Page 67 - Physics Coursebook 2015 (A level)

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.
If you draw a scale drawing be careful to:
■■ state the scale used
■■ draw a large diagram to reduce the uncertainty.
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.
8.0 N
Figure 4.4 Two forces act on this shuttlecock as it travels through the air; the vector triangle shows how to find the resultant force.
We add the forces by drawing two arrows, end-to-end, as shown on the right of Figure 4.4.
■■ First, a horizontal arrow is drawn to represent the 6.0 N push of the wind.
■■ Next, starting from the end of this arrow, we draw a second arrow, downwards, representing the weight of 8.0 N.
■■ Now we 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:
R2 = 6.02 +8.02 = 36+64 = 100 R = 10 N
tanθ = opp = 8.0 = 4
adj 6.0 3
θ = t a n − 1 43 ≈ 5 3 °
So the resultant force is 10 N, at an angle of 53° 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° because the downward force is greater than the horizontal force.
55
Chapter 4: Forces – vectors and moments
Direction of travel
6.0 N
6.0 N
θ
R
8.0 N
■■ ■■
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.
tension in thread
push of wind
weight
push of wind
θ
tension
triangle of forces
weight
Figure 4.5 Blowing in the wind – this spider is hanging in equilibrium.