Skiing Exercise

The magnitude of kinetic friction was given in to be 45.0 N. Kinetic friction is related to the normal force $N$ as $f_{\text{k}}={\mu }_{\text{k}}N$; thus, the coefficient of kinetic friction can be found if we can find 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. (See the skier and free-body diagram in Figure 5.4.)

Figure 5.4 The motion of the skier and friction are parallel to the slope and 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). $\mathbf{\text{N}}$ (the normal force) is perpendicular to the slope, and $\mathbf{\text{f}}$ (the friction) is parallel to the slope, but $\mathbf{\text{w}}$ (the skier’s weight) has components along both axes, namely ${\mathbf{\text{w}}}_{\text{⊥}}$ and ${\mathbf{\text{W}}}_{//}$. $\mathbf{N}$ is equal in magnitude to ${\mathbf{w}}_{\perp }$, so there is no motion perpendicular to the slope. However, $\mathbf{f}$ is less than ${\mathbf{W}}_{\text{//}}$ in magnitude, so there is acceleration down the slope (along the x-axis).

That is,

Substituting this into our expression for kinetic friction, we get

which can now be solved for the coefficient of kinetic friction ${\mu }_{\text{k}}$.

Solving for ${\mu }_{\mathrm{k}}$ gives

Substituting known values on the right-hand side of the equation,

This result is a little smaller than the coefficient listed in Table 5.1 for waxed wood on snow, but it is still reasonable since values of the coefficients of friction can vary greatly. In situations like this, where an object of mass $m$ slides down a slope that makes an angle $\theta$ with the horizontal, friction is given by $f_{\text{k}}={\mu }_{\text{k}}\text{mg}\text{cos}\theta$. All objects will slide down a slope with constant acceleration under these circumstances. Proof of this is left for this chapter’s Problems and Exercises.