Kinetic Friction Problem

1. Oct 16, 2009

r_swayze

A rescue worker pulls an injured skier lying on a toboggan (with a combined mass of 127 kg) across flat snow at a constant speed. A 2.43 m rope is attached to the toboggan at ground level, and the rescuer holds the rope taut at shoulder level. If the rescuer's shoulders are 1.65 m above the ground, and the tension in the rope is 148 N, what is the coefficient of kinetic friction between the toboggan and the snow?

Is the net force of the toboggan = Tension force - Friction force = ma = 0 ??

Is the Rope tension force = 148*cos(arcsin(1.65/2.43)) ??

I dont know if Im reading the problem about the rope part wrong . Should we be using trigonometry here?

2. Oct 16, 2009

ideasrule

Yes, but careful: the tension force is not the total tension in the rope, but the horizontal component of the total tension.

If you mean the horizontal component of the tension, yes.

I'm reading it the same way you are.

3. Oct 16, 2009

r_swayze

Yes, I meant the horizontal component of the rope tension.

So, that means that the Friction force is the negative of the rope tension's horizontal component?

I computed this and I got the wrong answer. I got a coefficient of kinetic friction of 0.0868, when the answer should be 0.0950.

Where have I erred?

4. Oct 18, 2009

cepheid

Staff Emeritus
The angle should be arctan(1.65/2.43), not arcsin.

EDIT: What I just said is totally wrong and arcsin was correct in the first place.

Last edited: Oct 18, 2009
5. Oct 18, 2009

cepheid

Staff Emeritus
I suspect that you used the full weight of the sled+skier as the normal force, and neglected the fact that the vertical component of the rope's pulling force would pull up on them slightly, reducing the normal force.