You stand on a bathroom scale inside a lift. At rest the scale reads 80 kg. While the lift accelerates upward the reading is 92 kg. Taking g = 9.8 m/s^2, what is the magnitude of the acceleration?

You stand on a bathroom scale inside a lift. At rest the scale reads 80 kg. While the lift accelerates upward the reading is 92 kg. Taking g = 9.8 m/s^2, what is the magnitude of the acceleration?

Approach: The scale reports the contact force it exerts, divided by g. Write the vertical force balance on the passenger and solve for the acceleration.

1.47 m/s^2. A scale displays the normal force N divided by g. Newton's second law in the vertical direction gives N - mg = ma, so N = m(g + a) and the apparent weight in kilograms is m(1 + a/g). Setting 92 = 80 * (1 + a/9.8) gives a/9.8 = 12/80 = 0.15, so a = 1.47 m/s^2. The reading depends on acceleration alone, so a lift travelling upward at a constant 10 m/s still shows 80 kg. In free fall a = -g and the display reads 0.

Follow-up: What does the scale read while the same lift decelerates at 2 m/s^2 on the way up?

Key concepts: normal force, newtons second law, apparent weight.