Kinetic Friction toboggan Problem

  • Thread starter r_swayze
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  • #1
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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?

I have two questions about this problem.

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?
 

Answers and Replies

  • #2
ideasrule
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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.
 
  • #3
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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
cepheid
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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:
  • #5
cepheid
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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.
 

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