Forces associated with levers

The force associated with a lever is a moment of a force (the turning effect of a force). The moment of a force, with respect to a given point, turning point, (or hinge or pivot or fulcrum) in the same plane, is the product of the force and the perpendicular distance from the point (hinge, pivot or fulcrum) to the line of action of the force.

The effect of the force must be considered in relation to the direction (sense) in which it tends to turn around.

We must make a distinction between a clockwise and an anticlockwise moment. Great attention must be given to the lever arm, which is always the perpendicular distance from the hinge or pivot or fulcrum to the line of action of the force. The calculation of moments is repeatedly used throughout the unit and it is important that this concept is fully understood.

Moment = Force x lever arm or

Moment = Force x distance

M = F x d

A moment is expressed in units of, newton-metres (Nm), or kilonewton-metres (kNm)

Where M = moment of the force about the hinge (F × d).
F = force (must be in newtons (N) or kilonewtons (kN)
d = perpendicular distance from the hinge to the line of action of the force

Figure 1

Hopefully Figure 1 makes sense to you. As you can see the point of application and direction (line of action) of the forces are significant in respect of the moments magnitude. By keeping the forces vertical the lever arm of F3 and F4 on the dashed member is lesser than on the solid member and by moving the member further down the distance to F3 and F4 will decrease even more. When the member reaches the vertical (turning 90° downwards) the distance will be zero. F3 creates a anticlockwise moment and F4 creates a clockwise moment.

Figure 2

Figure 2 is the same the dashed member shown in Figure 1. Here the member is placed in a horizontal position and the forces acting on the member are inclined. Remember that forces can be resolved into horizontal and vertical components. Whether we use of F3 or F4 times the perpendicular distance or the vertical components of F3 or F4 times their perpendicular distance the magnitude of the resulting moments will be the same. .

The horizontal component of F4 times the horizontal distance to the hinge equals F4 times the inclined distance to the hinge.

The following example will prove this statement.

Figure 3

As can be seen the vertical component of the 5 kN force times the perpendicular distance
3.54 kN × 5.00 m (horizontal component × distance is zero)
is equal to the inclined force × the perpendicular distance
5 kN × 3.54 m.

A see-saw is a common equipment found on playgrounds. We all know how to keep the see-saw levelled; children on either side must be of the same weight and sit from the support (fulcrum) at equidistant. If one child is heavier than the other the heavier child must move closer to the fulcrum. The see-saw works on the principle that the clockwise moment is equal to the anticlockwise moment.

When a force system, acting in the same plane, is in equilibrium, then the sum of the clockwise moments must equal the sum of the anticlockwise moments about a given point. We need to establish a sign convention for working problems involving moments.

Since an international convention does not exist, in this course, a clockwise rotation about the center of moments will be considered as positive; a counter-clockwise rotation about the center of moments will be considered as negative.

Use the following convention:

(The sign convention for a moment can be the other way round. What is important is to make a distinction between a clockwise and anti-clockwise moment. This may be confusing and students should stick to the forementioned convention.)

Equilibrium and equilibrants

A body is in a state of equilibrium when there is no tendency for it to be disturbed from its existing state of rest (or uniform motion). As all structural components of a building usually are at rest a state of equilibrium is preserved.

Force Couples

Figure 4
Two co-planar parallel forces of equal manitude, separated by a distance from each other and opposite sense, form a couple.

Since no resultant force exists a couple does not cause translation (i.e. motion in a straight line). However, a couple does cause rotation as can be seen in the opposite Figure.

We will now resolve a single force into a force acting at another point and a couple. This principle is often applied in beam design when we tranferred a force to a different position. If this is done then consideration must be given to the moment resulting from the transfer (i.e. force × tranferred distance). The following example will demonstrate this.

Figure 5
  • 5 kN is acting on point A at the cantilever.

  • At Ra add to equal and opposite forces, 5 kN (upwards & downwards) so that their lines of action are parallel to the line of action of the 5 kN force acting at A. Note that the 5 kN (acting at Ra) and the 5 kN (acting at A) make up a couple whose moment is 10 kNm.

  • Therefore the force system acting on the cantilever can be represented by the 5 kN (acting at Ra) and the moment of the couple (5 x 2 = 10 kNm).

    (This principle can be used for the calculation of the reactions of a beam when the beam has a cantilever.)


Determine the maximum mass in kg that the 12.5 ton lifting truck shown below can carry before the front wheels will be on the point of lifting off the roadway.


We use the principle of moments to find the answer.

In our example the fulcrum is the back wheel at 2.3 metre from the centre of gravity.

To keep the system in equilibrium the clockwise moment weight of truck times 2.3 metre (12.500 kg x 2.3 m) must equal the anticlockwise moment mass times the distance to the back wheel [kg x (6.0-2.3)].

The equation written mathematically

12.500 x 2.3 = mass x (6.0 - 2.3)
28750 = mass x 3.7
mass = 28750 / 3.7
= 7.770.3 kg