5 Projectiles on inclined planes

The forces acting on an object resting on a sloping surface are its weight $\underset{̲}{W}$ , the normal reaction $\underset{̲}{N}$ , and the frictional force $\underset{̲}{R}$ . Since all of the forces act in a vertical plane then they can be described with just two axes (indicated by the unit vectors $\underset{̲}{j}$ and $\underset{̲}{k}$ in Figure 15). If there is negligible friction ( $\underset{̲}{R}=\underset{̲}{0}$ ) then, of course, the object will slide down the plane. If we apply Newton’s second law to motion in the frictionless inclined plane, then the only remaining force to be considered is $\underset{̲}{W}$ , since $\underset{̲}{N}$ is normal to the plane. Resolving in the $\underset{̲}{j}$ -direction gives the only force as

$-\left|\underset{̲}{W}\right|cos\left(90-\alpha \right)\underset{̲}{j}=-\left|\underset{̲}{W}\right|sin\alpha \underset{̲}{j}.$

Figure 15 :

This will be the only in-plane force on an object projected across the inclined plane, moving so that it is always in contact with the plane, and projected at some angle $\theta$ above the horizontal in the plane (as in Figure 16 for Example 8). Newton’s second law for such an object may be written

$m\underset{̲}{a}=-mgsin\alpha \underset{̲}{j}$ or $\underset{̲}{a}=-gsin\alpha \underset{̲}{j}.$

The resulting acceleration vector differs from that considered in HELM booklet  34.2 Subsection 2 only by the constant factor $sin\alpha$ . In other words, the ball will move on the inclined plane as a projectile under reduced gravity (since $gsin\alpha <g$ ).

Suppose that the object has an intial velocity $\underset{̲}{u}$ . This is given in terms of the chosen coordinates by

$ucos\theta \underset{̲}{i}+usin\theta \underset{̲}{j}$

which is the same as that considered earlier in Section 34.1. So it is possible to use the result for the range obtained in Section 34.1 Equation (1.6), after remembering to replace $g$ by $gsin\alpha$ . Equations (1.9) to (1.11) may be applied as long as $g$ is replaced by $gsin\alpha$ .

Example 8

In a game a small disc is projected from one corner across a smooth board inclined at an angle $\alpha$ to the horizontal. The disc moves so that it is always in contact with the board. The speed and angle of launch can be varied and the object of the game is to collect the disc in a shallow cup situated in the plane at a horizontal distance $d$ from the point of launch. Calculate the speed of launch at an angle of $45^{\&compfn;}$ in the plane of the board that will ensure that the disc lands in the cup.

Solution

Figure 16

Consider a point along the disc’s path portrayed from the side and looking down on the plane of the board in Figure 16. We choose coordinates and unit vectors as shown in these figures, so that $y$ is up the plane and $x$ along it, while $z$ is normal to the plane. Equation (1.11) will apply as long as $g$ is replaced by $gsin\alpha$ . So the range is given by $R=\frac{u^{2}sin2\theta }{gsin\alpha }.$

When $R=d$ and $\theta =45^{\&compfn;}$ , this gives $d=\frac{u^2}{gsin\alpha }$ or $u=\sqrt{gdsin\alpha }$

Note that $d$ represents the maximum ‘horizontal’ , i.e. in-plane, range from the point of launch for this launch speed.

Task!

Suppose that, in the game featured in the last example, there is another cup in the plane with the centre of its open end at coordinates ( $3d∕5,d∕5$ ) with respect to the point of launch, and that a successful ‘pot’ in the cup will gain more points. What angle of launch will ensure that the disc will enter this second cup if the magnitude of the launch velocity is $u=\sqrt{gdsin\alpha }$ ?

Answer

Task!

Skateboarders have built jumps consisting of short ramps angled at about $30^{\&compfn;}$ from the horizontal. Assume that the speed on leaving the ramp is $10\text{m}\text{s}-1$ and ignore air resistance.

1. Write down appropriate position, velocity and acceleration vectors:

   Answer

   Ignoring air resistance, the relevant vectors are

   $\underset{̲}{r}\left(t\right)=\left[\begin{array}{c}\hfill \left(ucos\theta \right)t\hfill \\ \hfill \hfill \\ \hfill usin\theta -\frac{1}{2}g{t}^2\hfill \end{array}\right],\underset{̲}{v}\left(t\right)=\left[\begin{array}{c}\hfill ucos\theta \hfill \\ \hfill \hfill \\ \hfill -gt\hfill \end{array}\right]$ , $\underset{̲}{a}\left(t\right)=\left[\begin{array}{c}\hfill 0\\ \hfill \\ \hfill -g\end{array}\right].$

   In the skateboarders case, $u=10$ and $\theta =30^{\&compfn;}$ , so the vectors may be written

   $\underset{̲}{r}\left(t\right)=\left[\begin{array}{c}\hfill 8.67t\hfill \\ \hfill \hfill \\ \hfill 5-\frac{1}{2}g{t}^2\hfill \end{array}\right],\underset{̲}{v}\left(t\right)=\left[\begin{array}{c}\hfill 8.67\hfill \\ \hfill \hfill \\ \hfill -gt\hfill \end{array}\right]$ , $\underset{̲}{a}\left(t\right)=\left[\begin{array}{c}\hfill 0\\ \hfill \\ \hfill -g\end{array}\right].$

2. Predict the maximum length of jump possible at the level of the ramp exits:

   Answer

   The horizontal range $x$ m at the level of exit from the ramp is given by

   $x=\frac{u^{2}sin2\theta }{g}=\frac{100sin60}{9.81}=8.828.$

3. Predict the maximum height of jump possible measured from the ramp exits:

   Answer

   From Section 34.1 Equation (1.11), the maximum height $y_m$ m measured from the ramp exit is given by

   $y_m=\frac{u_v^2}{2g}=\frac{u^2{sin}^2\theta }{2g}=\frac{25}{2\times 9.81}=1.274.$

4. Comment on the choice of slope for the ramp:

   Answer

   For a fixed value of $u$ , the maximum range $\frac{u^{2}sin2\theta }{g}$ is given when $\theta =45^{\&compfn;}$ . On the other hand, the maximum height $\frac{u^2{sin}^2\theta }{2g}$ is given when $\theta =90^{\&compfn;}$ . The latter would require vertical ramps and these are not very practicable!