A block of wood has a mass of 3.67 kg and a density of 594 kg/m³. It is to be loaded with lead so that it will float in water with 0.883 of its volume immersed. What mass of lead is needed (a) if the lead is on top of the wood and (b) if the lead is attached below the wood? The density of lead is 1.14 X 10⁴ kg/m³.

To solve this problem, we need to apply the principles of buoyancy and the concept of density. Let's break it down step by step for both scenarios: when the lead is on top of the wood and when it's attached below the wood.

Understanding Buoyancy

According to Archimedes' principle, a floating object displaces a volume of fluid equal to the weight of the fluid it displaces. In this case, the block of wood will float with some portion of its volume submerged in water. The volume submerged can be calculated based on the percentage given (0.883) and the total volume of the wood.

Calculating the Volume of Wood

The volume of the wood can be calculated using its mass and density:

• Volume (V) = Mass (m) / Density (ρ)
• V = 3.67 kg / 594 kg/m³ = 0.00619 m³

Volume Immersed in Water

The volume of the wood that is submerged is:

• Volume submerged = 0.883 * V = 0.883 * 0.00619 m³ = 0.00546 m³

Weight of Water Displaced

The weight of the water displaced by the submerged volume helps us determine how much weight must be supported by the wood and lead combined. The density of water is approximately 1000 kg/m³.

• Weight of displaced water = Volume submerged * Density of water
• Weight = 0.00546 m³ * 1000 kg/m³ = 5.46 kg

Calculating Required Weight of Lead

Now, let's find out how much weight (including wood and lead) needs to equal the weight of the displaced water for the block to float.

• Weight of wood = 3.67 kg
• Let the mass of lead be m_lead.
• The combined weight must equal the weight of the water displaced:
• 3.67 kg + m_lead = 5.46 kg
• m_lead = 5.46 kg - 3.67 kg = 1.79 kg

Scenario A: Lead on Top of Wood

In this case, the lead sits directly on top of the wood. The total weight needed to float is still 5.46 kg, so we have already calculated that we need 1.79 kg of lead.

Scenario B: Lead Attached Below the Wood

When the lead is attached below the wood, the lead adds to the total weight, but it also contributes to the volume submerged. To find the volume of lead that corresponds to the required mass, we can use its density:

• Volume of lead (V_lead) = Mass (m_lead) / Density (ρ_lead)
• V_lead = 1.79 kg / (1.14 x 10⁴ kg/m³) = 0.000157 m³

When the lead is below the wood, it displaces additional water, which means the wood plus the lead must now displace the total weight of both to maintain buoyancy. The calculation remains the same for required total weight, as the buoyancy conditions don't change, but the effective weight supported by the water does include the submerged volume of lead. Fortunately, the weight needed remains the same as we calculated previously, so:

• m_lead = 1.79 kg for both cases.

Summary of Required Lead Mass

To summarize:

• (a) If the lead is on top of the wood, 1.79 kg of lead is required.
• (b) If the lead is attached below the wood, 1.79 kg of lead is still required, but it also contributes additional submerged volume.

This analysis shows how buoyancy works in relation to weight and density, helping us understand how different arrangements of materials affect floating behavior.