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    I'm trying to do question 2.

    Here is my workings. I get x = 2a/3 but the book tells me it's a/3
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    (Original post by maggiehodgson)
    I'm trying to do question 2.

    Here is my workings. I get x = 2a/3 but the book tells me it's a/3
    When the weight W is attached the total downward force is then 2W.

    You should end up with the equation 2W=\dfrac{3T}{2}.
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    (Original post by maggiehodgson)
    I'm trying to do question 2.

    Here is my workings. I get x = 2a/3 but the book tells me it's a/3
    For the second part you seem to have neglected to include the weight of the rod.

    Resolving vertically, W+W=T+T\cos 60

    etc.

    Edit: Too slow.
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    (Original post by ghostwalker)
    For the second part you seem to have neglected to include the weight of the rod.

    Resolving vertically, W+W=T+T\cos 60

    etc.

    Edit: Too slow.
    I thought, "Since ghostwalker isn't around, I'll do."
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    (Original post by BabyMaths)
    When the weight W is attached the total downward force is then 2W.

    You should end up with the equation 2W=\dfrac{3T}{2}.

    OMG If all else fails, read the question properly!!!
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    (Original post by ghostwalker)
    For the second part you seem to have neglected to include the weight of the rod.

    Resolving vertically, W+W=T+T\cos 60

    etc.

    Edit: Too slow.

    Yep, I neglected to read the question properly.

    Thanks
 
 
 
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