4.5 Normal, tension, and other examples of force  (Page 3/11)

Solution

The magnitude of the component of the weight parallel to the slope is $w_{\parallel }=\mathit{w}\text{sin}(\text{25º})=\mathit{mg}\text{sin}(\text{25º})$ , and the magnitude of the component of the weight perpendicular to the slope is $w_{\perp }=\mathit{w}\text{cos}(\text{25º})=\mathit{mg}\text{cos}(\text{25º})$ .

(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,

where $F_{\text{net}\parallel }=w_{\parallel }=\text{mg}\text{sin}(\text{25º})$ , assuming no friction for this part, so that

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, $a_{\parallel }=\frac{F_{\text{net}\parallel }}{m}$ , gives

We substitute known values to obtain

which yields

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 $a=g\text{sin}\theta$ , 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

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

$w_{\parallel }=w\text{sin}(\theta )=\text{mg}\text{sin}(\theta )$

and

$w_{\perp }=w\text{cos}(\theta )=\text{mg}\text{cos}(\theta ).$

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:

$\begin{array}{lll}\text{cos}(\theta )& =& \frac{w_{\perp }}{w}\\ w_{\perp }& =& w\text{cos}(\theta )=\text{mg}\text{cos}(\theta )\end{array}$

$\begin{array}{lll}\text{sin}(\theta )& =& \frac{w_{\parallel }}{w}\\ w_{\parallel }& =& w\text{sin}(\theta )=\text{mg}\text{sin}(\theta )\end{array}$