Equilibrium of a Rigid Body


Throughout our A-level course so far, we have used a particle model. We will now start to treat objects as a rigid body, extending the model to give objects size, but continuing to assume that they are inflexible so do not bend or deform when forces are applied to them.

We now consider the moment or turning effect of forces. To do this we need to calculate the centre of mass of standard shapes and composite bodies. We will also look at when objects are on the point of breaking equilibrium, causing them to slide or topple.

Moment of a Force (the word “moment” is interchangeable with “turning effect”)

The moment of a force F, about a point O, is F x d, where d is the perpendicular distance from the point O to the line of action of the force F. The moment of each force is measured in Newton metres (Nm). If the perpendicular distance between the force and the point O is zero, then it contributes no moment.

In the first diagram above, the moment of the 6N force about the point A is 6×2, so 12 Nm.

In the second diagram, we need to find the component of the force acting perpendicular to the line through B, which is 3cos20. So the moment is (3cos20) x 5, so 15cos20 or 14.1Nm.

The third diagram shows two forces applying a moment to point C. Both act in the same direction, i.e. clockwise, so they are combined by addition. The total moment on point C is 2×6 + 2.5×4, so 22Nm.

Worked Example

Calculate the moment in each of the examples below, specifying whether it acts in a clockwise or an anticlockwise direction:

Moment on a light rod

We can select a point on a rod and calculate the moment of forces on that point, modelling the rod as light and as a rigid body.

Worked example

In each of the diagrams below, a light rod is pivoted at a fixed point, A and we want to find the unknown values such that the rod will have a zero moment about A:

Moment on a rod that has a weight

Below we consider a uniform rod with length 2m and mass 2kg that rests in equilibrium, extending over the edge of a surface. The edge of the surface is point A, and a mass of 3kg holds the rod down at point B:

As the rod is modelled as uniform, so its weight acts through its centre, as marked on the diagram below. If there was no mass at B, then taking moments about A would give 2g \times x, giving a moment of 20x Nm clockwise, i.e. the force would not be in equilibrium.

The mass at B causes a moment of 3g \times (1-x) anticlockwise on A. If the force is in equilibrium, then 3g(1-x) = 2gx, so x=0.6m.

Worked Example

A uniform rod AB, of length 5m and mass 10kg, is placed over the edge of a cliff such that B hangs 4m over the edge of the cliff. A man of mass 80kg stands on the cliff side of the rod at a distance of xm from the edge of the cliff.

(a) Find the value of x so that the rod is on the point of tipping over the cliff;

(b) The man now stands at A and a boy of mass 35kg walks across the rod towards B. Can the boy walk all the way to the end of the rod?

Exercise 1


1.) 1.05kg

2.) 0.333kg

3.) 10/3

4.) (a) 6.3Nm anticlockwise (b) 22.1 Nm anticlockwise (c) 36Nm anticlockwise

5.) (a) 46.5 Nm clockwise (b) 12.8Nm anticlockwise (c) 2.89 Nm anticlockwise

6.) (a) 0.119m (b) 3.8kg

7.) m = \frac{6}{19}M - \frac{78}{19}

8.) (a) x = 9.8m (b) x = 40.2m (c) x = 3.20 m

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