Apparent Weight Is the Normal Force: m(g + a) Up, m(g − a) Down
In a lift, on a rope or inside a cage, what a scale reads is the contact force, not mg. That force changes with the acceleration of the support, so the stages of a motion graph translate straight into readings.
Motion and force questions become simpler when direction, starting condition, time and the acting forces are marked first.
Apparent weight
A 0.5 kg block resting on another block in a lift descending at 2 m/s² presses down with 0.5 x (10 - 2) = 4 N. A lift of mass 100 kg whose rope tension follows its speed graph has T = m(g + a): during acceleration upward T1 = m(g + a1), at constant speed T2 = mg, and while slowing T3 = m(g - a3). If the acceleration is g/5 upward, the speed is constant for a while, and then the deceleration is 0.4 g, the tensions are in the ratio 6 : 5 : 3.
A rope that can only take so much
A fireman sliding down a rope whose breaking strength is 3/4 of his weight needs the tension to stay at or below 3mg/4.
He must slide at least that fast; any slower and the rope carries more than it can bear.
Systems with something moving inside
When a parrot in a sealed cage accelerates upward, it pushes down on the cage and air with a force greater than its weight, so a boy holding the cage feels it heavier. The weight felt depends on the total external force needed to accelerate the centre of mass of the cage and its contents, not on whether the parrot touches the cage.
OlympIQ exam lens
- ✓Apparent weight = m(g + a) up-accelerating, m(g - a) down-accelerating, 0 in free fall
- ✓Rope tension in a lift follows T = m(g + a); read a from the slope of the v-t graph
- ✓Maximum slide with a weak rope: a ≥ g - (T_max / m)
- ✓A sealed cage with an upward-accelerating parrot weighs more to the holder
Do not decide whether the weight rises by the direction of motion. Apparent weight depends on the direction of the acceleration; a lift moving down but slowing down has an upward acceleration, so the reading is above mg.
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