A skier with a 65kg mass skies down a 30 degree incline hill. The coefficient of friction is 0.1.

a. Draw a free body diagram.

b. Determine the normal force.

c. Determine the acceleration of the skier down the hill.

Part a.

Refer to the diagram shown below.

m = 65 kg, the mass of the skier

θ = 30°, the angle of the incline

μ = 0.1, the coefficient of friction

W = mg, the weight of the skier

N = mg cosθ, the normal reaction

F = mg sinθ, the component of the weight acting down the hill

R = μN, the frictional force resisting motion down the plane

a = the acceleration down the plane.

The weight is

W = mg = (65 kg) * (9.8 m/s) = 637 N

Part b.

The normal force is

N = (637 N) * cos (30) = 551.658 N

The frictional force is

R = 0.1 (551.658 N) = 55.1658 N

Part c.

Calculate the component of the weight acting down the hill.

F = mg sin (30) = (637 N) * sin (30) = 318.5 N

The equation of motion is

ma = F - R

(65 kg) * (a m/s²) = 318.5 - 55.1658 N

a = 263.3342/65 = 4.05 m/s²

Answers:

N = 551.7 N

a = 4.05 m/s²