Moments

The moment of a force (or torque) is defined as the magnitude of the applied force, times the perpendicular distance of that force to the point of action.

The thing about that though is that it's not very helpful. You can get confused. Even the following:

M=Fs sin(theta)

Might not be too helpful (although if you do use the formula correctly, you'll be fine). Why is this? It's this rather loose talk of angles, when we don't have a diagram in front of us. So without further ado, here's the diagram:

I've alradey drawn on a few things here, so I'll quickly explain them. Firstly, the downwards arrow is a force that's being aplied to the thick black line and we're calculating a moment about the centre of the circle. I've extended the thick line and resolved the downwards force. This ought to look eerily familiar to anyone who has done M! dynamics- it looks just like an inclined plane problem!

Having seen this, it's quite obvious that the force times the cosine of 40 is our 'perpendicular' component, and we just need to multiply this by 0.25m to get the moment.

Just to further prove the fact that this angle is 40 degrees, here is another diagram:

I've drawn in all the right angles. Now, because the angles in a triangle add up to 180, and the angles on a line also add up to 180, we can fill in all remaining angles and show that the two angles marked 40 must be equal.

How else could we calculate the moment? Well, our origional definition mentioned "perpendicular distance" which we have only used implicitly in the previous calculation. Another approach is to use our origional value of F, but use the perpendicular distance- which is the horizontal dotted line in the above diagram. Conveniently, this is the adjacent side in a right angled triangle and a little trig tells us that

s_perpendicular=0.25 cos40

We can then state that

M=Fs cos40

That's both approaches to the issue at hand covered, you can pick your favourite of course. There are two more things to mention:

!) The last formula I gave has a cos in it instead of a sin. At the start I gave M=Fs sin(theta). That is because the general formula refers to the angle between the force and the displacement (in this case it's 50 degrees). We've been given (theta-90) in the diagram, and cos(x-90)=sin(x).

2) The moment is maximised when sin(theta)=1. Again, the angle given in the diagram is not the most helpful, but this means that the maximum moment is achieved when it is equal to 0, and the minimum when it is 90. You can see this from the shape of a cosine graph and the fact that we worked out the moment to be M=Fscos(x) where x is the size of the marked on angle.

There aren't too many difficult moments questions at A level, this should be about as hard as they are likely to be. I might have a look out for a more difficult example later.