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A man of mass 70 kg stands on a weighing scale in a lift which is movinga) upwards with a uniform speed of 10 m s–1 , b) downwards with a uniform acceleration of 5 m s–2 , c) upwards with a uniform acceleration of 5 m s–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? |
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Answer» Mass of the man, m = 70 kg Acceleration, a = 0 Using Newton’s second law of motion, we can write the equation of motion as: R – mg = ma Where, ma is the net force acting on the man. As the lift is moving at a uniform speed, acceleration a = 0 R = mg = 70 × 10 = 700 N Reading on the weighing scale = 700/g = 700/10 = 70 kg Mass of the man, m = 70 kg Acceleration, a = 5 m/s2 downward Using Newton’s second law of motion, we can write the equation of motion as: R + mg = ma R = m(g – a) = 70 (10 – 5) = 70 × 5 = 350 N Reading on the weighing scale = 350/g = 350/10 = 35 kg Mass of the man, m = 70 kg Acceleration, a = 5 m/s2 upward Using Newton’s second law of motion, we can write the equation of motion as: R – mg = ma R = m(g + a) = 70 (10 + 5) = 70 × 15 = 1050 N Reading on the weighing scale = 1050/g = 1050/10 = 105 kg When the lift moves freely under gravity, acceleration a = g Using Newton’s second law of motion, we can write the equation of motion as: R + mg = ma R = m(g – a) = m(g – g) = 0 Reading on the weighing scale = 0/g = 0 kg The man will be in a state of weightlessness. |
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