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A Level Maths- Moments

Can anybody do this question i'm confused:
A non uniform rod AB of length 20 cm rests horizontally on two supports at C and D, where AC=BD=4cm. The greatest mass that can be hung from A without disturbing equilibrium is 8g, and the greatest mass that can be hung from B is 10g (when no mass is hung from A) is 10g. Find the mass of the rod and the distance of its centre of mass from A.

The answer is a mass of 6g and 9.3cm. But not sure how to get it.
Reply 1
Original post by Bob7866
Can anybody do this question i'm confused:
A non uniform rod AB of length 20 cm rests horizontally on two supports at C and D, where AC=BD=4cm. The greatest mass that can be hung from A without disturbing equilibrium is 8g, and the greatest mass that can be hung from B is 10g (when no mass is hung from A) is 10g. Find the mass of the rod and the distance of its centre of mass from A.

The answer is a mass of 6g and 9.3cm. But not sure how to get it.


Take moments about each support separately.
About C, the rod will rotate when the mass at A means the reaction at D ibecomes zero. That means the COM moment equals the mass at A moment.
Similarly about D.
Two equations in two unknowns, so solve.
(edited 5 years ago)
Reply 2
Thanks, so will there be a reaction force at C and D on both supports when you take moments separately?
Original post by mqb2766
Take moments about each support separately.
About C, the rod will rotate when the mass at A means the reaction at D ibecomes zero. That means the COM moment equals the mass at A moment.
Similarly about D.
Two equations in two unknowns, so solve.
Reply 3
Original post by Bob7866
Thanks, so will there be a reaction force at C and D on both supports when you take moments separately?

With no weights at either end, the COM will lie between C and D. When you put a heavy weight at A say, the rod will start to rotate about C and the reaction at D will disappear (the rod will move off this support due to the rotation). The critical mass is when the reaction at D ibecomes zero.
Reply 4
I'm still not getting the answer. Can you work it out please so i know where i am going wrong?
Original post by mqb2766
With no weights at either end, the COM will lie between C and D. When you put a heavy weight at A say, the rod will start to rotate about C and the reaction at D will disappear (the rod will move off this support due to the rotation). The critical mass is when the reaction at D ibecomes zero.
Reply 5
Can you show me what you've tried?
Original post by Bob7866
I'm still not getting the answer. Can you work it out please so i know where i am going wrong?
Reply 6
20190326_210457.jpg
Original post by mqb2766
Can you show me what you've tried?
Reply 7
About right. In 2 it should be
12-x
That probably explains the negative mass
Original post by Bob7866
20190326_210457.jpg
Reply 8
oh yeahh thankss a lot for that
Original post by mqb2766
About right. In 2 it should be
12-x
That probably explains the negative mass
Reply 9
Apart from that slip, the moment expressions were correct.
Original post by Bob7866
oh yeahh thankss a lot for that
Reply 10
Thanks for that One quick question, to find the distance from point A do i take moments from the first diagram or the 2nd diagram?
Original post by mqb2766
Apart from that slip, the moment expressions were correct.
Neithet?
You've solved for x? Simply add 4.
Original post by Bob7866
Thanks for that One quick question, to find the distance from point A do i take moments from the first diagram or the 2nd diagram?
Reply 12
Naa, i've solved for m which is the mass now i need to find the distance from A
Original post by mqb2766
Neithet?
You've solved for x? Simply add 4.
Reply 13
Oh wait don't matter ive got it. Thanks for the help
Original post by mqb2766
Neithet?
You've solved for x? Simply add 4.
Yeah but ..... x = 32/m


Original post by Bob7866
Naa, i've solved for m which is the mass now i need to find the distance from A
Original post by Bob7866
20190326_210457.jpg


Hi sorry, Im stuck on the same question and looking at your working out I don't understand why you have ignored the reaction forces that should come out from each support
When you take a moment about C, the added mass at the end balances the COM of the bar. Hence, in equilibrium, the reaction at D must be zero. If the bar was not in equilibrium, either there would be a positive reaction at D and the support would do its job, or if the reaction was negative .the bar would start to rotate.
Original post by Mermaidqueen
Hi sorry, Im stuck on the same question and looking at your working out I don't understand why you have ignored the reaction forces that should come out from each support
Oh okay, now looking at both of your diagrams I get it but the only thing is.. on the second diagram the distance of the COM to D you wrote x-12, can you please explain why its not 12-x as you would subtract 2 x 4 from 20 to be left with 12cm in total between C and D and x must be some distance from D hence 12-x...

Im really sorry, I appreciate you answering, I have just learnt moments yesterday and am trying to wrap my head around it, thank you
Original post by mqb2766
When you take a moment about C, the added mass at the end balances the COM of the bar. Hence, in equilibrium, the reaction at D must be zero. If the bar was not in equilibrium, either there would be a positive reaction at D and the support would do its job, or if the reaction was negative .the bar would start to rotate.
It is 12-x, that was the mistake they made.
Original post by Mermaidqueen
Oh okay, now looking at both of your diagrams I get it but the only thing is.. on the second diagram the distance of the COM to D you wrote x-12, can you please explain why its not 12-x as you would subtract 2 x 4 from 20 to be left with 12cm in total between C and D and x must be some distance from D hence 12-x...

Im really sorry, I appreciate you answering, I have just learnt moments yesterday and am trying to wrap my head around it, thank you

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