Moments Of A Force

M Moments of a Force When we look at moments in everyday life three specific examples stand out as a greater way of explaining exactly what a moment of a force is; 1. When a child sits on a seesaw which is pivoted at its centre, the child can be balanced on it by a heavier adult sitting nearer to the centre of the seesaw. 2. It is easier to undo a nut with a longer spanner, when the force is applied to the end of the spanner than if you used a shorter spanner. 3. A door is easier to close if you push it at the edge of the door rather than if you push it at a point mid-way between the hinges and the edge of the door.

In general terms, suppose you have two forces of unequal magnitude but want them to produce the same turning effect, for example in the case of the seesaw above. To achieve this, the point of application of the smaller force must be at a greater distance than the application of the larger force from the pivot. So in our example above the child will sit further away from the pivot to produce the same turning effect of the force (or the moment of the force) as the adult who will sit closer to the pivot. The moment of a force is dependent on the magnitude of the force and where it is applied; “The moment of the force
about a point  is the product of the magnitude of the force
and the perpendicular distance from the point  to the line of action of
.”






moment of force


 The moment of the force is measured in newton-metres (Nm) as the force is measured in newtons and distance in metres. It is important that when you attempt a question involving moments to state whether the turning effect is either clockwise or anticlockwise, you can usually represent these two rotations using arrows. An  arrow would represent a clockwise motion, an  arrow would represent an anticlockwise motion.

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M Example One A light rod AB is 2m long and can rotate in a vertical plane about a fixed point B. A vertical force
of magnitude 5N acts in a direction perpendicular to AB at the point A. Calculate the moment of the force about B at A.

5N

$

#

When F is applied at A the moment of F about B is given by;

2m

Force perpendicular distance moment 8 2 16N Note that the full answer to this question is 16Nm anticlockwise. When multiple forces, that all lie in the same plane, act on a body the moments about any point can be added, as long as the sense of rotation for each moment is taken into account. This means that the algebraic sum of the moments is added. For example, a moment moving 16Nm anticlockwise is counteracted by a moment moving 20Nm clockwise, therefore the overall moment is 4Nm clockwise. Example Two Forces of magnitude 3N, 4N, 5N and 6N are applied to a light rod as shown in the diagram. Calculate the sum of the moments of these forces about A.

5N

1m

1m

1m

1m $

Our first step is to calculate all the individual moments; 6N

3N

4N

Moment of the 5N force about A is: 5 3 15N This is 15Nm in a clockwise !" direction. Moment of the 6N force about A is: 6 2 12N This is 12Nm in an anticlockwise !" direction. Moment of the 3N force about A is: 3 1 3N This is 3Nm in an anticlockwise !" direction. Moment of the 4N force about A is:

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M 4 1 4N

This is 4Nm in a clockwise !" direction. The next step is to add the clockwise and anticlockwise moments separately and find the difference between both. Therefore for the clockwise !" moments we get; 15N & 4N 19N

The anticlockwise !" moments equal; 12N & 3N 15N

Therefore the sum of the moments is; 19N ( 15N 4N

The full answer of the sum of moments is 4N clockwise !" . One of the most important things to note is that the distance used to calculate a moment must be the perpendicular distance from the point to the force. If the length given is not perpendicular to the force, you need to know the angle between that length and the force to calculate the moment of the force.


The moment of the force is found by the expression;

0

)*+,-
cos!90° ( 0" 



Basically this means that we must resolve force so that it is perpendicular to the object and then multiply the resolved force (
cos!90° ( 0") by the distance ( ).



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M Example Three Calculate the moment about X in the following diagram, and state the sense of each moment. In this question we have to resolve the force sp that it is perpendicular to the point from which we measure the distance, our resolved force would look something like this; 8 cos 40° 50°

10m

8N

50°

10m

8N

4 Remember that when we resolve a force we find the cosine of the angle through which we will resolve and then multiply it by the force. In this case we are resolving through an angle of 40°. Therefore our resolved force is;

4

8 cos 40° 6.13N

Therefore the moment of the force is; 6.13 10 61.3N

From the diagram we can see that the rotation is in an anticlockwise !" direction and has a moment of force about X of 61.3Nm. Example Four Find the moment about the origin of a force of !4ii & 2jj"N acting at the point which has a position vector !3ii & 2jj"m. Our first step in this question is to do a quick diagram of the forces. We can use the diagram to find the perpendicular distances from the origin; From 3 to 4N force is 2m From 3 to 2N force is 3m We now need to decide whether the forces are clockwise or anticlockwise. The moment of the 4N force is clockwise, and the moment of the 2N force is anticlockwise. Therefore the total moment about 3 is: !4 2" ( !2 3" 8 ( 6 2N clockwise

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