M2 AQA Geometry for Moments

I'm struggling with a moments question.

A rectangle ABCD is free to rotate in a vertical plane about its centre O. Weights of W, 5W, 2W and 3W are attached to A,B,C and D respectively.

If the system is in equilibrium find the inclination of AC to the horizontal.

I'm attaching my workings. I think I need help with the geometry as a starter.

Thanks.

Would you like a scan of the question?

Yes please.

Here is a scan of the original.

Hmm, I thought that you might have left out essential information but you haven't.

Imagine BC is very much longer than AB. You would effectively have a weight of 6W at one end and a weight of 5W at the other end. How would that look?

Now imagine a new scenario...

Nope. It looks almost just the same to me. Am I not left with the same problem!

I give in - next clue please.

How about trying these problems instead?

1) A light square lamina ABCD is free to rotate in a vertical plane about its centre O. Weights W, 5W, 2W and 3W are attached etc..

2) A light rectangular lamina ABCD (with AB:BC = 2:1) is free to rotate in a vertical plane about its centre O. Weights W, 5W, 2W and 3W are attached etc..

How about trying these problems instead?

1) A light square lamina ABCD is free to rotate in a vertical plane about its centre O. Weights W, 5W, 2W and 3W are attached etc..

2) A light rectangular lamina ABCD (with AB:BC = 2:1) is free to rotate in a vertical plane about its centre O. Weights W, 5W, 2W and 3W are attached etc..

OK

After goodness knows how many attempts at the geometry and subsequent moments I've now arrived at the correct number but the wrong sign.

HELP.

Work attached.

Good morning

Your diagram is rather complicated. I'm attaching another. See how it goes working from this one.

Good morning to you too.

Oh no! another day of question 9. Ah well, your diagram is a good start.

Good morning

Your diagram is rather complicated. I'm attaching another. See how it goes working from this one.

So am I still supposed to be finding the angle between the weight and the perpendicular to the diagonal?

I've tried it twice (I thought I was OK at geometry) and have still not got the correct answer.


Perhaps I should be doing a different way. But if not here are my workings

@A -wcosa
@B -5Wcos(90-a)
@C 2wcos(180-a)
@D 3wcos(90-a)

Could my original -ve value be anything to do with the fact that my diagonal is in a different position to yours and as you have your diagram the way round you have it would that "common sense" be needed?

I would put B at the lowest point but it won't change the answer if you don't.

I would write the equation as $\text{sum of clockwise moments = sum of anti-clockwise moments.}$

Then, checking the diagram, we can write $\cos a + 5 \sin a = 2 \cos a + 3 \sin a$.

I would put B at the lowest point but it won't change the answer if you don't.

I would write the equation as $\text{sum of clockwise moments = sum of anti-clockwise moments.}$

Then, checking the diagram, we can write $\cos a + 5 \sin a = 2 \cos a + 3 \sin a$.