Moments

In this example on the right, the distance is easily measured from the centre of the pivot point to the centre of the force acting on it.

In this example we have not considered the mass of the beam holding the object but we will look at this in other examples.

Sometimes we find examples where the rotating beam is at an angle to the horizontal, just as in the example on the right.

This time we are going to need to use some trigonometry to calculate the perpendicular distance from the pivot to the object:

\textcolor{aa57ff}{\text{Moment} = F \times d \cos θ}

Example: Calculate the clockwise moment produced in the diagram on the right.

[3 marks]

\text{Perpendicular distance} = d \cos θ = \boldsymbol{1.8 \cos 35 = 1.474} \: \textbf{m}

\text{Moment} = \text{force} \times \text{perpendicular distance} = \boldsymbol{1100 \times 1.474} = 1621.4 \: \text{Nm}

\text{Moment} = \boldsymbol{1600} \: \textbf{Nm} \left(2 \: \text{sf} \right)

The object on the left is causing an anti-clockwise turning effect whilst the object on the right is causing a clockwise turning effect.

If these two moments are equal, we can say that the moments are in equilibrium.

If the moments are in equilibrium, no rotation will occur.

For the forces to be a couple, the forces must be coplanar and equal and opposite in magnitude. For forces to be coplanar, they must be acting in the same plane (but may be acting opposite in direction).

This pair of forces can cause a rotation without the need of a pivot. As the forces are equal and opposite, the resultant force is zero. Therefore there is no net movement other than the rotation of the object.

For an object of uniform density (has the same density throughout), its centre of mass is at the centre of the object.

If the object is symmetrical this can easily be found. If you take more than one line of symmetry, the centre of mass will be at the point where the lines of symmetry intersect.

The centre of mass of this regular hexagon can be seen where all of its lines of symmetry meet.

In the diagram on the right, both objects above are currently stable. This is because their centre of mass (indicated by the arrows) are acting above the base of the object. However, we would consider the right hand object to be more stable as it would take more effort to move the object enough so that it’s centre of mass moved outside of its base:

Both objects are now unstable as their centre of mass is no longer above their base. However, it has taken a much greater movement to cause the left object to become unstable when compared to the right object. This is because of the large base of the right hand object.