A cylindrical block of wood 1 m in diameter and 1 m long has a specific weight of 7500 N/m^3. Will it float in water with its axis vertical?

The block will float with its axis vertical.

Explanation:

For it to float, the upward force on the cylindrical block must be equal to the weight of an equal volume of water. Also, this upward force must be greater than or equal to the weight of the cylindrical block for it to float.

So, weight of cylindrical block, W = specific weight × volume

specific weight = 7500 N/m³

volume = πd²h/4 where d = diameter of block and h = height of block

volume = π(1 m)² × 1 m/4 = π/4 m³ = 0.7854 m³

W = 7500 N/m³ × 0.7854 m³ = 5890.5 N

Since the density of water = 1000 kg/m³, its specific weight W' = 1000 kg/m³ × 9.8 m/s² = 9800 N/m³

Since the volume of the cylinder = volume of water displaced, the weight of water displaced W' = upward force = specific weight of water × volume of water displaced =  9800 N/m³ × 0.7854 m³ = 7696.92 N

Since W' = 7696.92 N > W = 5890.5 N, the block will float with its axis vertical since the upward force is greater than the weight of the cylindrical block.

answer:

true

c

explanation: