# 5.6 Common forces  (Page 2/11)

 Page 2 / 11
$\stackrel{\to }{N}=\text{−}m\stackrel{\to }{g}.$

In scalar form, this becomes

$N=mg.$

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?

## Strategy

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.

## Solution

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

${w}_{x}=w\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}25\text{°}=mg\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}25\text{°},$

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

${w}_{y}=w\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}25\text{°}=mg\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}25\text{°}.$

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,

${a}_{x}=\frac{{F}_{\text{net}\phantom{\rule{0.2em}{0ex}}x}}{m}$

where ${F}_{\text{net}\phantom{\rule{0.2em}{0ex}}x}={w}_{x}-mg\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}25\text{°},$ assuming no friction for this part. Therefore,

$\begin{array}{}\\ \\ \\ {a}_{x}=\frac{{F}_{\text{net}\phantom{\rule{0.2em}{0ex}}x}}{m}=\frac{mg\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}25\text{°}}{m}=g\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}25\text{°}\hfill \\ \left(9.80\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}\right)\left(0.4226\right)=4.14\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}\hfill \end{array}$

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

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