Another friction problem!

[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?

[Pyrrhus]

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

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 :biggrin:]

[Pyrrhus]

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

[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!

[Pyrrhus]

Y-axis:

F_{y} + N = mg

N = mg - F_{y}

X-axis:

F_{f} = F_{x}

\mu N = F_{x}

[Pyrrhus]

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

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

[pinky2468]

what is N, is that the normal force?

[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

[Pyrrhus]

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