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

    MS: https://8a40d6c38bafca75cc407741c0f3...%20Edexcel.pdf
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    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:

    Name:  Untitled.jpg
Views: 45
Size:  34.1 KB

    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.
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    (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:

    Name:  Untitled.jpg
Views: 45
Size:  34.1 KB

    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
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    (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

    Name:  Untitled.jpg
Views: 44
Size:  33.6 KB
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    (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

    Name:  Untitled.jpg
Views: 44
Size:  33.6 KB
    It is much clearer, thank you so much
 
 
 
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