Example 1. Weight on an Incline, a Two-Dimensional Problem
Consider the skier on a slope shown in Figure 2. Her mass including equipment is 60.0 kg. (a) What is her acceleration if friction is negligible? (b) What is her acceleration if friction is known to be 45.0 N?
Figure 2. Since motion and friction are parallel to the slope, 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). N
is perpendicular to the slope and f
is parallel to the slope, but w
has components along both axes, namely w⊥
. N is equal in magnitude to w⊥
, so that there is no motion perpendicular to the slope, but f is less than w∥, so that there is a downslope acceleration (along the parallel axis).
This is a two-dimensional problem, since the forces on the skier (the system of interest) are not parallel. The approach we have used in two-dimensional kinematics also works very well here. Choose a convenient coordinate system and project the vectors onto its axes, creating two
-dimensional problems to solve. The most convenient coordinate system for motion on an incline is one that has one coordinate parallel to the slope and one perpendicular to the slope. (Remember that motions along mutually perpendicular axes are independent.) We use the symbols ⊥ and ∥ to represent perpendicular and parallel, respectively. This choice of axes simplifies this type of problem, because there is no motion perpendicular to the slope and because friction is always parallel to the surface between two objects. The only external forces acting on the system are the skier’s weight, friction, and the support of the slope, respectively labeled w
, and N
in Figure 2. N is always perpendicular to the slope, and f is parallel to it. But w is not in the direction of either axis, and so the first step we take is to project it into components along the chosen axes, defining w∥
to be the component of weight parallel to the slope and w⊥
the component of weight perpendicular to the slope. Once this is done, we can consider the two separate problems of forces parallel to the slope and forces perpendicular to the slope.
The magnitude of the component of the weight parallel to the slope is
, and the magnitude of the component of the weight perpendicular to the slope is
(a) Neglecting friction. Since the acceleration is parallel to the slope, we need only consider forces parallel to the slope. (Forces perpendicular to the slope add to zero, since there is no acceleration in that direction.) The forces parallel to the slope are the amount of the skier’s weight parallel to the slope w∥
and friction f
. Using Newton’s second law, with subscripts to denote quantities parallel to the slope,
, assuming no friction for this part, so that
(9.98 m/s2)(0.4226) = 4.14 m/s2
is the acceleration.
(b) Including friction. We now have a given value for friction, and we know its direction is parallel to the slope and it opposes motion between surfaces in contact. So the net external force is now
and substituting this into Newton’s second law,
We substitute known values to obtain
which is the acceleration parallel to the incline when there is 45.0 N of opposing friction.
Since friction always opposes motion between surfaces, the acceleration is smaller when there is friction than when there is none. In fact, it is a general result that if friction on an incline is negligible, then the acceleration down the incline is a
, regardless of mass
. This is related to the previously discussed fact that all objects fall with the same acceleration in the absence of air resistance. Similarly, all objects, regardless of mass, slide down a frictionless incline with the same acceleration (if the angle is the same).