Problem: Spring with two hanging masses

Problem

A vertical spring ($K = 400\;\text{N/m}$, $\ell_0 = 0{,}3\;\text{m}$) has mass A ($3\;\text{kg}$) hanging from its upper end attached to a fixed support, and mass B ($5\;\text{kg}$) hanging from its lower end. At equilibrium, calculate the spring’s extension and the force the support exerts on A.

Solution

Spring extension. The spring holds only mass B, hanging from its lower end. At equilibrium the spring’s elastic force must balance the weight of B, so $F_s = P_B = m_B g$. By Hooke’s law $F_s = K\,\Delta\ell$, from which:

$\Delta\ell = \frac{m_B\, g}{K} = \frac{5 \cdot 9{,}81}{400} = \frac{49{,}05}{400}$

$\ev{\Delta\ell \approx 0{,}123\;\text{m}}$

Force of the support on A. We isolate mass A. Three vertical forces act on it: its own weight $\vv{P}_A = m_A g$ (downward), the spring force (the spring is stretched and pulls A downward, towards B, with magnitude $F_s = K\,\Delta\ell = 49{,}05\;\text{N}$) and the force $\vv{F}_{\text{sup}}$ from the support (upward). At equilibrium:

$F_{\text{sup}} = m_A g + F_s = 3 \cdot 9{,}81 + 49{,}05 = 29{,}4 + 49{,}05$

$\ev{F_{\text{sup}} \approx 78{,}5\;\text{N}}$

The support holds up the entire system: indeed $F_{\text{sup}} = m_A g + m_B g = (m_A + m_B)\,g = 8 \cdot 9{,}81 \approx 78{,}5\;\text{N}$, i.e. the combined weight of A and B, as must be the case for equilibrium of the whole hanging system.

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Topics: Dynamics Concepts: Elastic force and Hooke’s law · Tension · Weight force Objects: Spring