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# M3 IAL Jan2015 Q5(b)

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1. I am so stuck with this question, I couldnt understand even I read the Mark Scheme. Could anyone help me out please? Many thanks.
2. (Original post by Kamisama)
I am so stuck with this question, I couldnt understand even I read the Mark Scheme. Could anyone help me out please? Many thanks.
Can you post/link the question and markscheme?
3. (Original post by ghostwalker)
Can you post/link the question and markscheme?
Sure
QP: https://8a40d6c38bafca75cc407741c0f3...%20Edexcel.pdf

MS: https://8a40d6c38bafca75cc407741c0f3...%20Edexcel.pdf
4. So, S is suspended from a point on its circumference, and the line OA makes an angle of 12 degrees with the vertical.

Here's a diagram:

Green line is supposed to be vertical. Point on circumference is suppoed to be A, not S - my bad.

So, Tan 12 = "distance of CofM from O" / r

And using the result from part a, the rest follows.
5. (Original post by ghostwalker)
So, S is suspended from a point on its circumference, and the line OA makes an angle of 12 degrees with the vertical.

Here's a diagram:

Green line is supposed to be vertical. Point on circumference is suppoed to be A, not S - my bad.

So, Tan 12 = "distance of CofM from O" / r

And using the result from part a, the rest follows.
Not 5c, is 5b
Thank you
6. (Original post by Kamisama)
Not 5c, is 5b
Thank you
Sorry - I must be going mad.

OK. S is on the ground with it's shorter side resting on the ground.

It will remain that way as long as the CofM is vertically above the line V to the circumference. As k gets larger the CofM moves to the right, until the shape tips about the circumference. When k is the greatest possible value and we still have eqiulibrium, the Cof M will be over the circumference.

In diagram the two angles c are equal.

Then tan c = r/4r = x bar / r

7. (Original post by ghostwalker)
Sorry - I must be going mad.

OK. S is on the ground with it's shorter side resting on the ground.

It will remain that way as long as the CofM is vertically above the line V to the circumference. As k gets larger the CofM moves to the right, until the shape tips about the circumference. When k is the greatest possible value and we still have eqiulibrium, the Cof M will be over the circumference.

In diagram the two angles c are equal.

Then tan c = r/4r = x bar / r

It is much clearer, thank you so much

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