Calculate how much of a wood volume will remain under water when it floats

Explanation

To determine how much of the pine wood's volume remains underwater when it floats, we can follow these steps:

Given:

• Weight of the pine wood (W): 30 lb
• Buoyant force when fully submerged (B): 62.4 lb
• Density of water (ρ_water): 62.4 lb/ft³ (since 1 cubic foot of water weighs 62.4 lb)

Step 1: Determine the weight of water displaced for the wood to float

For the wood to float, the buoyant force must equal the weight of the wood. The buoyant force is equal to the weight of the water displaced by the submerged part of the wood.

Buoyant force (B) = Weight of water displaced = Weight of the wood

Weight of water displaced=30 lb

Answer to question 1:
The weight of water displaced is 30 lb.

Step 2: Calculate the volume of this displaced water

The volume of water displaced (V_displaced​) can be found using the density of water:

Volume of water displaced = \(\frac{\text{Weight of water displaced}}{\text{Density of water}}\)

V_displaced = \(\frac{30 \text{ lb}}{62.4 \text{ lb/ft}^3} \)

V_displaced ≈ 0.48 ft³

Answer to question 2:
The volume of water displaced is 0.48 ft³.

Step 3: Determine how much of the wood is submerged

The volume of the wood that is submerged is equal to the volume of water displaced, as this is the part of the wood that is underwater when it floats.

V_submerged = V_displaced ≈ 0.48 ft³

Answer to question 3:
The volume of the wood below water level is 0.48 ft³.

Summary:

• The wood displaces 30 lb of water to float.
• The volume of displaced water is 0.48 ft³.
• Therefore, 0.48 ft³ of the wood's volume remains underwater when it floats.