- SCALARS AND VECTORS
- ADDING FORCES
- RESOLUTION OF FORCES
- MOMENTS
- CENTRE OF MASS AND GRAVITY
- COUPLES
- EQUILIBRIUM OF COPLANAR FORCES

*A scalar quantity only has size or magnitude*. It is completely specified by a number of appropriate units. For example: distance, speed, mass, energy, work, volume, ...*A vector quantity has size and direction*. For example: displacement, velocity, force (weight), acceleration, momentum, ...

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:

- Weight = m*g (m = mass in kg; g
=10m/s
^{2}, and is called the 'acceleration due to gravity')

Note:

- In these notes, * is used to stand for for 'times' or 'multiply' (to avoid confusion with the letter 'x')
- The units used above are examples of SI units, which are the standard units used in science. (‘SI’ comes from the French, ‘Système International d'Unités’)

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:

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.

**Example**

Two forces of 40N and 60 N act at 60^{0} to each other at a point as represented
below:

What is their resultant force?

*Procedure:*

(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
60^{0} angle between
them:

2.

3.

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 = 23^{0}.

Thus, the resultant of the two original forces is a force of
size 87N acting at 23^{0} to
the 60N force.

**Example**

Two forces of 50N and 60N act at a point and at right angles
(90^{0}) 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 = 40

So the resultant force has a size or magnitude of 78N acting at 40

(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 F

**Example**

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:

- an arrowed line to represent the 280N applied force
- horizontal and vertical lines through the point where the 280N force acts (any lengths, but not too short)
- lines from the end of the 280N force perpendicular to the two lines just drawn
- arrowed lines to represent the required components

F

Check using
Pythagoras: (161^{2} + 229^{2} )^{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)

**Definition**

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**

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
90*^{0} is
used.

**Clockwise and anticlockwise moments**

Comparing with the clock, we say that:

- the moment produced by F
_{1}is anticlockwise - the moment produced by F
_{2}is clockwise

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

*The centre of mass of an object is the point at which all its mass may be considered to be concentrated**The centre of gravity of an object is the point through which the resultant force of gravity acts*

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

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)**

- Cut a card into an irregular shape
- Make a plumb line about 50 cm long – this is just a thread with a mass on the end
- Secure a pin in a clamp
- 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
- Suspend the plumb line from the pin. Put a mark (B) directly under the plumb line
- Remove the card and draw the line AB
- Make a second hole near the edge (C), well away from A
- Use the plumb line as before and mark the point (D) below the plumb line
- Draw the line CD

It must also be directly below point C, somewhere along line CD.

But the only point along AB

**COUPLES** (return
to start of page)

*A couple is a pair of parallel forces, equal in size but opposite in direction, whose lines of action do not coincide*

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’.

*A body is in equilibrium if it undergoes no translational or rotational acceleration*

- the sum of the forces on the body equals zero
- the sum of moments (torques) about any point equals zero
(this is called the
*principle of moments)*

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

**Consider**

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

In particular it is not accelerating in the x or the y directions.

This means that:

- the algebraic sum of the components of the forces in the x direction must add up to zero
- the algebraic sum of the components of the forces in the y direction must add up to zero

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:

- the algebraic sum of the components of the forces in
**any**two mutually perpendicular directions is respectively zero - the sum of the clockwise moments about
**an**y 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 F_{1
}and the reactions F_{2} and F_{3}?

By rule (1) for equilibrium:

- Vertical forces must balance, so: F
_{2}= 170N - Horizontal forces must balance, so: F
_{1}= F_{3}

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:

*sum of forces in one direction = sum of forces in opposite direction*

If the beam is in equilibrium:

- What is the weight W on the end of the beam?
- 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

But, R

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

Thus:

*If a body is in equilibrium under the action of three coplanar forces, then the forces can be represented by the sides of a triangle taken in order*

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/s^{2})

The ring is in equilibrium under the action of the three forces: T

Check using Pythagoras: (80

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