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N = m g .

In scalar form, this becomes

N = m g .

The normal force can be less than the object’s weight if the object is on an incline.

Weight on an incline

Consider the skier on the slope in [link] . 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 45.0 N?

Figure shows a person skiing down a slope of 25 degrees to the horizontal. Force f is up and parallel to the slope, force N is up and perpendicular to the slope. Force w is straight down. Its component wx is down and parallel to the slope and component wy is down and perpendicular to the slope. All these forces are also shown in a free body diagram. X axis is taken to be parallel to the slope.
Since the acceleration is parallel to the slope and acting down 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 to it (axes shown to the left of the skier). N is perpendicular to the slope and f is parallel to the slope, but w has components along both axes, namely, w y and w x . Here, w has a squiggly line to show that it has been replaced by these components. The force N is equal in magnitude to w y , so there is no acceleration perpendicular to the slope, but f is less than w x , so there is a downslope acceleration (along the axis parallel to the slope).


This is a two-dimensional problem, since not all forces on the skier (the system of interest) are parallel. The approach we have used in two-dimensional kinematics also works well here. Choose a convenient coordinate system and project the vectors onto its axes, creating two one-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. (Motions along mutually perpendicular axes are independent.) We use x and y for the parallel and perpendicular directions, respectively. This choice of axes simplifies this type of problem, because there is no motion perpendicular to the slope and the acceleration is downslope. Regarding the forces, friction is drawn in opposition to motion (friction always opposes forward motion) and is always parallel to the slope, w x is drawn parallel to the slope and downslope (it causes the motion of the skier down the slope), and w y is drawn as the component of weight perpendicular to the slope. Then, we can consider the separate problems of forces parallel to the slope and forces perpendicular to the slope.


The magnitude of the component of weight parallel to the slope is

w x = w sin 25 ° = m g sin 25 ° ,

and the magnitude of the component of the weight perpendicular to the slope is

w y = w cos 25 ° = m g cos 25 ° .

a. Neglect 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 component of the skier’s weight parallel to slope w x and friction f . Using Newton’s second law, with subscripts to denote quantities parallel to the slope,

a x = F net x m

where F net x = w x m g sin 25 ° , assuming no friction for this part. Therefore,

a x = F net x m = m g sin 25 ° m = g sin 25 ° ( 9.80 m/s 2 ) ( 0.4226 ) = 4.14 m/s 2

is the acceleration.

b. Include friction. We 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

Questions & Answers

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Practice Key Terms 3

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