A 70 kg person in an elevator with acceleration 2 m/s²: find the weight when accelerating up and down — Physics problems | PhysLabo
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Weight of a person in an accelerating elevator

Medium 9 grade Mechanics / Newton's Laws
Problem statement

A person of mass $m = 70$ kg stands on the floor of an elevator cabin. The elevator moves with an acceleration $a = 2$ m/s² directed: 1) upward; 2) downward (see the figure). Find the weight of the person in each case. Take $g = 10$ m/s².

Two elevator cabins with a person: in the first the acceleration a points up, in the second down
Given
  • $m = 70$ kg
  • $a = 2$ m/s²
  • $g = 10$ m/s²
  • $P_1 = ?$, $P_2 = ?$
Hints
Hint 1 / 2
The weight of a body is the force with which it presses on the support; by Newton's third law it equals the normal force $N$.
Hint 2 / 2
Take the axis along the acceleration. Acceleration upward: $N - mg = ma$; downward: $mg - N = ma$.
Solution

The weight $P$ is the force with which the person presses on the floor of the cabin. By Newton's third law it equals the normal force $N$ in magnitude, so it is enough to find $N$. The forces on the person are gravity $m\vec g$ (downward) and the normal force $\vec N$ (upward). Take the $y$ axis along the acceleration.

Forces on the person: the y axis points along the acceleration
Forces on the person: the y axis points along the acceleration

1. The acceleration points up:

$$N - mg = ma \;\Rightarrow\; P = N = m(g + a) = 70 \cdot (10 + 2) = 840 \text{ N}$$

The weight is larger than gravity ($mg = 700$ N) — an overload.

2. The acceleration points down:

$$mg - N = ma \;\Rightarrow\; P = N = m(g - a) = 70 \cdot (10 - 2) = 560 \text{ N}$$

The weight is smaller than gravity. At $a = g$ the weight would drop to zero — weightlessness.

Answer

1) $P = 840$ N; 2) $P = 560$ N

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