Problem — Scale in a lift

Problem

A person of $75\;\text{kg}$ stands on a scale in a lift that accelerates downward with $a = 4{,}9\;\text{m/s}^2$. What value does the scale show?

Solution

The scale measures the force $R$ (normal reaction) with which the person presses on its surface; by the third law, this equals the force the scale exerts on the person. Two vertical forces act on the person: the weight $\vv{P}$ (downward) and the reaction $\vv{R}$ (upward). We choose the $y$-axis positive upward. The acceleration is directed downward, so $a_y = -4{,}9\;\text{m/s}^2$. The second law along $y$:

$R - m g = m\,a_y = -m a \quad\Rightarrow\quad R = m(g - a)$

$R = 75 \cdot (9{,}81 - 4{,}9) = 75 \cdot 4{,}91$

$\ev{R \approx 368\;\text{N}}$

The scale, calibrated to convert force into mass by dividing by $g$, shows an apparent value:

$m_{\text{read}} = \frac{R}{g} = \frac{368}{9{,}81}$

$\ev{m_{\text{read}} \approx 37{,}5\;\text{kg}}$

The person apparently weighs half as much: accelerating downward, they press less on the scale.

Free fall and apparent weightlessness

If the lift were in free fall ($a = g$), we would have $R = m(g - g) = 0$: apparent weightlessness. This is exactly what happens to astronauts on the International Space Station, in perpetual “fall” around the Earth.

Links

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