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A 3.0-m rod is pivoted about its left end. A force of 6.0 N is applied perpendicular to the rod at a distance of 1.2m from the pivot causing a CCW torque, and a force of 5.2N is applied at the end of the rod 3.0m from the pivot. the 5.2N is at an angle of 30 degrees to the rod and causes a CW torque. What is the net torque about the pivot?

  • Physics -

    (6.0*1.2) - (5.2*3.0)*sin 30
    = 7.2 - 7.8 = -0.6 Newton meters

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