A 70- skier speeds down a trail, as shown in the figure (Figure 1) . The surface is smooth and inclined at an angle of 22 with the horizontal.
1 Answer
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Include the units !!!
m = mass of the skier = 70 kg
θ = angle of the incline = 22°
g = acceleration by gravity = 9.8 m/s²
The weight of tfe skier is:
W = m × g
W = (70 kg) × (9.8 m/s²) = 686 N
The component of that weight that is pointing perpendicularly into the incline is:
Wp = W × cos(θ)
Wp = (686 N) × cos(22°) = 636 N
The incline will react with an equal but opposite force, being the normal force, so
Fn = Wp
Fn = 636 N
