Friction Question?

Question

A sled is pulled at a constant velocity. The coefficient of static friction is μs=0.230. The coefficient of kinetic friction is μk=0.170. The sled is pulled at an angle θ=31.0° above the horizontal, and the combined mass of the sled and rider is M=58.0 kg. What is the force T?

So, here's what I have:
N = mgcosΘ
fk = uk*mg*cos Θ

Are those equations correct? Thanks for any help.

Answer 1

Those equations may be correct for some problem, but not this one. It looks like they're for an object on a ramp.

The trick to this problem is that the T is at an angle, so it needs to be resolved into horizontal (x) and vertical (y) components using Tx = Tcos31 and Ty = Tsin31.

The normal force needs to be large enough so that all the upward forces will equal the weight (mg) of the object. The weight is 568 N and there's already a force of Tsin31 pulling up, so the normal needs to be 568 N - Tsin31. That way the normal force plus the Tsin31 force are equal to the weight of the object.

Friction will be .17 of the normal, so friction = .17*(568 N - Tsin31).

Finally, the forward pull must equal the friction so the net force will be zero so the sled moves at a constant velocity. The forward pull is Tcos31, so Tcos31=.17*(568 N - T sin 31).

Using a calculator for the cos and sin gives, .857 T = 96.6 N - .088 T. Rearranging gives .95 T = 96.6 N, so T = 102 N.

Answer 2

You have it correct.

One minor point
use T=μk*m*g*cos Θ

Thomas G makes a valid point. I assumed that the T was pulling the sled up a ramp and was parallel to the ramp.

If the sled is on flat ground and T is pulling upward, the the normal force is reduced by the sin times T

or N=m*g*cosΘ-T*sinΘ
and the frictinal force is still
uk*N
which is equal to
T*cosΘ
or
T*cosΘ/uk=m*g*cosΘ-T*sinΘ


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