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  • IMAT 2013 Q54 Physics and Mathematics

  • Nicolas Sanchez

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    June 21, 2021 at 9:44 am
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    A uniform beam, 3.0m long, of weight 100N has a 300N weight placed 0.50m from one end. The beam is suspended by a string 1.0m from the same end.
    A diagram of the weights placed on the beam is given below:

    [See diagram below]

    How far from the other end must a weight of 80N be placed for the beam to be balanced?

    A. 0.75m

    B. 2.25m

    C. 1.25m

    D. 1.875m

    E. 0.125m

  • MANISH MULCHANDANI

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    July 15, 2021 at 10:51 pm
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    Answer= A

    You have to use concept of center of mass. You can also solve it by considering movement of beam like clockwise and anti clockwise.

    Let us consider x as distance for 80N from other end to be balanced.

    Concept of center of mass stats that-

    m1x1 + m2x2= m3x3

    Let m1= 100N and x1= 0.5m.

    The 100N weight will be balanced by string with length (1-0.5)m.

    Let m2= 80 N and x2= (2-x)

    The 80N weight will be balanced by (2-x)m because its placed at that end.

    The 300N weight will be balanced by string with length (1-0.5)m because its placed on that end.

    So, Let m3= 300N and x3= 0.5m

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