in the experimental setup shown in the figure, a beam, B1, of mass M1=0.6887kg a
ID: 2114283 • Letter: I
Question
in the experimental setup shown in the figure, a beam, B1, of mass M1=0.6887kg and lenght L1=1m is pivoted about its lowest point at P1. A second beam, B2, of mass M2=0.200kg and length L2=0.200m is suspended (pivoted) from B1 at a point P2, which is a horizontal distanced d=0.550m from P1. To keep the system at equilibrium, a mass m=0.500 kg has to be suspended from a massless string that runs horizontaly from P3, at the top of beam B1, and passes over a frictionless pulley. the string runs at a vertical distance y=0.707 m above the pivot point P1. Determine the net tourqe at P1, P2, and P3
Explanation / Answer
As the system is in equillibrium .. so the net torque through every point in the system is zero
so torque trhorugh P1 = P2 = P3 = 0
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