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The weightless horizontal bar in the figure below is in equilibrium. Scale B reads 4.00 kg (N.B. as you know, the scale should read N, but no one told the manufacturer). The distances in the figure (which is not to scale) are: D1 = 9.0 cm, D2 = 8.0 cm, and D3 = 6.5 cm. The mass of block X is 0.96 kg and the mass of block Y is 1.93 kg.

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2 answers

  1. without the figure, I have no idea what the distances mean.

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  2. I believe that I know this figure.
    If it is so, then...
    Let's examine the torques. The torque on A must equal the torque on B
    since the system is at rest. The torque on B can be found by summing the
    torques from mass x and mass y. Since mass z is on the axis of rotation (at B),it produces no torque.
    M(B) = (D2 + D3) •m(x) + D3•m(y),
    The torque on A can be calculated by referencing it from B. In other words,
    what torque must be applied at A to counteract the torques from m(x) and m(y)? This can by found by denoting the variable m(A) to represent the mass (the scale reading) at A to counteract the torque calculated above. That is,

    m(A)•(D1+D2+D3) = (D2 + D3) •m(x) + D3•m(y),

    Thus, the scale reading at A, that is, m(A)

    m(A) = [(D2 + D3) •m(x) + D3•m(y)]/ (D1+D2+D3)

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