0.2 Force, momentum and impulse  (Page 9/35)

A body of mass M is at rest on an inclined plane.

What is the magnitude of the frictional force acting on the body?

1. Mg
2. Mg cos $\theta$
3. Mg sin $\theta$
4. Mg tan $\theta$

1. Analyse the situation

   The question asks us to identify the frictional force. The body is said to be at rest on the plane, which means that it is not moving and therefore there is no resultant force. The frictional force must therefore be balanced by the force F up the inclined plane.

2. Choose the correct answer

   The frictional force is equal to the component of the weight (Mg) parallel to the surface, which is equal to Mg sin  $\theta$ .

A force T = 312 N is required to keep a body at rest on a frictionless inclined plane which makes an angle of 35 ${}^{\circ }$ with the horizontal. The forces acting on the body are shown. Calculate the magnitudes of forces P and R, giving your answers to three significant figures.

1. Find the magnitude of P

   We are usually asked to find the magnitude of T, but in this case T is given and we are asked to find P. We can use the same equation. T is the force that balances the component of P parallel to the plane (P ${}_x$ ) and therefore it has the same magnitude.

   $$\begin{array}{ccc}\hfill T& =& P\sin \theta \hfill \\ \hfill 312& =& P\sin {35}^{\circ }\hfill \\ \hfill P& =& 544\mathrm{N}\hfill \end{array}$$
2. Find the magnitude of R

   R can also be determined with the use of trigonometric ratios. The tan or cos ratio can be used. We recommend that you use the tan ratio because it does not involve using the value for P (for in case you made a mistake in calculating P).

   $$\begin{array}{ccc}\hfill \tan {55}^{\circ }& =& \frac{R}{T}\hfill \\ \hfill \tan {55}^{\circ }& =& \frac{R}{312}\hfill \\ \hfill R& =& \tan {55}^{\circ }\times 312\hfill \\ \hfill R& =& 445,6\mathrm{N}\hfill \\ \hfill R& =& 446\mathrm{N}\hfill \end{array}$$

   Note that the question asks that the answers be given to 3 significant figures. We therefore round 445,6 N up to 446 N.

Lifts and rockets

So far we have looked at objects being pulled or pushed across a surface, in other words motion parallel to the surface the object rests on. Here we only considered forces parallel to the surface, but we can also lift objects up or let them fall. This is vertical motion where only vertical forces are being considered.

Let us consider a 500 kg lift, with no passengers, hanging on a cable. The purpose of the cable is to pull the lift upwards so that it can reach the next floor or to let go a little so that it can move downwards to the floor below. We will look at five possible stages during the motion of the lift.

Stage 1:

The 500 kg lift is stationary at the second floor of a tall building.

Because the lift is stationary (not moving) there is no resultant force acting on the lift. This means that the upward forces must be balanced by the downward forces. The only force acting down is the force of gravity which is equal to (500 x 9,8 = 4900 N) in this case. The cable must therefore pull upwards with a force of 4900 N to keep the lift stationary at this point.

Stage 2:

The lift moves upwards at an acceleration of 1 m $·$ s ${}^{-2}$ .

If the lift is accelerating, it means that there is a resultant force in the direction of the motion. This means that the force acting upwards is now greater than the force of gravity F ${}_g$ (down). To find the magnitude of the force applied by the cable (F ${}_c$ ) we can do the following calculation: (Remember to choose a direction as positive. We have chosen upwards as positive.)