A piece of wood of sp gr 0.65 is 80mm square and 1.5m long. How many Newtons of lead weighing 120KN/m3 must be fastened at one end of the stick so that it will float upright with 0.3m out of water?

Floating Wood with Lead Attachment

Problem Statement

A wooden stick with a square cross-section and specific gravity of 0.65 is floating upright in water. Given:

Width: 80 mm (0.08 m)
Length: 1.5 m
Length of wood submerged in water: 1.2 m
Specific weight of lead: 120 kN/m³
Target: To determine the weight of lead required to be attached at the bottom so that the stick remains upright with 0.3 m out of water.

Solution

1. Define the Equilibrium Condition

The total weight of the wood and lead must equal the buoyant force: \[ \gamma_{\text{wood}} V_{\text{wood}} + \gamma_{\text{lead}} V_{\text{lead}} = \gamma_{\text{water}} V_{\text{displaced water}} \]

2. Calculate the Weight of Wood

\[ W_{\text{wood}} = 0.65 \times 9810 \times (0.08)^2 \times 1.5 \] \[ = 61.214 \text{ N} \]

3. Calculate the Buoyant Force

\[ F_B = 9810 \times \left[(0.08)^2 \times 1.2 + V_{\text{lead}}\right] \] \[ = 75.34 + 9810 V_{\text{lead}} \]

4. Solve for the Volume of Lead

\[ 61.214 + 120000 V_{\text{lead}} = 75.34 + 9810 V_{\text{lead}} \] Solving for \( V_{\text{lead}} \), \[ V_{\text{lead}} = 0.000128 \text{ m}^3 \]

5. Calculate the Weight of Lead

\[ W_{\text{lead}} = \gamma_{\text{lead}} V_{\text{lead}} \] \[ = 120000 \times 0.000128 \] \[ = 15.38 \text{ N} \]

Final Result:

Weight of lead required: 15.38 N

Explanation

1. Floating Equilibrium:
The piece of wood floats upright because the buoyant force exerted by the water balances the total weight of the wood and attached lead. Since part of the wood remains above the water, we calculate the submerged volume and use Archimedes’ principle.

2. Importance of the Lead Weight:
The wood itself is not heavy enough to stay upright with the required portion out of the water. By attaching lead at the bottom, we lower the center of gravity, making the floating object more stable and achieving equilibrium.

3. Volume of Lead:
The small volume of lead found here indicates that a high-density material (lead) is effective for adjusting buoyancy with minimal material.

4. Archimedes’ Principle Application:
The total weight of the system must equal the buoyant force exerted by the displaced water. Since fresh water has a density of 9810 N/m³, we use this value to determine the necessary lead volume.

Physical Meaning

1. Stability of Floating Objects:
In real-world applications, boats, ships, and floating structures need to maintain stability. By adjusting the weight distribution, such structures can achieve a desired floating position.

2. Use of High-Density Materials:
Lead is often used in ballast applications because it has a high density and allows for fine-tuning the buoyancy of a floating object without adding too much volume.

3. Engineering Considerations:
Engineers must consider the weight distribution and buoyancy of floating structures to prevent tipping and ensure they remain upright in water.