# Another friction problem!

1. Sep 17, 2004

### pinky2468

Ok here is another problem that I am stuck on!
A 20.0kg sled is being pulled across a horizontal surface at a constant velocity. The pulling force has a magnitude of 80.0N and is directed at an angle 30.0 degrees above the horizontal. Determine the coefficient of friction.
I think that Fn=(20.0)(9.80)(cos 30) and I know that uk=Fk/Fn and a=0m/s(sqr) but I am stuck on how to go about finding Fk?
Am I on the right track?

2. Sep 17, 2004

### Pyrrhus

You're using Newton's First Law $$\sum^n_{i=0} \vec{F_{i}} = 0 \Rightarrow \vec{V} = constant$$ that's good .

Now, Remember you got a weight pointing down, the normal force pointing up, friction that always points against movement, so in this case is pointing left, and you got a force pointing at angle 30 degrees, so put the force in its components Fy and Fx.

[Edits: I was just checking LaTex ]

Last edited: Sep 17, 2004
3. Sep 17, 2004

### Pyrrhus

If Fn is normal force then you are not calculating it right.

4. Sep 17, 2004

### pinky2468

So is it W=20(9.80)=196N and then 80(cos 30)-Ff=0 and Ff=69N
I know that something is still wrong because I am still getting the wrong answer!

5. Sep 17, 2004

### Pyrrhus

Y-axis:

$$F_{y} + N = mg$$

$$N = mg - F_{y}$$

X-axis:

$$F_{f} = F_{x}$$

$$\mu N = F_{x}$$

6. Sep 17, 2004

### Pyrrhus

By the way your problem is asking:
not the force of friction!

If you notice Coefficient of Friction has no units and its represented as $$\mu$$

7. Sep 17, 2004

### pinky2468

what is N, is that the normal force?

8. Sep 17, 2004

### K.J.Healey

From what I did, your Ff is right, but you need to solve for Fn too.

Sum forces in each direction, set it equal to Ma

Sum in the Y = -m*g + Fn + 80sin(30) = ma = 0 // NOT ACCELERATING

Sum in the X = 90cos(30) - Ff = 0 // Not accelerating again.

So solve top for Fn, solve bottom for Ff, then use Ff = mu*Fn

I get 0.44 = mu_k

9. Sep 17, 2004

### Pyrrhus

N is the normal force, yes, i represented it as such

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