Example Illustrating the Normal Force

Example: Weight on an Incline, a Two-Dimensional Problem

Consider the skier on a slope shown in the figure below. 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?

A skier is skiing down the slope and the slope makes a twenty-five degree angle with the horizontal. Her weight W, shown by a vector vertically downward, breaks into two components—one is W parallel, which is shown by a vector arrow parallel to the slope, and the other is W perpendicular, shown by a vector arrow perpendicular to the slope in the downward direction. Vector N is represented by an arrow pointing upward and perpendicular to the slope, having the same length as W perpendicular. Friction vector f is represented by an arrow along the slope in the uphill direction. IIn a free-body diagram, the vector arrow W for weight is acting downward, the vector arrow for f is shown along the direction of the slope, and the vector arrow for N is shown perpendicular to the slope.

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). \(\mathbf{\text{N}}\) is perpendicular to the slope and f is parallel to the slope, but \(\mathbf{\text{w}}\) has components along both axes, namely \({\mathbf{\text{w}}}_{\perp }\) and \({\mathbf{\text{w}}}_{\parallel }\). \(\mathbf{\text{N}}\) is equal in magnitude to \({\mathbf{\text{w}}}_{\perp }\), so that there is no motion perpendicular to the slope, but \(f\) is less than \({w}_{\parallel }\), so that there is a downslope acceleration (along the parallel axis).

Strategy

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 connected 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. (Remember that motions along mutually perpendicular axes are independent.) We use the symbols \(\perp \) and \(\parallel \) 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 \(\mathbf{\text{w}}\), \(\mathbf{\text{f}}\), and \(\mathbf{\text{N}}\) in the figure above. \(\mathbf{\text{N}}\) is always perpendicular to the slope, and \(\mathbf{\text{f}}\) is parallel to it.

But \(\mathbf{\text{w}}\) 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 \({w}_{\parallel }\) to be the component of weight parallel to the slope and \({w}_{\perp }\) 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.

Solution

The magnitude of the component of the weight parallel to the slope is \({w}_{\parallel }=\mathit{w}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(\text{25º}\right)=\mathit{mg}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(\text{25º}\right)\), and the magnitude of the component of the weight perpendicular to the slope is \({w}_{\perp }=\mathit{w}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\left(\text{25º}\right)=\mathit{mg}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\left(\text{25º}\right)\).

(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 \({w}_{\parallel }\) and friction \(f\). Using Newton’s second law, with subscripts to denote quantities parallel to the slope,

\({a}_{\parallel }=\cfrac{{F}_{\text{net}\parallel }}{m}\)

where \({F}_{\text{net}\parallel }={w}_{\parallel }=\text{mg}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(\text{25º}\right)\), assuming no friction for this part, so that

\({a}_{\parallel }=\cfrac{{F}_{\text{net}\parallel }}{m}=\cfrac{\text{mg}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(\text{25º}\right)}{m}=g\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(\text{25º}\right)\)

\(\left(9.80\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}\right)\left(0.4226\right)=4.14\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}\)

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

\({F}_{\text{net}\parallel }={w}_{\parallel }-f,\)

and substituting this into Newton’s second law, \({a}_{\parallel }=\cfrac{{F}_{\text{net}\parallel }}{m}\), gives

\({a}_{\parallel }=\cfrac{{F}_{\text{net}\mid \mid }}{m}=\cfrac{{w}_{\parallel }-f}{m}=\cfrac{\text{mg}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(\text{25º}\right)-f}{m}.\)

We substitute known values to obtain

\({a}_{\parallel }=\cfrac{\left(\text{60}\text{.}\text{0 kg}\right)\left(9\text{.}{\text{80 m/s}}^{2}\right)\left(0\text{.}\text{4226}\right)-\text{45}\text{.}\text{0 N}}{\text{60}\text{.}\text{0 kg}},\)

which yields

\({a}_{\parallel }=3\text{.}{\text{39 m/s}}^{2},\)

which is the acceleration parallel to the incline when there is 45.0 N of opposing friction.

Discussion

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 \(a=g\phantom{\rule{0.25em}{0ex}}\text{sin}\theta \), 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).

Resolving Weight into Components

Vector arrow W for weight is acting downward. It is resolved into components that are parallel and perpendicular to a surface that has a slope at angle theta to the horizontal. The coordinate direction x is labeled parallel to the sloped surface, with positive x pointing uphill. The coordinate direction y is labeled perpendicular to the sloped surface, with positive y pointing up from the surface. The components of w are w parallel, represented by an arrow pointing downhill along the sloped surface, and w perpendicular, represented by an arrow pointing into the sloped surface. W parallel is equal to w sine theta, which is equal to m g sine theta. W perpendicular is equal to w cosine theta, which is equal to m g cosine theta.

An object rests on an incline that makes an angle θ with the horizontal.

When an object rests on an incline that makes an angle \(\theta \) with the horizontal, the force of gravity acting on the object is divided into two components: a force acting perpendicular to the plane, \({\mathbf{\text{w}}}_{\perp }\), and a force acting parallel to the plane, \({\mathbf{\text{w}}}_{\parallel }\).

The perpendicular force of weight, \({\mathbf{\text{w}}}_{\perp }\), is typically equal in magnitude and opposite in direction to the normal force, \(\mathbf{\text{N}}\). The force acting parallel to the plane, \({\mathbf{\text{w}}}_{\parallel }\), causes the object to accelerate down the incline. The force of friction, \(\mathbf{f}\), opposes the motion of the object, so it acts upward along the plane.

It is important to be careful when resolving the weight of the object into components. If the angle of the incline is at an angle \(\theta \) to the horizontal, then the magnitudes of the weight components are

\({w}_{\parallel }=w\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(\theta \right)=\text{mg}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(\theta \right).\)

and

\({w}_{\perp }=w\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\left(\theta \right)=\text{mg}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\left(\theta \right).\)

Instead of memorizing these equations, it is helpful to be able to determine them from reason. To do this, draw the right triangle formed by the three weight vectors. Notice that the angle \(\theta \) of the incline is the same as the angle formed between \(\mathbf{\text{w}}\) and \({\mathbf{\text{w}}}_{\perp }\). Knowing this property, you can use trigonometry to determine the magnitude of the weight components:

\(\text{cos}(θ)=\cfrac{{w}_{\perp }}{w}\)
\({w}_{\perp }=w\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\left(\theta \right)=\text{mg}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\left(\theta \right)\)
\(\text{sin}(θ) = \cfrac{{w}_{\parallel }}{w}\)
\({w}_{\parallel }=w\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(\theta \right)=\text{mg}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(\theta \right)\)

Optional Take-Home Experiment: Force Parallel

To investigate how a force parallel to an inclined plane changes, find a rubber band, some objects to hang from the end of the rubber band, and a board you can position at different angles. How much does the rubber band stretch when you hang the object from the end of the board?

Now place the board at an angle so that the object slides off when placed on the board. How much does the rubber band extend if it is lined up parallel to the board and used to hold the object stationary on the board? Try two more angles. What does this show?

Example Illustrating the Normal Force

Example: Weight on an Incline, a Two-Dimensional Problem

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