5.1 – Moment of a Force

Definition: Moment of a force
Moment of a force (or torque) about a point is defined as the product of a force and the perpendicular distance from the line of action of the force to the point.

Moments of a force = Force x Perpendicular distance from the line of action of the force to the pivot

moment = F x d

  • SI unit for moment is newton-metre (N m).
    We may also use newton-centimetres N cm
    (where 1 N m = 100 N cm)
  • It is a vector quantity with magnitude and direction.
  • The direction is either clockwise or anticlockwise.
  • The point about which the object rotates may be referred to as a pivot or turning point or fulcrum.

 

Clockwise Moments about the pivot = F1 x d1

 

Anticlockwise Moments about the pivot = F2 x d2

Taking moments about a point
Although we will usually be taking moments about a turning point, they can in fact be taken about any point. It is thus important to state where we are taking moments about.
Example 1A
A light metre rule is freely pivoted on a triangular prism at the 50.0 cm mark.

A force of 20 N is applied on metre rule at the 30.0 cm mark. Calculate the moment of the force on the metre rule about the pivot and state its direction.

Taking moments about the 50.0 cm mark of the rule:

moments = Force x Distance

moments = 20 N x (0.500 – 0.300) m

moments = 20 N x 0.200 m

moments = 4.0 N m anticlockwise

The rule will begin to rotate anticlockwise about the pivot.
Example 1B
The 20.0 N force is now applied to the 80.0 cm mark but in an upwards direction.

Calculate the new moment of the force on the metre rule about the pivot and state its direction.

Taking moments about the 50.0 cm mark of the rule:

moments = Force x Distance

moments = 20 N x (0.800 – 0.500) m

moments = 20 N x 0.300 m

moments = 6.0 N m anticlockwise

The rule will begin to rotate anticlockwise about the pivot.
Example 1C
The 20.0 N force is now applied to the centre of the rule as shown.

Calculate the new moment of the force on the metre rule about the pivot and state its direction.

Taking moments about the 50.0 cm mark of the rule:

moments = Force x Distance

moments = 20 N x 0.00 m

moments = 0 N m

The rule will not begin to rotate.
Zero Moments
Whenever the line of action of a force passes directly through the pivot there will be zero moments and the object will not begin to rotate.
Is it a ‘light’ object
If we are told an object is ‘light’ it means we can assume it has negligible weight/mass and so we do not need to take this into consideration when calculating moments.
Example 2
Which method of applying a force, F, 2.0 m away from a pivot on a 4.0m plank will produce the larger turning effect about the pivot?

method 1

method 2

For method 1:As F is perpendicular to the plank, shortest distance is along the plank.

moments about the pivot = F x d1 anticlockwise

For method 2:

As F is not perpendicular to the plank, we need to find the shortest distance from the pivot to the line of action of the force. This is shown on the diagram as d2.

moments about the pivot = F x d2 anticlockwise

As d1 > d2  => method 1 produces the greatest moment about the pivot.

 

When two (or more) forces act on an object we add the individual moments produced by each force.

Example 3
Two 10 N forces act on a 4.0 m plank as shown below.

Calculate the resultant moment about the pivot.

Anticlockwise moment due to F1 = 10 N x 1.0 m = 10 N m

Anticlockwise moment due to F2 = 10 N x 2.0 m = 20 N m

Thus,

Resultant anticlockwise  = 10 N m + 20 N m = 30 N m
moment about the pivot

If the moments are in opposing directions then we need to subtract one from the other.

Example 4
Two 10 N forces act on a 4.0 m plank as shown below.

Calculate the resultant moment about the pivot.

Anticlockwise moment due to F3 = 10 N x 2.0 m = 20 N m

Clockwise moment due to F4 = 10 N x 1.0 m = 10 N m

Thus, as anticlockwise moments are larger than clockwise moments there will be a net anticlockwise moment

Resultant anticlockwise  = 20 N m – 10 N m = 10 N m
moment about the pivot

The plank will begin to rotate clockwise due to the net moments.
Example 5
A 1 m long uniform beam of mass 2 kg is being lifted by a vertical force, F, at the 100 cm mark.

What if the minimum force, F, that will lift the beam?

Weight = mg =2 x 10 = 20 N (this acts at the 50 cm mark)

When lifting the 0 cm end of the ruler will remain on the floor and acts as the pivot.

Tanking moments about the 0 cm mark;

Anticlockwise moment due to F = F x 100 cm= 100F

Clockwise moment due to weight = 20 N x 1.0 m = 10 N m
Turning Effects of a Force Next >>

 

2025 Physics Lessons