SCALARS AND VECTORS (return to start of page) Scalars are added arithmetically (i.e. just add the numbers, so long as the units are the same). E.g. 2kg + 4kg = 6kg.

Vectors are added geometrically, which takes into account their directions as well as their sizes. This is described shortly using forces as an example. A familiar force is that produced by gravity - the weight of an object is defined as the force of gravity acting on the object, so weight is a vector. Like all forces, weight is measured in newtons, N. We see later that:

For example,  a person of mass 70kg has a weight of  70*10 = 700N (acting straight downwards).


ADDING FORCES (return to start of page)

Two or more forces acting on an object have a single equivalent force, called their resultant.

Forces acting on the same straight line

A force is a vector quantity, i.e. it has a size (or magnitude) and a direction. Both of these must be taken into account when combining forces to produce their resultant. For forces acting on the same straight line, this is quite easy:

Drawing a scale diagram to find the resultant

When forces do not act on the same straight line, we can still find their resultant by using the 'Parallelogram Rule'. This can be done by drawing a scale diagram or sometimes by calculation.


Two forces of 40N and 60 N act at 600 to each other at a point as represented below:

What is their resultant force?


(The following diagrams are not actually drawn to scale - but you could do so as a check.)

1. Choose a scale:

Suppose we let 2cm represent 10N. Then an 8cm line represents the 40N force and a 12cm line represents the 60N force. Draw these lines with a 600 angle between them:

2. Complete the parallelogram:
3. Draw the diagonal from the point of application of the forces:
The diagonal labelled R represents the resultant force - measure this and convert its length to newtons:

Diagonal R = 17.4cm, so the resultant force R = (17.4/2)*10 = 87N

4. Measure a suitable angle:

The angle A = 230.

Thus, the resultant of the two original forces is a force of size 87N acting at 230 to the 60N force.


Two forces of 50N and 60N act at a point and at right angles (900) to each other. The 60N acts horizontally to the right. Find the resultant force (a) graphically (i.e. by drawing a scale diagram), and (b) by calculation.

(a) Scale: 2cm represents 10N (the diagram below is not actually drawn to scale - you could do so as a check)

Diagonal R = 15.6cm, so resultant force R = (15.6/2)*10 = 78N. Also, angle A = 400.
So the resultant force has a size or magnitude of 78N acting at 400 to the 60N force.

(b) We can represent the set-up as:

This diagram need not now be drawn to scale, since we will calculate R and A this time. The diagram is just to indicate relative positions. Since opposite sides of a parallelogram are the same length, the side opposite the applied 50N is also 50N ‘long’. We can apply Pythagoras’ Theorem:

RESOLUTION OF FORCES (return to start of page)

We have seen that two forces can be added to produce a single resultant force.

In the above diagram, R is the resultant of F1 and F2. Looking at this the other way round, we can say R is made up of F1and F2, or that F1 and F2 are the components of the force R. The process of taking a single force and finding components for it is called ‘resolving the force into its components’.


A garden roller is pulled with a force of 280N acting along its handle, as represented below.

How much force is acting directly to the right, pulling the roller along the ground, and how much is trying to lift it straight upwards?

To obtain a diagram representing the original force and the desired components, we can draw:

F1 is the vertical component of the 280N. F2 is the horizontal component, which is pulling the roller along the ground. The above could be drawn to scale and the components found from the scale diagram. However, we can also calculate them:

Check using Pythagoras: (1612  +  2292 )1/2  =  280   (the power 1/2 indicates 'take square root of')

MOMENTS (return to start of page)

When we open a door, we apply a force to produce a turning effect. Looking top down on a door:

The turning effect or moment or torque of a force depends on the size of the force and how far it is applied from the hinge (the hinge may also be referred to as the pivot or fulcrum)



The ‘line of action of a force’ means exactly what it says – i.e. the ‘line along which the force acts’ – denoted by the left hand dotted line in the above diagram. To get the ‘perpendicular distance from the line of action of the force to the fulcrum’ put a ruler at right angles to the line of action in the diagram and slide it along, keeping the ruler at right angles, until it passes through the hinge – in the above case, the required distance is 3m.

Therefore, the moment of the force = 5N x 3m = 15 Nm

Moving the point of application of the force nearer the hinge

moment = 5N x 2m = 10 Nm

So, reducing the distance between the force and the hinge, reduces the size of the moment, i.e. reduces the turning effect. This is what we would expect by experience.

Changing the angle of the force

We know the size of the force but, from the definition of ‘moment’, we also need to find the perpendicular distance from the line of action of the force to the fulcrum’ - as can be done in steps (1), (2) and (3) below:
The diagram need not be particularly accurate, unless AB is to be determined by using a scale diagram.

When the force was applied at right angles to the gate, the moment was 15Nm, so the turning effect is smaller if any angle less than 900 is used.

Clockwise and anticlockwise moments

Comparing with the clock, we say that: In calculations we can choose, say, clockwise moments as positive and anticlockwise moments as negative.

CENTRE OF MASS AND GRAVITY (return to start of page)

Symmetric objects, of uniform density (i.e. the the same density throughout), have their centre of mass and gravity C at their geometric centre:
In most cases the centre of mass and the centre of gravity of an object coincide. A uniform metre ruler has its centre of mass at its 50cm point, and in the uniform gravitational field near to the Earth, it balances at this point, so this is also its centre of gravity. However, if the gravitational field were not uniform, it may balance at a different point, in which case, while its centre of mass is still in the middle, its centre of gravity is not.

For unsymmetric objects we can find C by experiment.

Experiment to find the centre of gravity of an irregular lamina (i.e. a thin sheet)

  1. Cut a card into an irregular shape
  2. Make a plumb line about 50 cm long – this is just a thread with a mass on the end
  3. Secure a pin in a clamp
  4. Make a small hole near the edge of the card (A in the diagram above) and hang the card on the pin – the card should be able to swing freely
  5. Suspend the plumb line from the pin. Put a mark (B) directly under the plumb line
  6. Remove the card and draw the line AB
  7. Make a second hole near the edge (C), well away from A
  8. Use the plumb line as before and mark the point (D) below the plumb line
  9. Draw the line CD
The centre of gravity must be directly below the point A, and so somewhere along line AB.
It must also be directly below point C, somewhere along line CD.
But the only point along AB and along CD is where the two lines cross – this point is therefore the centre of gravity of the card. The card should balance on your finger at this point.

COUPLES (return to start of page)

E.g. Top view of a tap handle:
The effect of the couple is to produce pure rotation (i.e. no translation - the forces have equal size but opposite directions, and so their vector sum is zero).

moment about O = (F * OA) + (F * OB) = F * (OA + OB) = F * AB

EQUILIBRIUM OF COPLANAR FORCES (return to start of page)

Note: ‘coplanar’ means ‘in the same plane’.

Thus, a body is in equilibrium if:
  1. the sum of the forces on the body equals zero
  2. the sum of moments (torques) about any point equals zero (this is called the principle of moments)
Note: The 'sums' in the above are 'vector sums' - i.e. they take into account the directions of forces and moments as well as their magnitudes.

We can express the conditions in a slightly different, but equivalent way.


The body is in equilibrium, so the forces all add up to zero (according to condition (1) above).
Now, each force has components in the x and y direction as shown.
Since the body is in equilibrium, it is not accelerating in any direction.
In particular it is not accelerating in the x or the y directions.

This means that:

Note: 'Algebraic' in the above means that we take one x (or y) direction as positive, and the other as negative.

There is nothing special about the x and y directions. We could use any two mutually perpendicular directions, and it would be true that the components would add up to zero in each direction respectively.

In condition (2) for equilibrium, to say that the sum of moments about any point is zero, is equivalent to saying that the sum of clockwise moments about the point equals the sum of anticlockwise moments about the same point.

Thus, we can restate the conditions for equilibrium for a body subject to coplanar forces as:

  1. the algebraic sum of the components of the forces in any two mutually perpendicular directions is respectively zero
  2. the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about the same point

A ladder of weight 170N rests against a vertical wall as shown. What are the values of the frictional force F1 and the reactions F2 and F3?

By rule (1) for equilibrium: By rule (2) for equilibrium:

We can determine or ‘take’ moments about any point. We choose a point where an unknown force acts, since its moment about that point is zero (as it is zero distance away from it), and so it does not produce an extra unknown term in the equation below.

Thus, taking moments about B:

Notice that we do not need to know the length, AB, of the ladder, since it cancels out in the calculation.

Equilibrium under the action of parallel forces

If a body is in equilibrium under the action of several parallel forces, then the first condition for equilibrium can be expressed as:

If the beam is in equilibrium:
  1. What is the weight W on the end of the beam?
  2. What is the reaction at the pivot?

Check: Since the beam is in equilibrium, the sum of clockwise moments should equal the sum of anticlockwise moments about any point. For example, at the point at which W acts:

Sum of antoclockwise moments = 320*6 + 540*4 = 4080
Sum of clockwise moments = 1360*3 = 4080

Triangle of forces rule

Suppose that a body is in equilibrium under the action of coplanar force P, Q and R:

R/ is the resultant of P and Q, i.e. P + Q = R/.
But, R/ must be exactly equal in size and opposite in direction to R if the body is in equilibrium.

We can represent this relationship as a ‘triangle of forces’, as below, with the arrows following each other ‘nose to tail’:

Thus: Example

An 8.0kg mass is supported by a cord attached to a hook in a ceiling. Another cord is pulled horizontally to produce the set-up represented below. What are the tensions in the cords? (g=10m/s2)

The ring is in equilibrium under the action of the three forces: T1, T2 and W. We can represent the 3 forces by:
Check using Pythagoras: (802  +  462 )1/2  = 92 (to 2 significant figure)

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A Level Physics - Copyright © A C Haynes 1999 & 2004