A 75kg person stands on a weighing scale in an elevator. 3 seconds after the motion starts from rest, the tension in the hoisting cable was found to be 8300N. Find the reading of the scale, in kg during this interval. Also find the velocity of the elevator at the end of this interval. The total mass of the elevator, including mass of the person and the weighing scale, is 750kg. If the elevator is now moving in the opposite direction, with same magnitude of acceleration, what will be the new reading of the scale?

#### Solution

t = 3 sec

u = 0 m/s a

T = 8300 N a

ΣF = ma

T – W = 750 x a

8300 – 7359 = 750 x a

a = 1.255 m/s^{2}

v = u + at

v = 0 + (1.255 x 3)

v = 3.765 m/s

For upward motion,

N_{1} – mg = ma

N_{1} = ma + mg

N_{1} = m(a+g)

N_{1} = 75(1.255+9.812)

N_{1} = 830.025 N

N_{1} = 84.59 kg

For downward motion, mg

N_{2} – mg = - ma

N_{2} = mg - ma

N_{2} = m(g-a) a

N_{2} = 75(9.812 – 1.255)

N_{2} = 641.775 N

N_{2} = 65.407 kg

**In upward motion the reading on the weighing scale is 84.59 kg, final velocity ****at the end = 3.765 m/s and the reading on the weighing scale is 65.407 kg in ****the downward direction.**