Lift with a scale and a spring

Problem

In a lift going up with acceleration $a = 2{,}0\ \text{m/s}^2$, a mass $m = 3{,}0\ \text{kg}$ hangs from a vertical spring of constant $k = 200\ \text{N/m}$. (a) By how much does the spring stretch? (b) At what reading in kgf would a scale placed under the mass support it?

Solution

The lift accelerates upward at $a = 2{,}0\ \text{m/s}^2$. The mass hanging from the spring accelerates together with the cabin. Acting on the mass are the elastic force $\vv{T}$ (upward) and the weight $\vv{P} = m\vv{g}$ (downward). We choose the positive vertical axis upward.

Part (a): spring extension.

We apply Newton’s second law to the mass: $T - mg = ma$

The elastic force must therefore equal: $T = m\,(g + a) = 3{,}0 \cdot (9{,}81 + 2{,}0) = 3{,}0 \cdot 11{,}81 = 35{,}43\ \text{N}$

Since the spring obeys Hooke’s law, $T = k\,\Delta L$, we obtain the extension: $\Delta L = \frac{T}{k} = \frac{m\,(g + a)}{k} = \frac{35{,}43}{200}$

$\ev{\Delta L \approx 0{,}177\ \text{m} = 17{,}7\ \text{cm}}$

Part (b): scale reading.

A scale placed under the mass would need to support it with the same force $T$ found above. The reading in kilograms-force is obtained by dividing the force by $g$: $m_{\text{read}} = \frac{T}{g} = \frac{m\,(g + a)}{g} = \frac{35{,}43}{9{,}81}$

$\ev{m_{\text{read}} \approx 3{,}61\ \text{kgf}}$

In a lift accelerating upward the mass “weighs more”: the spring stretches more than at rest (where it would be $\Delta L_0 = mg/k \approx 14{,}7\ \text{cm}$) and the scale reads above the real $3{,}0\ \text{kg}$.

Links

Topics: Dynamics Concepts: Elastic force and Hooke’s law · Newton’s second law · Normal force Objects: Spring · Lift · Scale