# TURNING EFFECT OF A FORCE - Form 2 Physics Notes

## Moment of a Force

• Basically, moment of a force refers to the turning effect of the force.
• It is defined as the product of the force and the perpendicular distance between the point of support (pivot or fulcrum) and the line of action of the force.

Moment of a force = force × perpendicular distance
• SI unit of moment of a force is the newton meter (Nm)

### Factors Affecting Moment of a Force

1. Amount of force – moment of force is directly proportional to the amount of force applied.
2. Perpendicular distance between line of action of force and point of support– momentis directly proportional the distance 900.

### Examples of Activities in Which Force Produces a Turning Effect

1. Opening and closing a door
2. Closing a lid of a container e.g.(geometrical instrument box)
3. A pair of scissors or garden shears in use
4. Children playing on “seesaw”
5. A wheelbarrow being used to lift heavy loads.
6. A screw driver being used to tighten or loosen a screw.
7. Beam balance in use.

Examples

Find the moment of the force about the pivot in the figure below

Moment of a force = force × perpendicular distance

∴ moment of force about pivot=20N×0.4m = 8Nm

## The Principle of Moments (The Law of the Lever)

• Consider a meter rule balanced (at equilibrium) on a pivot at its centre by weights W1,W2,W3 and W4 as shown below.
• The forces W1 and W2 tend to make the rule turn in the anticlockwise direction about the pivot. Therefore, the moments due to these weights are referred to as an anticlockwise moments.
• Similarly, the forces Wand W4 tend to make the rule turn in a clockwise direction and therefore, their moments about the pivot are clockwise moments.
Sum of clockwise moment =W3d+ W4d4
Sum ofanticlockwise moment =W1d+ W2d2
At equilibrium (balance), Sum of clockwise moment= Sum of anti-clockwise moment
W3d3+W4d4=W1d1+W2d2
• This can be summarized by the principle of moments which states “for a system in equilibrium the sum of clockwise moments about a point must be equal to the sum of anticlockwise moments about the same point”.
N/B:A body is said to be at equilibrium when it is balanced under the action of a number of forces.

Examples

1. State the law of the lever.
2. A uniform meter rule pivoted at its centre is balanced by a force of 4.8N at 20cm mark and some other two forces, F and 2.0N on the 66cm and 90cm marks respectively. Calculate the force F.

At equilibrium (balance),
Sum of clockwise moment = Sum of anticlockwise moment
F×0.16+2.0×0.40=4.8×0.30
0.16F+0.80=1.44
0.16F=0.64
F = 0.64/0.16
F=4.0N
3. A boy of mass 40kg sits at a point 2.0m from the pivot of a see saw. Find the weight of a girl who can balance the see-saw by sitting at a distance of 3.2m from the pivot.(Takeg=10nkg)
solution

At equilibrium (balance),
Sum of clockwise moment= Sum of anticlockwise moment
Wg×3.2m = (40kg×10N/kg )×2.0m
3.2Wg=800
Wg=800/3.2
Wg = 250N

Exercise

A half meter rule is suspended vertically from a pivot at the 0cm mark. It is maintained in the vertical position by four horizontal forces acting in the directions shown in the figure below.

The 10.0N force acts through the 15cm mark, 4.0N force through the 20cm mark and 5.0N force through the 40cm mark. Calculate F which acts through the 30 cm mark.

### Parallel Forces

• Consider a uniform rod below balanced by the forces F1,F2,F3,F4,Fand R which is the normal reaction on pivot.
• The forces F1,F2,F3,F4,F5 and R are parallel.
• For parallel forces:
1. The sum of forces acting on one side of the system is equal to the sum of forces acting on opposite direction i.e. the algebraic sum of parallel forces is zero.
2. The sum of clockwise moments is equal to the sum of anticlockwise moments i.e. the algebraic sum of the moments of parallel forces is zero.

Exercise

A uniform metal rod of length 80cm and mass 3.2kg is supported horizontally by two vertical spring’s balances C and D balance C is also from one end while balance D is 30cm from the other end. Find the reading on each balance.

### Anti–parallel Forces(Couples)

• Anti– parallel forces or a couple refers to equal, parallel but opposite forces.
• The total moment of anti-parallel forces is the product of one of the forces and the perpendicular distance between the forces.

Example

Two vertical equal and opposite forces act on a meter rule at 20cm and 90cm marks respectively. If each of the forces has a magnitude of 4.0N, calculate their moment on the meter rule about the 40cm mark.

solution

Total moment = one of the force,F x perpendicular distance between the forces,d

=4.0N×(0.9-0.2)m
=4.0N ×0.7
=2.8Nm

#### Examples of couples

1. Forces applied on a wheel spanner when tightening or loosening a nut
2. Forces applied when opening a water tap
3. Forces applied on the steering wheel of a car when going round a bend
4. Forces applied on bicycle handle

## Revision Exercise

1. Explain why the handle of a door is placed as far as possible from the hinges.
2. Explain why it is easier to loosen a tight nut using a spanner with along handle than the one with a short handle.

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Read 23577 times Last modified on Friday, 14 January 2022 11:57

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