Question

A sled weighing 60.0 N is pulled up a snowy hill inclined at 10 degrees to the horizontal at a constant speed so that the coefficient of kinetic friction between sled and snow is 0.070. A penguin weighing 70.0 N rides on the sled. If the coefficient of static friction between penguin and sled is 0.950, find a). the force of friction between the sled and the snow, b). the force of static friction between the penguin and the sled, and c). the maximum force that can be exerted on the sled before the penguin begins to slide off.

Answer 1

a)
Total downward weight = weight of the sled + weight of the penguin
= 70 + 60 = 130 N
Normal force, N = 130 x cos(10)
= 128.025 N

Frictional force, Ff = _kN
Where, _k is the coefficient of kinetic friction.
Ff = 0.07 x 128.025
= 8.962 N.

b)
The normal force on the penguin, Np = 70 x cos(10)
= 68.94 N

Frictional force, Fp = _sNp
Where _s is the coefficient of static friction.
Fp = 0.95 x 68.94
= 65.49 N

c)
The maximum force that can be applied = Ff + Fp
= 8.962 + 65.49
= 74.45 N