A wooden plank of length 1 m and uniform cross-section is hinged at one end to the bottom of a tank

Q: Why does the upthrust act at the midpoint of the submerged portion rather than the center of the entire plank?

The buoyant force is the equivalent force of the pressure exerted by the fluid on the submerged surface. By Archimedes' principle, this force acts through the center of gravity of the displaced fluid, which corresponds to the geometric center (midpoint) of only the submerged part of the uniform plank, not the entire plank.

Q: Why did we exclude the case where $\theta = 0^\circ$?

If $\theta = 0^\circ$, the plank is perfectly vertical. While this is mathematically a state where torques balance (since the lever arms are zero), it is an unstable equilibrium in this setup because any slight perturbation would cause the buoyant force to create a restoring torque that tilts the plank to its stable equilibrium angle of $45^∘$.

Q: What would happen if the specific gravity of the plank was greater than 1?

If the specific gravity of the plank was greater than 1, the density of the plank would be greater than that of water. The plank would sink completely and rest at the bottom of the tank, making $\theta = 0^∘$ (pointing straight down) as its stable equilibrium position.

Q: Does the cross-sectional area of the plank affect the equilibrium angle?

No, as we saw in the derivation, the cross-sectional area $A$ cancels out from both sides of the torque balance equation. The equilibrium angle depends only on the length of the plank, the height of the water, and the relative density (specific gravity) of the wood.

Q: What is the physical significance of the hinge force, and why did we ignore it?

The hinge exerts a reaction force to keep the end of the plank anchored at the bottom of the tank. We did not ignore its physical existence, but by choosing the hinge itself as our pivot point for calculating torques, the line of action of this reaction force passes directly through the pivot, making its lever arm zero and its torque contribution zero.

Question

A wooden plank of length $1	ext{ m}$ and uniform cross-section is hinged at one end to the bottom of a tank as shown in figure. The tank is filled with water upto a height $0.5	ext{ m}$. The specific gravity of the plank is $0.5$. Find the angle $	heta$ that the plank makes with the vertical in the equilibrium position (exclude the case $	heta = 0^\circ$).

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a beautiful, sunny day where you are observing a wooden plank floating in a water tank. The plank is not completely free; it is **hinged** at one of its ends to the bottom of the tank. Because the wood is lighter than water (with a specific gravity of $0.5$), it wants to float up. However, the hinge keeps it anchored, causing it to tilt at an angle $	heta$ with the vertical.

To find this equilibrium angle, we must dive into the world of **rotational mechanics** and **fluid statics**. For any rigid body in rotational equilibrium, the sum of all torques acting on it about any point must be zero. In this case, the most convenient point to calculate torque is the hinge $O$, because the unknown hinge forces will have a lever arm of zero and thus produce no torque.

Let's identify the active forces that *do* produce torque:
1. The **weight** of the plank ($W$), acting downwards at its center of gravity.
2. The **buoyant force** or **upthrust** ($F_B$), acting upwards at the center of buoyancy of the submerged portion.

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### The Torque due to Weight

Let's define the physical properties of our plank. Let $A$ be its uniform cross-sectional area and $L = 1	ext{ m}$ be its total length.

The specific gravity of the wood is given as $0.5$. This means the density of the plank, $ho$, is exactly half the density of water, $ho_w$:
$$ho = 0.5 ho_w$$

The mass of the plank is its volume multiplied by its density:
$$M = A L ho = 0.5 A L ho_w$$

Therefore, the weight of the plank is:
$$W = M g = 0.5 A L ho_w g$$

This weight acts vertically downwards at the **center of gravity** $G$, which lies at the geometric center of the uniform plank, at a distance of $L/2$ from the hinge $O$.

The perpendicular distance from the line of action of the weight to the hinge is $\frac{L}{2} \sin 	heta$. This creates a **clockwise torque**:
$$	au_W = W \cdot \left(\frac{L}{2} \sin 	hetaight) = (0.5 A L ho_w g) \cdot \left(\frac{L}{2} \sin 	hetaight)$$

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### The Torque due to Upthrust

Now, let's look at the submerged portion of the plank. Let $x$ be the length of the plank that is underwater. The water level is at a height $h = 0.5	ext{ m}$. From the geometry of the right-angled triangle formed by the submerged plank, the vertical depth, and the horizontal water surface, we have:
$$x \cos 	heta = h \implies x = h \sec 	heta$$

The volume of this submerged portion is:
$$V_{	ext{sub}} = A x = A h \sec 	heta$$

According to Archimedes' principle, the upthrust force $F_B$ is equal to the weight of the displaced water:
$$F_B = V_{	ext{sub}} ho_w g = A h \sec 	heta ho_w g$$

This upward force acts at the **center of buoyancy** $B$, which is the midpoint of the submerged portion of the plank. The distance of $B$ from the hinge $O$ along the plank is:
$$d_{OB} = \frac{x}{2} = \frac{h \sec 	heta}{2}$$

The perpendicular distance from the line of action of the upthrust to the hinge is $d_{OB} \sin 	heta$. This creates a **counter-clockwise torque**:
$$	au_{F_B} = F_B \cdot \left(\frac{x}{2} \sin 	hetaight) = (A h \sec 	heta ho_w g) \cdot \left(\frac{h \sec 	heta}{2} \sin 	hetaight)$$

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### Balancing the Torques

For rotational equilibrium, the clockwise torque must exactly balance the counter-clockwise torque:
$$	au_W = 	au_{F_B}$$

Substituting our expressions into this equation:
$$(0.5 A L ho_w g) \cdot \left(\frac{L}{2} \sin 	hetaight) = (A h \sec 	heta ho_w g) \cdot \left(\frac{h \sec 	heta}{2} \sin 	hetaight)$$

Notice the beautiful cancellations! The area $A$, the density of water $ho_w$, the acceleration due to gravity $g$, and the term $\sin 	heta$ (since $	heta 
eq 0^\circ$) cancel out from both sides. We are left with:
$$0.5 L \cdot \frac{L}{2} = h \sec 	heta \cdot \frac{h \sec 	heta}{2}$$

$$\frac{L^2}{4} = \frac{h^2 \sec^2 	heta}{2}$$

Rearranging this to solve for $\cos^2 	heta$ (since $\cos 	heta = 1/\sec 	heta$):
$$\sec^2 	heta = \frac{L^2}{2 h^2} \implies \cos^2 	heta = \frac{2 h^2}{L^2}$$

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### Final Calculation

Now, let's substitute the given numerical values: $L = 1	ext{ m}$ and $h = 0.5	ext{ m}$:
$$\cos^2 	heta = \frac{2 (0.5)^2}{1^2} = \frac{2 \cdot 0.25}{1} = 0.5 = \frac{1}{2}$$

Taking the square root on both sides (since $	heta$ is an acute angle, $\cos 	heta$ must be positive):
$$\cos 	heta = \frac{1}{\sqrt{2}}$$

This corresponds to an angle of:
$$	heta = 45^\circ$$

Thus, the plank settles at a perfect **$45^\circ$** angle with the vertical in its equilibrium state!

Enter Numerical Value:

Visualized Solution

Visualizing the Forces on the Plank

Calculating the Weight of the Plank

Torque due to Weight

Determining the Submerged Length

Calculating the Upthrust Force

Center of Buoyancy and Torque of Upthrust

Setting up the Equilibrium Equation

Simplifying the Master Equation

Solving for $cos θ$

Substituting Numerical Values

Finding the Angle $θ$

Exploring Further Scenarios

The Sigma Insight: Buoyancy and Archimedes' Principle

Analyzing the Setup

The Torque due to Weight

The Torque due to Upthrust

Balancing the Torques

Final Calculation

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