# Kinetic Friction toboggan Problem

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?

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

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

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

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

I dont know if Im reading the problem about the rope part wrong .

I'm reading it the same way you are.

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?

cepheid
Staff Emeritus
Gold Member
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:
cepheid
Staff Emeritus