# Answer to Question #21963 in Mechanics | Relativity for rahav

Question #21963

A man weighs himself inside a lift. He weighs 72kg-wt when the lift goes up with a constant acceleration ‘a’ , while he weighs 69kg-wt when the lift goes down with the same acceleration. The mass of the man is?

Expert's answer

According to the second Newton's law, the force acting on a man when the lift goes up with a constant acceleration is

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.

Therefore,

M = (F1 + F2) / (2·g)

Taking into account that

F1 = 72[kg-wt] ≈ 72·g[N]

and

F2 = 69[kg-wt] ≈ 69·g[N],

we have

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.

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.

Therefore,

M = (F1 + F2) / (2·g)

Taking into account that

F1 = 72[kg-wt] ≈ 72·g[N]

and

F2 = 69[kg-wt] ≈ 69·g[N],

we have

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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