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The torques acting on the crane boom can be computed about anaxis passing through the top of the boom. (this is the place wherethe spring scale and mass hanger connect to the boom) apply thesecond condition for equilibrium to the crane boom and derivethe resulting equation for the sum of torques about this newpoint

  • physics - ,

    What equation?

  • physics - ,

    exactly. This is all I was given. Possibly

    SumOf t = T(exp)*length* sin(theta) - W(hang) * length * sin(phi) - W(boom) * L(exp) * sin(phi) = 0

    but im not sure.

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