Moments

When looking at forces acting on a rigid body, as opposed to objects modelled as particles, it is possible to study the turning effects of that rigid body as well as forces acting at different points on the rigid body. Moments measure the turning effect of a given force and so we must specify the point at which we are assessing that turning effect. The moment about a given point is calculated as follows:

\text{ Moment }=\text{ Force }\times\text{ Perpendicular Distance}

or

\text{ Moment }=\text{ Perpendicular Component of Force }\times\text{ Distance}

Moment = F\times D\sin(\theta)
Moment = F\sin(\theta)\times D

This can also be thought of as the direct distance to the point where the force is acting multiplied by the component of the force acting perpendicularly to that distance. For a force acting completely at right angles to the distance from the point, the formula simplifies to \text{Moment }=\text{ Force }\times \text{ Distance} – this is because in this case \sin(\theta)=\sin(90^\circ)=1.

Force is perpendicular, Moment = F\times D

Since we are talking about turning effect we must specify whether the moment is a clockwise moment or an anti-clockwise moment. We also give the units of a moment which is Newton metres (Nm) since they are calculated from the product of a force and a distance – see Example 1.

Equilibrium

As with a resultant force, there is also a resultant moment at a given point which is the sum of all moments acting at that point. In order to calculate the resultant moment of a rigid body at a given point, the forces acting on the body must be coplanar. This means that the forces are all acting in the same plane. When a body is in equilibrium, the resultant moments are 0 around any given point as well as the resultant forces – this means that, as well as the total upwards forces being equal to the total downwards forces, all clockwise moments balance anticlockwise moments:

\text{Equilibrium }\Leftrightarrow \text{ Total Clockwise Moments = Total Anti-clockwise Moments}

In many equilibrium cases, a careful selection of the point around which to calculate moments can simplify the problem vastly – see Example 2.

object in equlibrium

Centres of Mass and Tilting

For rigid bodies that are uniform, their mass is spread evenly throughout the body and the centre of mass (the point at which the weight of the body acts) is at the centre. For non-uniform rigid bodies, the mass is spread unevenly and the centre of mass does not act at the centre. Calculating moments around certain points in a rigid body can reveal the location of the body’s centre of mass – see Example 3. Questions like this may also include bodies that are at the point of tilting about a pivot. In these cases, the reaction force at any other support point that isn’t the pivot is equal to 0 – see Example 4.

Examples

The diagram shows 3 forces acting on a light rod. Find the exact resultant moment about the point P.

The clockwise moments are given by 12\times2+13\sin(60^\circ)\times 1=24+\frac{13\sqrt{3}}{2}. The anticlockwise moment is given by 15\sin(75^\circ)=\frac{15\sqrt{6}+15\sqrt{2}}{4}. The resultant moment is 24+\frac{13\sqrt{3}}{2}- \frac{15\sqrt{6}+15\sqrt{2}}{4}=\frac{96+26\sqrt{3}-15\sqrt{6}-15\sqrt{2}}{4} in the clockwise direction.

A uniform rod AB of length 5m and mass 80kg rests on two supports located 1m from A and 2m from B. A woman of mass 55kg is standing on top of the rod 2m from A. Find the reaction forces at each of the supports.

We being by sketching the information:

We first find the reaction force at one of the supports (it doesn’t matter which one first). We find the reaction force at the right support by calculating the moments around the left support. All forces are perpendicular so they are straight forward to calculate. The clockwise moments are given by 1\times 55g+1.5\times 80g=175g. The anticlockwise moment is given by 2\times R2. The rod is in equilibrium and so the clockwise and anticlockwise moments balance. Setting them equal we have 2R2=67g and so R2=\frac{175g}{2}. The remaining reaction force can be found by setting upward forces equal to downward forces: R1+R2=55g+80g. It follows that the reaction at the left support is R1=\frac{95g}{2}.

Two children sit at either end of a seesaw that is 4 metres long and is balancing horizontally at a pivot at the centre. The children weigh 25kg and 30kg respectively. Modelling the seesaw as a non-uniform rod, find the location of the centre of mass of the rod.

Again we sketch the information labelling the location of the centre of mass of the rod at an unknown distance of x from one end:

There are multiple ways to answer this question but the straight forward way would be to take moments around the leftmost point on the seesaw noting that the reactive force at the pivot is equal to the total of the downward forces, that is R=70g. The pivot is at the centre of seesaw and so the anticlockwise moment is 2\times R=140g. The clockwise moments are given by 4\times 30g +x\times 15g. The moments balance and so 140g=120g+15gx. Solving for x gives x=\frac{4}{3}. The centre of mass of the seesaw is \frac{4}{3}m from the left of the seesaw in the diagram above.

A non-uniform metal rod PQ of mass 1.2kg and length 1.6 metres is hanging horizontally from the ceiling by two strings at a distance 0.4 metres and x metres from P at the point R. The centre of mass of the rod acts at a distance of 0.6xm from P. An ornament of mass 400g is hung 30cm from Q and, as a result, the rod is about to tilt about R. Find the value of x and hence the location of the centre of mass of the rod.

Sketch the information in a diagram being careful to write all units consistently:

Since the rod is about to tilt about R, the reactive force at the other string is zero, that is, R1=0. We can then find the value of x by taking moments about R. We first note that the centre of mass has a distance of 0.4x from R. To find the distance of the ornament from R we first see that the distance RQ is (1.2-x)m, the ornament is 0.3m from Q and so the ornament is a distance of (0.9-x)m from R. It follows that the clockwise moment about R is (0.9-x)\times 0.4g and the anticlockwise moment is 0.4x\times 1.2g. Equating the clockwise and anticlockwise moments we have (0.9-x)\times 0.4g=0.4x\times 1.2g which simplifies to 0.9-x=1.2x. Solving for x we get x=\frac{9}{22}m. The centre of mass is \frac{9}{22}\times 0.6\approx 0.25 metres from P (closer to P than the first string).

AS Mechanics Moments