Look back to Figure 4.5. The spider is in equilibrium, even though three forces are acting on it. We can think of the tension in the thread as having two effects. It is:
- pulling upwards, to counteract the downward effect of gravity 
- pulling to the left, to counteract the effect of the wind.
We can say that this force has two effects or components: an upwards (vertical) component and a sideways (horizontal) component. It is often useful to split up a vector quantity into components like this, just as we did with velocity in Chapter 2. The components are in two directions at right angles to each other, often horizontal and vertical. The process is called resolving the vector.
Then we can think about the effects of each component separately; we say that the perpendicular components are independent of one another. Because the two components are at ${90^ \circ }$ to each other, a change in one will have no effect on the other. Figure 4.8 shows how to resolve a force F into its horizontal and vertical components. These are:

$horizontal\,component\,of\,F,{F_x} = F\,\cos \,\theta $
$vertical\,component\,of\,F,{F_y}\, = F\sin \,\theta $

Making use of components

When the trolley shown in Figure 4.9 is released, it accelerates down the ramp. This happens because of the weight of the trolley. The weight acts vertically downwards, although this by itself does not determine the resulting motion. However, the weight has a component that acts down the slope. By calculating the component of the trolley’s weight down the slope, we can determine its acceleration.

Figure 4.10 shows the forces acting on the trolley. To simplify the situation, we will assume there is no friction. The forces are:
- the weight of the trolley, W, which acts vertically downwards
- the contact force of the ramp, N, which acts at right angles to the ramp.

You can see at once from Figure 4.10 that the forces cannot be balanced, since they do not act in the same straight line.
To find the component of W down the slope, we need to know the angle between W and the slope. The slope makes an angle θ with the horizontal, and from the diagram we can see that the angle between the weight and the ramp is $({90^ \circ } - \theta )$. Using the rule for calculating the component of a vector given previously, we have:

$component\,of\,W\,down\,the\,slope\,\, = \cos \,({90^ \circ } - \theta ) = W\,\sin \,\theta $

(It is helpful to recall that $\cos \,({90^ \circ } - \theta ) = \sin \,\theta $; you can see this from Figure 4.10.)
Does the contact force N help to accelerate the trolley down the ramp? To answer this, we must calculate its component down the slope. The angle between N and the slope is ${90^ \circ }$. So:

$component\,of\,N\,down\,the\,slope\,\, = N\cos \,{90^ \circ } = 0$

The cosine of ${90^ \circ }$ is zero, and so N has no component down the slope. This shows why it is useful to think in terms of the components of forces; we don’t know the value of N, but, since it has no effect down the slope, we can ignore it.
(There’s no surprise about this result. The trolley runs down the slope because of the influence of its weight, not because it is pushed by the contact force N.)

Changing the slope

If the students in Figure 4.9 increase the slope of their ramp, the trolley will move down the ramp with greater acceleration. They have increased θ, and so the component of W down the slope will have increased.
Now we can work out the trolley’s acceleration. If the trolley’s mass is m, its weight is mg. So the force F making it accelerate down the slope is:

$F = mg\,\sin \,\theta $

Since from Newton’s second law for constant mass we have , the trolley’s acceleration a is given by:

$a = \frac{{mg\,\sin \,\theta }}{m} = g\,\sin \,\theta $

We could have arrived at this result simply by saying that the trolley’s acceleration would be the component of g down the slope (Figure 4.11). The steeper the slope, the greater the value of $\sin \,\theta $, and hence the greater the trolley’s acceleration.

Questions

 

4) The person in Figure 4.12 is pulling a large box using a rope. Use the idea of components of a force to explain why they are more likely to get the box to move if the rope is horizontal (as in a) than if it is sloping upwards (as in b).

5) A crate is sliding down a slope. The weight of the crate is $500 N$. The slope makes an angle of ${30^ \circ }$ with the horizontal.
a: Draw a diagram to show the situation. Include arrows to represent the weight of the crate and the b: contact force of the slope acting on the crate.
Calculate the component of the weight down the slope.
c: Explain why the contact force of the slope has no component down the slope.
d: What third force might act to oppose the motion? In which direction would it act?

Solving problems by resolving forces

A force can be resolved into two components at right angles to each other; these can then be treated independently of one another. This idea can be used to solve problems, as illustrated in Worked example 1.