XyeThesis asked:
A portion of the beginner’s slope in a ski resort enables the skier’s speed to remain constant. The coefficient of static and kinetic friction between skis and snow is 0.12 and 0.07 respectively.
A portion of the beginner’s slope in a ski resort enables the skier’s speed to remain constant. The coefficient of static and kinetic friction between skis and snow is 0.12 and 0.07 respectively.
Based on the above information, the slope angle should be _____ degree.
.07
6.84277341263094
angle = atan(.12)
Answer has been edited.
Slope angle should be 4° as explained below.
Slope angle should be more than arc tan (0.12) = 6.84° for the motion to start.
Let the slope angle = θ for constant speed.
Then the component of weight down the slope
= frictional force due to kinetic friction up the slope
=> mg sin θ = μ mg cos θ
=> θ = arc tan (μ) = arc tan (0.07)
=> θ = 4°.
This means that initially for the motion to start, slope angle should be 6.84°, then as the kinetic friction is less, the skier will have acceleration and for constant velocity, the angle should then be reduced to 4°. If the angle is not reduced and is anywhere between 4° and 6.84° as many have answered, then the skier will have an accelerated motion and not motion with constant velocity.
Really though provoking question!
Force pulling down mg sin θ equals
Friction acting up {μ mg cos θ}
mg sin θ = μ mg cos θ
tan θ = μ [static] = 0.12.
θ = 6.84°
He cannot slide but he will be at rest unless some force acts for a brief time.
tan θ = μ [sliding] = 0.07.
θ = 4°
At this angle if he has already a speed then he will maintain that speed even at this angle.
Hence if he is already in motion the slope can be any thing between 4 to 6.87.