Question

A uniform cylinder of length $L$ and mass $M$ having cross-sectional area $A$ is suspended, with its length vertical, from a fixed point by a massless spring, such that it is half-submerged in a liquid of density at the equilibrium position. The extension ${x}_{0}$ of the spring when it is in equilibrium is?

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Solution

**Extension of spring:**

Extension springs are used to absorb and store energy while also providing resistance to a pulling force.

**Step 1: Given:**

- Length of cylinder = L
- Mass of cylinder = M
- Cross Sectional area of cylinder = A

**Step 2: Formula Used:**

$k{x}_{o}+{F}_{B}=Mg$

Here, $k{x}_{o}$ is stress on spring, ${F}_{B}$ is buoyant force, $M$ is mass, and $g$ is acceleration due to gravity.

**Step 3: Calculating extension x _{o}**

The figure shows the various forces acting on the cylinder.

$Mg$ is the weight of the cylinder, ${F}_{B}$ is buoyant force, and $k{x}_{o}$ is stress on spring, ${x}_{o}$ is the extension in the string, and $k$ is spring constant.

At equilibrium,

Upward forces = Downward forces

$k{x}_{o}+{F}_{B}=Mg$ ------------- (i)

${F}_{B}$ is the force due to buoyancy which is equal to the weight of the liquid displaced.

${F}_{B}=\sigma \xb7\frac{L}{2}\xb7A\xb7g$

Putting the value of ${F}_{B}$ in equation (i)

$k{x}_{o}+\sigma \xb7\frac{L}{2}\xb7A\xb7g=Mg\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{x}_{o}=\frac{Mg-\sigma \xb7\frac{L}{2}\xb7A\xb7g}{k}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{x}_{o}=\frac{Mg}{k}\left(1-\frac{\sigma LA}{2M}\right)$

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