Scale in vertical motion

Problem

A person of $70\ \text{kg}$ steps onto a scale inside a lift: the reading shows $80\ \text{kg-force}$. What is the acceleration of the cabin and in which direction is it directed?

Solution

The scale does not measure mass directly, but the force with which the person presses on the platform, i.e. the normal reaction $\vv{N}$. The reading of $80\ \text{kg-force}$ corresponds to a force: $N = 80 \cdot g \quad \text{(in newtons, with } g \text{ in m/s}^2\text{)}$

Only two vertical forces act on the person: the weight $\vv{P} = m\vv{g}$ downward and the normal $\vv{N}$ upward. We choose the positive vertical axis upward and apply Newton’s second law: $N - mg = ma$

Since the scale reads more than the real weight ($80 > 70$), the normal force is greater than the weight: the cabin must be accelerating upward. We obtain the acceleration: $a = \frac{N - mg}{m} = \frac{(80 - 70)\,g}{70}$

Substituting $g = 9{,}81\ \text{m/s}^2$: $a = \frac{10 \cdot 9{,}81}{70} = \frac{98{,}1}{70}$

$\ev{a \approx 1{,}40\ \text{m/s}^2 \quad \text{directed upward}}$

The positive sign confirms the intuition: when the scale reads a value greater than the real weight, the person feels heavier because the cabin is accelerating upward (or braking while going down).

Links

Topics: Dynamics Concepts: Newton’s second law · Normal force Objects: Lift · Scale