Problem — The Goods Lift (Scale Graph)

Problem

The goods lift. A goods lift rises and then stops. A person stands inside it on a set of scales. Which graph of the reading $W(t)$ corresponds to the sequence: starting from rest $\to$ rising at constant speed $\to$ decelerating $\to$ stopping? (O’Kuma, Maloney and Hieggelke 2004; see Riferimenti bibliografici)

Solution

Step 1 — What the scales read. The scales measure the normal reaction $W$ they exert on the person (the “apparent weight”). Applying Newton’s second law vertically, with the axis pointing upward:

$W - mg = m\,a \quad\Longrightarrow\quad W = m(g + a)$

where $a$ is the acceleration (positive if directed upward). Hence:

• $a > 0$ (acceleration upward) $\Rightarrow W > mg$;
• $a = 0$ (at rest or constant velocity) $\Rightarrow W = mg$;
• $a < 0$ (acceleration downward) $\Rightarrow W < mg$.

Step 2 — Reconstructing the sequence.

1. Starting from rest, moving upward: to begin rising, the lift accelerates upward, $a > 0$, so $W > mg$ (the scales read more than the true weight).
2. Rising at constant speed: $a = 0$, so $W = mg$ (back to the resting value).
3. Decelerating (slowing to stop while still rising): the acceleration is now directed downward, $a < 0$, so $W < mg$ (the scales read less).
4. Stopped: again $a = 0$, so $W = mg$.

Step 3 — Comparing with the graphs. The required sequence is therefore: $mg \to$ above $mg \to$ back to $mg \to$ below $mg \to$ back to $mg$. This is graph B, the only one that rises above $mg$ in the initial phase, returns to the resting value during the uniform rise, drops below $mg$ during deceleration, and finally returns to $mg$ at the stop. Graphs A and C do not respect the correct order of the phases (in particular they do not bring the reading back to $mg$ during the constant-speed stretch).

$\ev{\text{Graph B}}$

Links

Topics: Dynamics Concepts: Newton’s second law · Normal force Skills: Reading graphs Objects: Lift · Scales