A uniform solid cylinder of density 0.8 g/cm^3 floats in equilibrium in a combination of two non-mix

Q: Why is the net force exerted by Liquid A on the cylinder zero?

Liquid A completely surrounds the cylinder horizontally. At any given depth, the pressure is uniform all around, so the horizontal forces from opposite sides are equal and opposite, canceling out completely.

Q: Does Liquid A contribute to the vertical buoyant force?

Yes, Liquid A contributes to the vertical buoyant force because it has a vertical pressure gradient. The upthrust from Liquid A is equal to the weight of Liquid A displaced by the cylinder, which is $s h_A \rho_A g$.

Q: Why does the portion of the cylinder in Liquid B increase by exactly $h$ when depressed?

Since Liquid A has a fixed height $h_A$ and the top of the cylinder is pushed to align with the top of Liquid A, the length of the cylinder in Liquid A remains exactly $h_A$. Therefore, the entire depressed length $h$ must push down into Liquid B.

Q: What is the physical meaning of the net restoring force $s h \rho_B g$?

It represents the extra buoyant force gained because a portion of the cylinder of length $h$ (which was in air) is now submerged in the denser Liquid B instead of air.

Q: Will the cylinder perform Simple Harmonic Motion (SHM) after release?

Yes, for small displacements where the cylinder remains within the boundaries of the liquids, the restoring force is proportional to the displacement, leading to SHM.

Question

A uniform solid cylinder of density $0.8	ext{ g/cm}^3$ floats in equilibrium in a combination of two non-mixing liquids A and B with its axis vertical. The densities of the liquids A and B are $0.7	ext{ g/cm}^3$ and $1.2	ext{ g/cm}^3$, respectively. The height of liquid A is $h_A = 1.2	ext{ cm}$. The length of the part of the cylinder immersed in liquid B is $h_B = 0.8	ext{ cm}$.
(a) Find the total force exerted by liquid A on the cylinder.
(b) Find $h$, the length of the part of the cylinder in air.
(c) The cylinder is depressed in such a way that its top surface is just below the upper surface of liquid A and is then released. Find the acceleration of the cylinder immediately after it is released.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a solid cylinder floating vertically at the interface of two immiscible liquids. This is a classic problem in fluid mechanics that tests our understanding of **buoyancy**, **hydrostatic pressure**, and **equilibrium**.

We are given:
- Density of the cylinder: $ho_{	ext{cylinder}} = 0.8	ext{ g/cm}^3$
- Density of Liquid A: $ho_A = 0.7	ext{ g/cm}^3$
- Density of Liquid B: $ho_B = 1.2	ext{ g/cm}^3$
- Height of Liquid A: $h_A = 1.2	ext{ cm}$
- Submerged length in Liquid B: $h_B = 0.8	ext{ cm}$

Let's tackle each part of the problem step-by-step.

---

### Part (a): Force Exerted by Liquid A

Liquid A completely surrounds the vertical curved surface of the cylinder. At any given depth within Liquid A, the pressure is uniform all around the perimeter of the cylinder.

Since pressure forces act normal to the surface, the horizontal forces from opposite sides are equal in magnitude and opposite in direction. Therefore, they **completely cancel each other out**.

Thus, the net force exerted by Liquid A on the cylinder is **zero**.

---

### Part (b): Finding the Length in Air ($h$)

In equilibrium, the total downward force (weight of the cylinder) must be balanced by the total upward buoyant force (upthrust) exerted by the displaced fluids.

Let $s$ be the cross-sectional area of the cylinder. The total length of the cylinder is:
$$L = h + h_A + h_B$$

The weight of the cylinder is:
$$W = s(h + h_A + h_B)ho_{	ext{cylinder}}g$$

The total upthrust is the sum of the buoyant forces from Liquid A and Liquid B:
$$U = s h_A ho_A g + s h_B ho_B g$$

Equating weight and upthrust:
$$s(h + h_A + h_B)ho_{	ext{cylinder}}g = s h_A ho_A g + s h_B ho_B g$$

Notice how the area $s$ and gravity $g$ cancel out beautifully from both sides:
$$(h + h_A + h_B)ho_{	ext{cylinder}} = h_A ho_A + h_B ho_B$$

Now, let's substitute the given values:
$$(h + 1.2 + 0.8)(0.8) = 1.2(0.7) + 0.8(1.2)$$
$$(h + 2.0)(0.8) = 0.84 + 0.96 = 1.80$$

Solving for $h$:
$$h + 2.0 = \frac{1.80}{0.8} = 2.25$$
$$h = 0.25	ext{ cm}$$

So, the length of the cylinder in air is **$0.25	ext{ cm}$**.

---

### Part (c): Acceleration Immediately After Release

When the cylinder is depressed by $h = 0.25	ext{ cm}$, its top surface is just below the upper surface of Liquid A.

Let's analyze the new state:
- The portion of the cylinder in Liquid A remains exactly $h_A = 1.2	ext{ cm}$ because the height of Liquid A is fixed.
- The portion of the cylinder in Liquid B increases from $h_B$ to $h_B + h = 0.8 + 0.25 = 1.05	ext{ cm}$.
- The portion in air is now $0$.

The new upthrust is:
$$U' = s h_A ho_A g + s(h_B + h)ho_B g$$

The net upward restoring force is the difference between the new upthrust and the weight:
$$F_{	ext{net}} = U' - W$$

Since $W = U$ (the initial upthrust), we can write:
$$F_{	ext{net}} = U' - U = s h ho_B g$$

This is a beautiful result! The extra upthrust comes entirely from the volume of length $h$ that was previously in air and is now submerged in the denser Liquid B.

Let's calculate the acceleration $a$:
$$a = \frac{F_{	ext{net}}}{M} = \frac{s h ho_B g}{s L ho_{	ext{cylinder}}} = \frac{h ho_B}{L ho_{	ext{cylinder}}} g$$

Substitute the values:
$$a = \frac{0.25 	imes 1.2}{2.25 	imes 0.8} g = \frac{0.3}{1.8} g = \frac{g}{6}$$

Thus, the acceleration of the cylinder immediately after release is **$\frac{g}{6}$ upwards**.

Visualized Solution

Understanding the Multi-Fluid System

Part (a) - Hydrostatic Force by Liquid A

Part (b) - Equilibrium Condition

Expressing Weight and Upthrust

Solving for $h$

Finalizing $h$

Part (c) - Analyzing the Depressed State

Calculating Net Upward Force

Calculating Acceleration

The Sigma Insight: Buoyancy and Archimedes' Principle

Analyzing the Setup

Part (a)

Part (b)

Part (c)

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