Friction and the Inclined Plane One situation where friction plays an obvious role is that of an object on a slope. It might be a crate being pushed up a ramp to a loading dock or a skateboarder coasting down a mountain, but the basic physics is the same. We usually generalize the sloping surface and call it an inclined plane but then pretend that the surface is flat. Let's look at an example of analyzing motion on an inclined plane with friction. Downhill Skier A skier with a mass of 62 kg is sliding down a snowy slope at a constant acceleration. Find the coefficient of kinetic friction for the skier if friction is known to be 45.0 N. Strategy The magnitude of kinetic friction is given as 45.0 N. Kinetic friction is related to the normal force NV by fx= H&N; thus, we can find the coefficient of kinetic friction if we can find the normal force on the skier. 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. (See figure below.) Free-body diagram y 12 Ñ 1 7 W. ?, ?x 25 3 £25º W. ? ? 25° The motion of the skier and friction are parallel to the slope, so it is most convenient to project all forces onto a coordinate system where one axis is parallel to the slope and the other is perpendicular (axes shown to left of skier). The normal force N is perpendicular to the slope, and friction f is parallel to the slope, but the skier's weight w has components along both axes, namely Wy and W2. The normal force N is equal in magnitude to Wy, so there is no motion perpendicular to the slope. However, f is equal to W2 in magnitude, so there is a constant velocity down the slope (along the x-axis). We have N = Wy = w cos 25* = mg cos 25". Substituting this into our expression for kinetic friction, we obtain fx = "_mg cos 25", which can now be solved for the coefficient of kinetic friction Jak.