Question 5.13:
A man of mass 70 kg stands on a weighing scale in a lift which is moving
(a) upwards with a uniform speed of ,
(b) downwards with a uniform acceleration of ,
(c) upwards with a uniform acceleration of 5 ms-2.What would be the readings on the scale in each case?
(d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?
Given,
m = 70 kg, g = 10 ms-2
( a ) When the elevator moves up at a uniform speed, its acceleration is 0.
According to Newton’s second law of motion, the equation can be written as:
R – mg = ma
R = mg = 70 x 10 = 700N
Thus, the reading = 700/g = 70 kg
( b ) Lift goes down at, a = 5 ms-2
According to Newton’s second law of motion, the equation can be written as:
R +ma = mg
R = m (g –a) = 70 (10 – 5)
= 350 N
Thus, the reading = 350 / 10 = 35 kg
( c ) Lift goes up at, a = 5 ms -2
According to Newton’s second law of motion, the equation can be written as:
R – mg = ma
R = m (g + a) = 70 (10 + 5)
= 1050 N
Thus, the reading = 1050/10 = 105kg
( d ) When the lift crashes down freely, a = g
According to Newton’s second law of motion, the equation can be written as:
R + mg = ma
R = m (g – g) = 0. Thus, under a free-fall condition the lady will be in a state of weightlessness.
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