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What force would the scale exert on a 53 kg person in a elevator and its moving up at a constant speed?
Wiki User
∙ 10y ago
Scale already reads 70 kg when elevator is at rest. This is Gravity Force down (Fg), plus Normal force (n) up. Now, force on the man in the upward direction from accelerating up (force E) = mass of the man x acc. of the lift in the upward direction = 70 x5.5 N = 385 N. Now force of gravity (Fg) = 70 x 9.8 N = 686 N. Plus the normal force (n) -which keeps him falling through floor of elevator - is also acting on him = 686 N (same as gravity in up direction).
So, F(y) = Elevator (Up) 385 (up) + Normal (up) 686 N - Gravity (down) 686 N. Therefore net force in the upward direction (against bottom of scale-floor) = (+385 + 686 - 686) N = 385 N. Hence the reading of the scale has extra = (385N). Scale is calibrated for "g = 9.8 m/ss". So, (385 / 9.8m/ss) kg = 39.3 kg extra. So, 39.3 kg (extra) + 70 kg (original) = 109.3 kg. This is why you feel heavier when elevator goes up.
Wiki User
∙ 9y ago
Wiki User
∙ 10y agoIf the elevator is moving at constant speed, up or down, then the force is
the same as it would be if the elevator were not moving at all. The force
on the bottom of a 53-kg mass is about 519.77 Newtons (116.84 pounds).
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