Bob7866
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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.
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mqb2766
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(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.
Last edited by mqb2766; 1 year ago
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Bob7866
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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.
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mqb2766
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(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.
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Bob7866
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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.
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mqb2766
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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?
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Bob7866
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(Original post by mqb2766)
Can you show me what you've tried?
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mqb2766
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About right. In 2 it should be
12-x
That probably explains the negative mass
(Original post by Bob7866)
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Bob7866
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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
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mqb2766
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Apart from that slip, the moment expressions were correct.
(Original post by Bob7866)
oh yeahh thankss a lot for that
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Bob7866
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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.
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mqb2766
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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?
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Bob7866
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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.
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Bob7866
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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.
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mqb2766
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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
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Mermaidqueen
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(Original post by Bob7866)
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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
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mqb2766
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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
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Mermaidqueen
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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.
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mqb2766
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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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