SAILING AHEAD

Force is a vector quantity. Sailors know a lot about the vector nature of forces. For example, they can sail ‘into the wind’. The sails of a yacht can be angled to provide a ‘component’ of force (in other words, an effect of the force in the forward direction) and the boat can then sail at almost ${45^ \circ }$ to the wind. The boat tends to ‘heel over’ and the crew sit on the side of the boat to provide a turning effect in the opposite direction (Figure 4.1). If the wind has an effect forwards, what stops the boat from moving sideways due to the ‘component’ of the wind sideways? (Hint: find out about the shape of the bottom of the boat.)

You will have learned that a vector quantity has both magnitude and direction. An object may have two or more forces acting on it and, since these are vectors, we must use vector addition (Chapter 1) to find their combined effect (their resultant). 
There are several forces acting on the car (Figure 4.2) as it struggles up the steep hill. They are:
- its weight $W (= mg)$
- the normal contact force N of the road
- air resistance D
- the forward force F caused by friction between the car tyres and the road.
If we knew the magnitude and direction of each of these forces, we could work out their combined effect on the car. Will it accelerate up the hill? Or will it slide backwards down the hill?

The combined effect of several forces is known as the resultant force. To see how to work out the resultant of two or more forces, we will start with a relatively simple example.

Two forces in a straight line 

We saw some examples in Chapter 3 of two forces acting in a straight line. For example, a falling tennis ball may be acted on by two forces: its weight mg, downwards, and air resistance D, upwards (Figure 4.3). The resultant force is then:

$resultant force = mg − D = 1.0 − 0.2 = 0.8 N$

When adding two or more forces that act in a straight line, we have to take account of their directions. A force may be positive or negative; we adopt a sign convention to help us decide which is which. In setting up the sign convention you decide for yourself which direction is positive. In Figure 4.3, for example, we have taken the direction downwards as positive so the weight is $+1.0 N$, a positive force, and the force upwards is $−0.2 N$, a negative force. The resultant is $+0.8 N$, which tells us the resultant is downwards.
You might choose the upwards direction as positive, but if you apply a sign convention correctly, the sign of your final answer will tell you the direction of the resultant force (and hence acceleration).

Two forces at right angles

Figure 4.4 shows a shuttlecock falling on a windy day. There are two forces acting on the shuttlecock: its weight vertically downwards, and the horizontal push of the wind. (It helps if you draw the force arrows of different lengths, to show which force is greater.) We must add these two forces together to find the resultant force acting on the shuttlecock.
We add the forces by drawing two arrows, head-to-tail, as shown on the right of Figure 4.4.
- First, draw a horizontal arrow to represent the $6.0 N$ push of the wind.

- Next, starting from the end of this arrow, draw a second arrow, downwards, representing the weight of $8.0 N$.
- Now, draw a line from the start of the first arrow to the end of the second arrow. This arrow represents the resultant force R, in both magnitude and direction.
The arrows are added by drawing them end-to-end; the end of the first arrow is the start of the second arrow. Now we can find the resultant force either by scale drawing or by calculation. In this case, we have a 3–4–5 right-angled triangle, so calculation is simple:

$\begin{array}{l}
{R^2} = {6.0^2} + {8.0^2} = 36\\
 = 100\\
R = \sqrt {100} \\
 = 10N
\end{array}$
$\begin{array}{l}
\tan \,\theta  = \frac{{opp}}{{adj}} = \frac{{8.0}}{{6.0}} = \frac{4}{3}\\
\theta  = {\tan ^{ - 1}}\frac{4}{3} \approx {53^ \circ }
\end{array}$

So the resultant force is $10 N$, at an angle of ${53^ \circ }$ below the horizontal. This is a reasonable answer; the weight is pulling the shuttlecock downwards and the wind is pushing it to the right. The angle is greater than ${45^ \circ }$ because the downward force is greater than the horizontal force.

Three or more forces

The spider shown in Figure 4.5 is hanging by a thread. It is blown sideways by the wind. The diagram shows the three forces acting on it:
- weight acting downwards
- the tension in the thread
- the push of the wind.
The diagram also shows how these can be added together. In this case, we arrive at an interesting result.
Arrows are drawn to represent each of the three forces, end-to-end. The end of the third arrow coincides with the start of the first arrow, so the three arrows form a closed triangle. This tells us that the resultant force R on the spider is zero, that is, $R = 0$. The closed triangle in Figure 4.5 is known as a triangle of forces.

So there is no resultant force. The forces on the spider balance each other out, and we say that the spider is in equilibrium. If the wind blew a little harder, there would be an unbalanced force on the spider, and it would move off to the right.
We can use this idea in two ways:
- If we work out the resultant force on an object and find that it is zero, this tells us that the object is in equilibrium.
- If we know that an object is in equilibrium, we know that the forces on it must add up to zero. We can use this to work out the values of one or more unknown forces.

Questions

 

1) A parachutist weighs $1000 N$. When she opens her parachute, it pulls upwards on her with a force of $2000 N$.
a: Draw a diagram to show the forces acting on the parachutist.
b: Calculate the resultant force acting on her.
c: What effect will this force have on her?

2) The ship shown in Figure 4.6 is travelling at a constant velocity.
a: Is the ship in equilibrium (in other words, is the resultant force on the ship equal to zero)? How do you know?
b: What is the upthrust U of the water?
c: What is the drag D of the water?

3) A stone is dropped into a fast-flowing stream. It does not fall vertically because of the sideways push of the water (Figure 4.7).
a: Calculate the resultant force on the stone.
b: Is the stone in equilibrium?