Example from Physics, 5.4 Inclined Planes

Finding the Coefficient of Kinetic Friction on an Inclined Plane

A skier, illustrated in Figure 5.35 (a), with a mass of 62 kg is sliding down a snowy slope at an angle of 25 degrees. Find the coefficient of kinetic friction for the skier if friction is known to be 45.0 N.

A skier is moving down a slope with an angle measure of twenty-five degrees to the ground. Force vectors are shown: N points upward from the skier, f points to the right of the skis, w points at a downward left angle from the skiers' feet, forming an angle with vector w that has a measure of twenty-five degrees. A free-body diagram illustrates the right triangle formed by these vectors.

Figure 5.35 Use the diagram to help find the coefficient of kinetic friction for the skier.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy
The magnitude of kinetic friction was given as 45.0 N. Kinetic friction is related to the normal force N as $f_{k} = μ_{k} N$ . Therefore, we can find the coefficient of kinetic friction by first finding the normal force of the skier on a slope. The normal force is always perpendicular to the surface, and since there is no motion perpendicular to the surface, the normal force should equal the component of the skier’s weight perpendicular to the slope.
That is,

$N = w_{⊥} = w cos (25^{\circ}) = m g cos (25^{\circ}) .$
Substituting this into our expression for kinetic friction, we get

$f_{k} = μ_{k} m g {cos 25}^{\circ},$
which can now be solved for the coefficient of kinetic friction μ_k.
Solution
Solving for $μ_{k}$ gives

$μ_{k} = \frac{f_{k}}{w {cos 25}^{\circ}} = \frac{f_{k}}{m g {cos 25}^{\circ}} .$
Substituting known values on the right-hand side of the equation,

$μ_{k} = \frac{45.0 N}{(62 kg)(9 {.80 m/s}^{2}) (0.906)} = 0.082 .$

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Strategy

Solution

Strategy

Solution