# Hi, need some help with friction

1. Dec 28, 2003

### I_BeLeaf87

I have this assignment to do and I am stuck on finding the force of friction acting on a curling rock.

The mass is 19Kg and the acceleration is -0.2m/s^2

I calculated the force applied using F = ma

This gave me -3.8N

I let negative rep force applied
positive rep force of friction

Therefore I have 3.8N [forward]

I am not sure if I am right from here on. I think I calculated the normal force or so.

I used

Fn = ma + mg

My answer for the force of gravity was 190N so then I added the 2 together and got 186.2 since the Force applied was negative. What I find awkward is that 186.2 N is the actual Force of gravity but due to sig digits, I rounded to 2 sig digits. It is here I am lost. How will I be able to calculate the force of friction?

I don't know the co-effecient of friction either. We also need to solve for that but I think I know how to once I solve for friction. Any help will be appreciated.

2. Dec 28, 2003

### jamesrc

You didn't say this explicity, but it seems like the rock is sliding and no other forces are acting on it. If that's the case, then the only force on the body is friction: Fnet = ma = f. The negative acceleration implies force opposing the motion, so the magnitude of the friction force is 3.8 N, the force you calculated.

The normal force in this problem is equal to the weight; check a free body diagram if you don't see this: Fnet = N - mg = ma = 0 (it doesn't move up or down). (Looking over what you wrote, your equation is right, you just have to recognize that a = 0 in the vertical direction.)

Modeling this as kinetic friction, use the fact that f = &mu;N to solve for &mu;, the coefficient of friction.

3. Dec 28, 2003

### I_BeLeaf87

OK I don't get the part where you are explaining

Fnet = N - mg = ma = 0

N signifies Newtons, right? As in 3.8N, right?

I did this then

Fnet = 3.8N - 186.2 = -182.4 = 0

How does it equal zero?

Or is it

Fnet = 3.8N - 190 = -186.2 = 0 and that means that the 186.2 will cancel each other out.

I also tried to figure out the coeffecient of friction, I came up with 0.02, I am not sure if I am correct. I am not too good at this stuff. Dynamics is my weakness.

BTW, thank you very much for your help

4. Dec 28, 2003

### jamesrc

My apologies for the ambiguity. N is the normal force. You called it FN and I should have followed suit (or at least explained the notation). Does that clear it up?

5. Dec 28, 2003

### I_BeLeaf87

Oh ok, yes it does help. I can see where you get the acceleration to be equal to zero.

So

186.2 - 186.2 = ma = 0

So how do I solve for friction?

OK hold on. To calculate the coeffecient of friction I will have to divide 3.8N/186.2N to give me 0.02 as the coeffecient. It's the friction part thats confusing me.

I'm sorry if I seem like a pain.

6. Dec 28, 2003

### jamesrc

No problem.

Your whole last paragraph is right. Summarizing the milestones of the problem:

Friction force = f = (19 kg)*(0.2 m/s/s) = 3.8 N

Normal force = N = mg = (19 kg)*(9.8 m/s/s) = 186 N

coefficient of friction = &mu; = f/N = 3.8/186 = 0.02

How's that?

There are 3 forces acting on the body: the weight, the normal force (that balances the weight) and the friction, opposing the motion (decelerating).

7. Dec 28, 2003

### I_BeLeaf87

Oh I was not even thinking that the friction force was 3.8 N. I was thinking that is was the force applied or something. I get it now. Thank you very much for your help, I really really appreciate it