Question

An elevator, in a building, starts from rest; accelerates uniformly upwards. It takes a time t₁ to go up to second floor, situated at a height H, above the ground. The corresponding time, for its downward journey, is t₂ .

If a person were to hold a mass M with the help of a massless string, the ratio of the respective tensions in the string, in the two cases, would be

(1) \(\frac{(gt^2_1+2H)}{(g^2_2-2H)}\frac{(t^2_2)}{(t^2_1)}\)

(2) \(\frac{(gt^2_1-2H)}{(g^2_2+2H)}\frac{(t^2_2)}{(t^2_1)}\)

(3) \(\frac{(2gt^2_1+2H)}{(2g^2_2-2H)}\frac{(t^2_2)}{(t^2_1)}\)

(4) \(\frac{(2gt^2_1-2H)}{(2g^2_2+2H)}\frac{(t^2_2)}{(t^2_1)}\)

Answer 1

 (1) \(\frac{(gt^2_1+2H)}{(g^2_2-2H)}\frac{(t^2_2)}{(t^2_1)}\)

For the upward journey; if a₁ is the (uniform) acceleration of the elevator, we have

The apparent weight, of the mass M, during the upward journey, would be

For the downward journey, if a₂ is the (uniform) acceleration, of the elevator, we have

The apparent weight, of the mass M, during the downward journey, would be

The tensions, in the string, in the two cases, equal these apparent weights. Hence, the ratio of the tensions, in the two cases, is