Floating Cube Physics Problem

This problem involves calculating the side length of a cubic wooden block floating in water, using principles of density and buoyancy.

Applying Buoyancy Principle

For an object to float, the upward buoyant force ($F_B$) exerted by the fluid must equal the object's downward weight ($W$).

Weight ($W$) = Density of block ($\rho_{block}$) $\times$ Volume of block ($V_{block}$) $\times$ Acceleration due to gravity ($g$)

Buoyant Force ($F_B$) = Density of water ($\rho_{water}$) $\times$ Submerged volume ($V_{sub}$) $\times$ Acceleration due to gravity ($g$)

Equating these:

$ \rho_{block} \times V_{block} \times g = \rho_{water} \times V_{sub} \times g $

We can cancel $g$ from both sides:

$ \rho_{block} \times V_{block} = \rho_{water} \times V_{sub} $

Calculating Side Length

Let the side length of the cube be $L$ (in cm).

• The total volume of the cube is $V_{block} = L^3$.
• The density of the block is given as $\rho_{block} = 0.8 \text{ gm/cc}$.
• The density of water is $\rho_{water} = 1.0 \text{ gm/cc}$ (standard value).
• The block has $2 \text{ cm}$ exposed above water. Therefore, the submerged depth is $(L - 2) \text{ cm}$.
• The submerged volume is $V_{sub} = \text{Area of base} \times \text{Submerged depth} = L^2 \times (L - 2)$.

Substitute these values into the buoyancy equation:

$ 0.8 \times L^3 = 1.0 \times L^2 \times (L - 2) $

Simplify the equation:

$ 0.8 L^3 = L^3 - 2 L^2 $

Rearrange the terms to solve for $L$:

$ 2 L^2 = L^3 - 0.8 L^3 $

$ 2 L^2 = 0.2 L^3 $

Since $L$ must be greater than 0 (it's a physical dimension), we can divide both sides by $L^2$:

$ 2 = 0.2 L $

Solve for $L$:

$ L = \frac{2}{0.2} = \frac{2}{1/5} = 2 \times 5 = 10 \text{ cm} $

Conclusion

The side length of the cubic wooden block is $10 \text{ cm}$.