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?

Answer 1

Extension of spring:

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

Step 1: Given:

1. Length of cylinder = L
2. Mass of cylinder = M
3. 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.

$F_B=\sigma ·\frac{L}{2}·A·g$

Putting the value of $F_B$ in equation (i)

$$\begin{gathered} k{x}_o+\sigma ·\frac{L}{2}·A·g=Mg \\ {x}_o=\frac{Mg-\sigma ·\frac{L}{2}·A·g}{k} \\ {x}_o=\frac{Mg}{k}\left(1-\frac{\sigma LA}{2M}\right) \end{gathered}$$