A weight weighing 20 kg is attached to the left end of the weightless rod. The rod is placed on a support that is 0.4 times

| January 13, 2021 | education

A weight weighing 20 kg is attached to the left end of the weightless rod. The rod is placed on a support that is 0.4 times the length of the weight, which weight must be suspended from the right end so that the rod is in equilibrium.

ml = 20 kg.
g = 9.8 m / s2.
Ll = 0.4 * L.
mp -?
When the rod is in equilibrium, the moments of forces that act on the right Mп and left Mп sides of the rod are equal to each other: Mп = Ml.
The moment of force M is the product of the force F on the shoulder L, on which this force acts: M = F * L.
Fp * Lp = Fl * Ll.
Weights act on the rod with the force of its weight: Fп = mp * g, Fl = ml * g.
mp * g * Lp = ml * g * Ll.
Since L = Lp + Ll, then Lp = 0.6 * L.
mp * 0.6 * L = ml * 0.4 * L.
mp = 0.4 * ml / 0.6.
mp = 0.4 * 20 kg / 0.6 = 13.3 kg.
Answer: the right weight has a mass mp = 13.3 kg.



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