0.2 Force, momentum and impulse  (Page 8/35)

A man is pulling a 20 kg box with a rope that makes an angle of 60 ${}^{\circ }$ with the horizontal. If he applies a force of 150 N and a frictional force of 15 N is present, calculate the acceleration of the box.

1. Draw a force diagram

   The motion is horizontal and therefore we will only consider the forces in a horizontal direction. Remember that vertical forces do not influence horizontal motion and vice versa.

   

2. Calculate the horizontal component of the applied force

   The applied force is acting at an angle of 60 ${}^{\circ }$ to the horizontal. We can only consider forces that are parallel to the motion. The horizontal component of the applied force needs to be calculated before we can continue:

   $$\begin{array}{ccc}\hfill F_x& =& 150\cos {60}^{\circ }\hfill \\ \hfill F_x& =& 75\mathrm{N}\hfill \end{array}$$
3. Calculate the acceleration

   To find the acceleration we apply Newton's Second Law:

   $$\begin{array}{ccc}\hfill F_R& =& ma\hfill \\ \hfill F_x+F_f& =& \left(20\right)\left(a\right)\hfill \\ \hfill 75+(-15)& =& 20a\hfill \\ \hfill a& =& 3\mathrm{m}·{\mathrm{s}}^{-2}\mathrm{to\; the\; right}\hfill \end{array}$$

A 2000 kg truck pulls a 500 kg trailer with a constant acceleration. The engine of the truck produces a thrust of 10 000 N. Ignore the effect of friction.

1. Calculate the acceleration of the truck.
2. Calculate the tension in the tow bar T between the truck and the trailer, if the tow bar makes an angle of 25 ${}^{\circ }$ with the horizontal.

1. Draw a force diagram

   Draw a force diagram indicating all the horizontal forces on the system as a whole:

   

2. Find the acceleration of the system

   In the absence of friction, the only force that causes the system to accelerate is the thrust of the engine. If we now apply Newton's Second Law:

   $$\begin{array}{ccc}\hfill F_R& =& ma\hfill \\ \hfill 10000& =& (500+2000)a\hfill \\ \hfill a& =& 4\mathrm{m}·{\mathrm{s}}^{-2}\mathrm{to}\mathrm{the}\mathrm{right}\hfill \end{array}$$
3. Find the horizontal component of T

   We are asked to find the tension in the tow bar, but because the tow bar is acting at an angle, we need to find the horizontal component first. We will find the horizontal component in terms of T and then use it in the next step to find T.

   

   The horizontal component is T cos 25 ${}^{\circ }$ .

4. Find the tension in the tow bar

   To find T, we will apply Newton's Second Law:

   $$\begin{array}{ccc}\hfill F_R& =& ma\hfill \\ \hfill F-T\cos {25}^{\circ }& =& ma\hfill \\ \hfill 10000-T\cos {25}^{\circ }& =& \left(2000\right)\left(4\right)\hfill \\ \hfill T\cos {25}^{\circ }& =& 2000\hfill \\ \hfill T& =& 2206,76\mathrm{N}\hfill \end{array}$$

Object on an inclined plane

When we place an object on a slope the force of gravity (F ${}_g$ ) acts straight down and not perpendicular to the slope. Due to gravity pulling straight down, the object will tend to slide down the slope with a force equal to the horizontal component of the force of gravity (F ${}_g$  sin  $\theta$ ). The object will 'stick' to the slope due to the frictional force between the object and the surface. As you increase the angle of the slope, the horizontal component will also increase until the frictional force is overcome and the object starts to slide down the slope.

The force of gravity will also tend to push an object 'into' the slope. The vertical component of this force is equal to the vertical component of the force of gravity (F ${}_g$ cos $\theta$ ). There is no movement in this direction as this force is balanced by the slope pushing up against the object. This “pushing force” is called the normal force (N) and is equal to the force required to make the component of the resultant force perpendicularly into the plane zero, F ${}_g$ cos $\theta$ in this case, but opposite in direction.