Coefficient of friction between the sled and the snow is?

Question

It's a snowy day and you're pulling a friend along a level road on a sled. You've both been taking physics, so she asks what you think the coefficient of friction between the sled and the snow is. You've been walking at a steady 1.5 m/s, and the rope pulls up on the sled at a 24.0 degrees. You estimate that the mass of the sled, with your friend on it, is 65.0 kg and that you're pulling with a force of 75.0 N.

Coefficient of friction between the sled and the snow is?

Answer 1

The forces on the sled in the horizontal are

friction=N*u
where N is the normal force and u is the friction coefficient

the horizontal component of the rope
cos(24)*75

Since the speed is constant, these are in balance
N*u=cos(24)*75

The Normal force
is the weight of the sled and rider minus the vertical component of the rope force

65*9.81-sin(24)*75

u=cos(24)*75 /(65*9.81-sin(24)*75)
=.113

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Answer 2

F cos(theta) = kN = f; where F = the tension on the rope = 75 Newtons at angle theta = 24 deg with the ground, k = the coefficient of friction, and N = mg = the normal weight of the sled and your friend with combined mass (m = 65 kg) and g = 9.81 m/sec^2.

Thus, k = F cos(theta)/N = F cos(theta)/mg = 75 cos(24)/(65*9.81)...you can do the math.

Physics lesson: the net force on that sled is zero because it is not accelerating (v = constant 1.5 m/sec). As consequence, the friction force and the horizontal (cos) pulling force exactly cancel out. Thus, kN = F cos(theta).

NOTE: The answer you get for k is not correct. In reality, N would be less than mg because there is a vertical force (F sin(theta)) from pulling that would lighten the load (N) of the sled on the snow.That is, the vertical force pulling up on the nose of the sled would pull up on its center of mass (CM) sitting some distance back from the nose. As we do not know the distance (d) for the CM from the nose, we cannot calculate how much N would be lightened by that upward pull on the rope. The only thing we can say from k = F cos(theta)/N is that less load (N) means the actual k will be higher than previously calculated while ignoring the upward force F sin(theta).

Also NOTE: The previous answerer failed to take the distance d between the CM and the point on the nose where the rope is attached to the sled into account. The previous answer is valid if and only if that rope were attached in some way so that the vertical force F sin(theta) passed through the CM of the combined mass (m).