Answer to Question #21963 in Mechanics | Relativity for rahav
F1 = M·(g+a), (1)
where M is the mass of a man. From the other side, the force acting on a man when the lift goes down with the same acceleration is
F2 = M·(g-a). (2)
Let's add equations (1) and (2):
F1 + F2 = M·(g+a) + M·(g-a),
F1 - F2 = M·g + M·a + M·g - M·a,
F1 + F2 = 2·M·g.
M = (F1 + F2) / (2·g)
Taking into account that
F1 = 72[kg-wt] ≈ 72·g[N]
F2 = 69[kg-wt] ≈ 69·g[N],
M = (72·g[N] + 69·g[N]) / (2·g[m/s²]) = (72[N] + 69[N]) / (2[m/s²]) = 70.5[kg].
So, the mass of a man is 70.5 kg.
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