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    Seem to be struggling with this question, could anyone help out?
    The centre of mass of a non-uniform beam AB, of weight W and length 4m, is at the point G. The beam rests horizontally on two supports X and Y, where AX = 1m and YB = 2m.
    A weight W/4 is suspended from the beam at B. Given that the magnitude of the force exerted on the beam by support Y is three times the magnitude of the force exerted on the beam by support X, calculate the distance AG.

    Thanks
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    (Original post by Rashina)
    Seem to be struggling with this question, could anyone help out?
    The centre of mass of a non-uniform beam AB, of weight W and length 4m, is at the point G. The beam rests horizontally on two supports X and Y, where AX = 1m and YB = 2m.
    A weight W/4 is suspended from the beam at B. Given that the magnitude of the force exerted on the beam by support Y is three times the magnitude of the force exerted on the beam by support X, calculate the distance AG.

    Thanks
    What have you tried? Post your working.
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    Where did you get this question from?
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    (Original post by Keira Larkin)
    Where did you get this question from?
    Why are you asking?
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    (Original post by Rashina)
    Why are you asking?
    Oh I just thought I might be able to find a worked solution if it is from the textbook.
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    (Original post by notnek)
    What have you tried? Post your working.
    I don't even have an idea of how to do it. Explain me how to do it and I will figure it out.
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    (Original post by Keira Larkin)
    Oh I just thought I might be able to find a worked solution if it is from the textbook.
    Its from a revision guide which does not have any solution banks or worked solutions.
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    I got these equations so far:

    W + W/4 = x + 3x
    2W/4 + x = W*z
    W(1-z) + 3W/4 = 3x

    z is the distance between support y and the centre of mass
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    (Original post by Rashina)
    Its from a revision guide which does not have any solution banks or worked solutions.
    Oh okay.
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    (Original post by Keira Larkin)
    Oh okay.
    Anyway thanks for trying to help me
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    (Original post by Rashina)
    Anyway thanks for trying to help me
    Well, now I want to know how to solve this as well. There are too many variables in my equations. Have you made any progress?
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    (Original post by Keira Larkin)
    Well, now I want to know how to solve this as well. There are too many variables in my equations. Have you made any progress?
    just calculated the resultant force upwards:
    R+3R=(W+W/4)g
    Wg= 16x/5
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    (Original post by Rashina)
    just calculated the resultant force upwards:
    R+3R=(W+W/4)g
    Wg= 16x/5
    Is the answer 1.1875m?

    WORKING:

    So I don't think you multiply W by g because W is weight, which is a force. Therefore, W= 16x/5
    Take moments about A to give the equation:
    x + (2 * 3x) = Wz + W
    7x = W * (z + 1)
    7x = 16x/5 (z + 1)
    7x * 5/16x = (z + 1)
    19/16 = z
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    (Original post by Keira Larkin)
    Is the answer 1.1875m?
    Yes! How did you find it?
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    (Original post by Rashina)
    Yes! How did you find it?
    Does your textbook say the answer is 1.1875?
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    (Original post by Keira Larkin)
    Does your textbook say the answer is 1.1875?
    The book only has final answers so I found this answer on the book. How did you find it?
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    Basically, the trick was not to take moments about support x or y. I took the moments about point A. Obviously not B because there is a force acting on point B. Refer to the second worked example in the M1 textbook in the section about non uniform planks.

    When I do tricky moment questions, I just try taking moments about a point and if that doesn't work, I try another point.
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    (Original post by Keira Larkin)
    Is the answer 1.1875m?

    WORKING:

    So I don't think you multiply W by g because W is weight, which is a force. Therefore, W= 16x/5
    Take moments about A to give the equation:
    x + (2 * 3x) = Wz + W
    7x = W * (z + 1)
    7x = 16x/5 (z + 1)
    7x * 5/16x = (z + 1)
    19/16 = z
    I wrote the working.
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    (Original post by Keira Larkin)
    I wrote the working.
    Thanks a lot for helping me.
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    (Original post by Rashina)
    Thanks a lot for helping me.
    Welcome
 
 
 
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