# A Level Maths- Moments

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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.

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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#2

(Original post by

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.

**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.

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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Thanks, so will there be a reaction force at C and D on both supports when you take moments separately?

(Original post by

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.

**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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#4

(Original post by

Thanks, so will there be a reaction force at C and D on both supports when you take moments separately?

**Bob7866**)Thanks, so will there be a reaction force at C and D on both supports when you take moments separately?

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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

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.

**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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#6

Can you show me what you've tried?

(Original post by

I'm still not getting the answer. Can you work it out please so i know where i am going wrong?

**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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#8

About right. In 2 it should be

12-x

That probably explains the negative mass

12-x

That probably explains the negative mass

(Original post by

**Bob7866**)
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oh yeahh thankss a lot for that

(Original post by

About right. In 2 it should be

12-x

That probably explains the negative mass

**mqb2766**)About right. In 2 it should be

12-x

That probably explains the negative mass

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#10

Apart from that slip, the moment expressions were correct.

(Original post by

oh yeahh thankss a lot for that

**Bob7866**)oh yeahh thankss a lot for that

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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

Apart from that slip, the moment expressions were correct.

**mqb2766**)Apart from that slip, the moment expressions were correct.

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#12

Neithet?

You've solved for x? Simply add 4.

You've solved for x? Simply add 4.

(Original post by

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?

**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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Naa, i've solved for m which is the mass now i need to find the distance from A

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Oh wait don't matter ive got it. Thanks for the help

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#15

Yeah but ..... x = 32/m

(Original post by

Naa, i've solved for m which is the mass now i need to find the distance from A

**Bob7866**)Naa, i've solved for m which is the mass now i need to find the distance from A

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#16

(Original post by

**Bob7866**)
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#17

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

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

**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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#18

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

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

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.

**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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#19

It is 12-x, that was the mistake they made.

(Original post by

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

**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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