# Finding the coefficient of friction

1. Mar 19, 2009

### alexas

1. The problem statement, all variables and given/known data
It's a snowy day and you're pulling a friend along a level road on a sled. You've both been taking physics, so she asks what you think the coefficient of friction between the sled and the snow is. You've been walking at a steady 1.5 m/s, and the rope pulls up on the sled at a 22 (DEGREE) Angle. You estimate that the mass of the sled, with your friend on it, is 57.0 kg and that you're pulling with a force of 78.0 N.

What is the coefficient of friction?

2. Relevant equations

F = ma
F = Umg

3. The attempt at a solution

Umg = mg

U = (a/g)

but in the description of the problem acceleration is 0 since the speed is constant right? So wouldn't U (friction) = 0? since (0/9.8) = 0?

2. Mar 19, 2009

### myxomatosii

Your problem is that you are approaching this with the view that the only force acting on the sled is the friction force, but it is stated that there is tension coming from a rope with a force of 78N @ 22 degrees above the x-axis.

So if you draw a force diagram you will see the forces acting on the sled are a bit more than that.

[positive-y] UP: normal force , Fsin(θ)

[negative-y] DOWN: mg

[positive-x]RIGHT: Fcos(θ)

[negative-x]LEFT: fk

You would be right if friction was the only force acting on the object. For an object to move at a constant speed with no outside force other than friction acting on it, friction would have to be zero. However in this case you have a Tension coming from the rope which is counteracting the friction in order to keep the sled at a constant speed.

Last edited: Mar 19, 2009
3. Mar 19, 2009

### alexas

The force i am pulling in the y direction would be: 21.35 N
Down: 558.6 <<<<<<< isn't this irrelevant since we are moving in the x direction or does the angle make it nessecary for us to know this?

The force i am pulling in the right x direction is: 52.84

The force of friction going to the left is what we want to find, right?

4. Mar 19, 2009

### alexas

I was thinking of using the formula of:

Fy = MAy
and
Fx = MAx

But when i solve that out im getting the acceleration as basically
Ay = sin(22)
Ax = cos(22)

Which doesn't seem to be correct to me?

5. Mar 19, 2009

### myxomatosii

It is not the angle that makes it important to know "Down", it is the force of friction.

It may have slipped your mind, which is completely understandable but mg is the force which the object pushes down on the earth, at the same time whatever the object is sitting on always pushes back [perpendicular to the surface, which is more important on problems where you are on a slope and not so much this one] with a normal force, n.

So in the case of the sled, where you are not on a slope, normal force n is equal to mg.

n=mg

This allows you to substitute mg into the

fkkn

equation, otherwise how would you know what n is?

It does make sense, think briefly. You have two crates in your garage with very rough, torn up bottoms. One weighs 100lbs and the other weighs 5lbs, assuming you push from the base the 5lb crate will slide much easier because it is not pressing down so hard on the earth like the 100lb crate.

Last edited: Mar 19, 2009
6. Mar 19, 2009

### myxomatosii

I recommend scribbling the "up,down,left,right" forces on a sheet of paper for your visual understanding of what I did below.

Your goal is to create an equation in which all of your known variables are on one side and μk is on the other. See what you can do with this.

I may be wrong, but here are what I came up as the sum of your forces in the X and Y directions.

ΣFx=Fcos(Θ)-fk

ΣFy=n+Fsin(Θ)-mg

since the object is not accelerating in the y-direction, we can write

ΣFy=n+Fsin(Θ)-mg=0

therefore

n=mg-Fsin(Θ)

Last edited: Mar 19, 2009
7. Mar 19, 2009

### alexas

I'm a little confused on how to proceed. I get what your talking about with the object not accelerating in the y-direction, but i don't understand why we need to factor it in still? I mean i know we're pulling it at an angle so there is a Y component in the Force. I guess i'm just a little confused on how to proceed can you please clarify things?

8. Mar 19, 2009

### sArGe99

What equation do you use to compute the force of friction?

9. Mar 19, 2009

### alexas

Ff = μFN or Ff = μmg ?

10. Mar 19, 2009

### sArGe99

The friction force is f = μ. N
N is the normal reaction. What is the normal reacton in this case?
I tell you, it's not mg.

11. Mar 19, 2009

### alexas

N = mg - Fsin(Θ) ?

12. Mar 19, 2009

### sArGe99

Yea, exactly.
So what is the frictional force now and what balances it?

13. Mar 19, 2009

### alexas

So we know N is equal to: 537.25N
But, how are we supposed to find μ???

14. Mar 19, 2009

### sArGe99

We know, f = μ N
A component of the applied force F balances this frictional force f.

15. Mar 19, 2009

### alexas

huh?

16. Mar 19, 2009

### hsarp71

The x-component of the force (with which the guy is pulling the sled) must balance the frictional force in the opposite direction right?

17. Mar 19, 2009

### alexas

Yes, so that it moves the sled.

18. Mar 19, 2009

### sArGe99

Yea.
Equate the forces F cos(angle) and frictional force since the body is in dynamic equilibrium (moving with constant velocity).

19. Mar 19, 2009

### myxomatosii

Thats my fault most likely, I am trying to think of a way to explain this so you can be on the same page with me... let me see..

So you are pulling your friend on the sled, the reason you want to know the forces in the y-direction is because all of those y forces effect friction. (And in this case, that is ALL they affect, everything you are doing in the y-direction is to make sure you have your friction right) Normal force pushes up from the earth, counteracting the weight, so IF there was no rope n would just equal mg.

So, in a simple case. n=mg. But in your case [forget the x-component for now] the rope is pulling your sled up, complicating things by lifting up on the sled.

So the friction is going to be different, you are pulling up on the sled so at that moment it does not weigh "mg", it weighs

"mg=n+Fsin(Θ)"

This equation means that the weight of the object equals 1)The force that the earth is pressing back against your sled PLUS(literally "+" sign) 2)the force that you are pulling vertically on the rope. This makes logical sense, therefore if you stopped pulling up, mg would simply equal n. And we know this.

But we are not looking for "mg", we want "n". Which I have already given you, but I would like you to understand it as I am sure you do as well.

You need n because of the equation..

fkkn

But wait... you are puzzle. By substituting in the y-value we found for "n", that would just give you..

fkk(mg - Fsin(Θ))

Wait a minute, that doesn't help! If I solve μk in that it just gives me a stupid equation with fk and I don't know that! Bah!

But then you remember that you have not yet added up the forces in the x-direction...

ΣFx

Oh yea.. what would those be? Is friction the ONLY force acting on the sled in the x-direction? If so then perhaps we can truly not solve this problem, if not, then perhaps there is a force acting in the opposite direction which would have to be equal to maintain constant velocity... perhaps... let me know what you think.

Last edited: Mar 19, 2009