realize what that is, then you can write the force equation using the second law and solve for the friction force.

8. Oct 23, 2006

### rofljohn

Since it's moving at a constant speed the acceleration is 0. So does that mean the friction force is 20.4?

So the mu is .09?

9. Oct 23, 2006

### Dorothy Weglend

20.4 is the acceleration without any friction. F = ma, so a = F/m is how I got that.

You are on the right track, though, with a = 0.

Consider:

Sum of forces = ma, a = 0, so Sum of forces = 0.

What are the two forces that add to zero?

10. Oct 23, 2006

### rofljohn

Friction force and the force being pushed on the object?

11. Oct 23, 2006

### Dorothy Weglend

Bingo.........

12. Oct 23, 2006

### rofljohn

Sooo does that mean the friction force is equal to 470? That doesn't make any sense though because wouldn't it have to be less in order for the object to move?

13. Oct 23, 2006

### Dorothy Weglend

That's right. 470 N is the force of friction.

There's two kinds of friction involved in these types of problems. One is called static friction. This is the friction that keeps an object from moving. Then there is kinetic friction, which operates when an object is moving. This is an example of kinetic friction.

They work very similarly. With static friction, you apply a force, and the frictional force opposes that with an equal amount of force. As you increase the applied force, the static friction increases to match it. Eventually, you apply enough force that static friction can't increase, and the object begins to move.

At that point, it's all about kinetic friction. Since the mass doesn't change, the kinetic friction reduces the acceleration of the object. If there is enough kinetic friction (or not enough applied force), then the acceleration is reduced to zero, but that doesn't mean the object stops moving.

But I have to say you are doing very well with these friction problems.

Good luck,
Dorothy

14. Oct 23, 2006

### rofljohn

Well for most of them it's a matter of finding the formula and plugging in the numbers. The ones in which I don't do that I'm for the most part lost.

On this next problem, I am supposed to find what force would be needed for a 35 kg mass to go up a 25 degree incline at 3 m/s/s with a mu of .11.

The net force is 105 N. What do I do with this? I've found the normal force to be 95 and the friction force to be 10.

15. Oct 23, 2006

### Dorothy Weglend

I hope you are drawing free body diagrams. Even if these problems seems easy, later ones won't. The practice of drawing the diagrams, and translating them into mathematics will help later, even if it seems silly now.

In any case, in the current problem, you are stuck at the same place you were in the first one. You just apply the Newton's second law, as you did when you succesfully solved the first problem, and solve for the unknown force.

16. Oct 23, 2006

### rofljohn

I drew a free body diagram for this one.

Here's what I got.

...........|
...........| <- Fn = 95
........o---o
???---|35 |---- <- Ff = 10
........o---o
...........|
...9.8-> |

I am trying to find the ???.... Do I take the net force minus the Ff or what? If i'm doing something wrong or have wrong calculations please tell me or else I'll get nowhere.

17. Oct 23, 2006

### Dorothy Weglend

Isn't this on an incline? Maybe you just left that out because it's hard.

I think your value for the Normal force is wrong, probably because you don't quite have your FBD right yet.

Draw a block on an incline. Then attach all the forces. You should have four forces (so four arrows). Assuming you have the X axis running parallel to the incline (tilted at 25 degrees, in other words), one of these will have to be decomposed into components parallel to your tilted axis.

Phew. Well, I hope that helps. I am getting sleepy, but will try to stay up a little longer.

18. Oct 23, 2006

### rofljohn

I don't know, it's still confusing to me but I think I'm going to go to bed and talk to the teacher about it in the morning. Thanks for all the help though.