AQA A2 Mathematics MM2B Mechanics 2 - Monday 22nd June 2015 [Exam Discussion Thread]

Taking a stab in the dark here. I believe it was 8kg? Uniform 6m?


Posted from TSR Mobile

Haha okayy, unfortunately I don't think you can take moments at A, because there will then be no clockwise moments, and you need to equate the clockwise moments about a point to the anti-clockwise moments about that point. Please correct me if I'm misunderstanding your method

It shouldn't matter which point you take moments about - the distance AC calculated will be the same. Even if they're all clockwise/anti-clockwise etc.


Posted from TSR Mobile

yes but when moved to A the overall system is a couple,(causes it to rotate) you could use this, but then you'd need the angular acceleration etc

The system is still in equilibrium though, regardless of where you take moments about.

All you're doing is choosing a different pivot. In equilibrium the resultant moment about any point is zero so it doesn't matter where you pivot around.

Posted from TSR Mobile

For equilibrium, clockwise moments=anti-clockwise moments and resultant force=0 . At A, there are no clockwise moments. The system is in equilibrium but there are different ways of representing it. You can have it as it is usually or have a single force and a couple, for which the moments cancel each other out to keep it in equilibrium. But this doesn't help you solve this question as you need more information. I'm sorry I can't explain it very well as I did this in M4 and as you can tell I'm not very good at it 😅

Equilibrium is completely irrelevant in that question. The rod could be hanging somewhere, it could be lying on the floor; no matter what it's doing, the centre of mass would always be the same, and that's what the question was asking. So no, if the rod were to be just in mid-air, it won't be in equilibrium.

EDIT: In short, it wasn't a moments question, it was a centre of mass question (x bar)

Hmm. Idk - I didn't equate anti-clockwise and clockwise moments. I just said that the turning effect of the centre of mass around A was the same as the sum of all masses multiplied by their distance from A. :/


Posted from TSR Mobile

I'm not sure really, for what you've done you're using the centre of mass of the whole system, which I don't know if we were given, as there were other masses attached to the rod. It may have been that and in that case you should be right

I'm actually not sure what I did was right, it would be an x bar question if we were given the centre of mass of the whole system, but I think we were only given the centre of mass of the rod, which then had other masses attached to it; in which I think you have to use moments

We were given the centre of mass of the system.

The rod we were told is uniform, and the centre of mass (of the system) acted 4.3m from A.

Oh dear 😅😓

Thoughts?




Posted from TSR Mobile

High, as that exam was really difficult, but not to be surprised as a lot of further Maths students took it.

So happy!

Nice one, I got 595