A cylindrical wooden float whose base area S and the height H drift on the water surface. Density of wood d and density of water is `rho`. What minimum work must be performed to take the float out of the water ?

Explanation

To find the minimum work required to take a cylindrical wooden float out of the water, we can follow these steps:

### Step 1: Understand the Forces Acting on the Float
When the wooden float is in equilibrium on the water surface, the buoyant force acting on it is equal to its weight. The buoyant force can be expressed as:
\[ F_b = \rho \cdot V_{submerged} \cdot g \]
where:
- \( \rho \) is the density of water,
- \( V_{submerged} \) is the volume of the submerged part of the float,
- \( g \) is the acceleration due to gravity.

The weight of the float can be expressed as:
\[ W = d \cdot V_{float} \cdot g \]
where:
- \( d \) is the density of wood,
- \( V_{float} \) is the total volume of the float.

### Step 2: Relate the Submerged Volume to the Height
The volume of the submerged part of the float can be expressed as:
\[ V_{submerged} = S \cdot h \]
where:
- \( S \) is the base area of the float,
- \( h \) is the height of the submerged portion.

Setting the buoyant force equal to the weight of the float gives us:
\[ \rho \cdot (S \cdot h) \cdot g = d \cdot (S \cdot H) \cdot g \]
This simplifies to:
\[ \rho \cdot h = d \cdot H \]
From this, we can solve for \( h \):
\[ h = \frac{d \cdot H}{\rho} \]

### Step 3: Calculate the Work Done Against Gravity
The work done by the external agent to lift the float out of the water must overcome the weight of the float and the buoyant force. The work done against gravity when lifting the float is:
\[ W_{gravity} = W \cdot H = d \cdot (S \cdot H) \cdot g \cdot H \]

### Step 4: Calculate the Work Done by the Buoyant Force
The buoyant force does negative work as the float is lifted. The work done by the buoyant force can be calculated as:
\[ W_{buoyant} = \text{Average Buoyant Force} \cdot \text{Distance} \]
The average buoyant force while lifting the float from \( h \) to \( 0 \) is:
\[ \text{Average Buoyant Force} = \frac{1}{2} \cdot F_b = \frac{1}{2} \cdot \rho \cdot (S \cdot h) \cdot g \]
The distance moved is \( h \), so:
\[ W_{buoyant} = \frac{1}{2} \cdot \rho \cdot (S \cdot h) \cdot g \cdot h \]

### Step 5: Calculate the Total Work Done
The total work done by the external agent \( W_{external} \) to lift the float out of the water is given by:
\[ W_{external} + W_{gravity} + W_{buoyant} = 0 \]
Thus,
\[ W_{external} = - (W_{gravity} + W_{buoyant}) \]

### Step 6: Substitute Values and Simplify
Substituting the expressions for \( W_{gravity} \) and \( W_{buoyant} \):
\[ W_{external} = - \left( d \cdot (S \cdot H) \cdot g \cdot H + \frac{1}{2} \cdot \rho \cdot (S \cdot h) \cdot g \cdot h \right) \]

### Final Expression
After substituting \( h = \frac{d \cdot H}{\rho} \) into the equation, we can simplify to find the minimum work required to take the float out of the water.