Question

A cube of wood having an edge dimension of 19.7 cm and a density of 655 kg/m3 floats on water.

(a) What is the distance from the horizontal top surface of the cube to the water level? __________cm

(b) What mass of lead should be placed on the cube so that the top of the cube will be just level with the water surface? ____kg

Answer 1

wood density 655 kg/m³
water density 998 kg/m³
The total volume of the wood is (19.7 cm)³ = (0.197 m)³ = 0.00764 m³
The total mass of the wood is 655 kg/m³ x 0.00764 m³ = 5.007 kg

The question is what volume of water will weigh 5.007 kg.
5.007 kg / 998 kg/m³ = 0.00502 m³

since it takes only 0.00502 m³ of the wood to support the total weight, then we can calculate the percent in the water as 0.00502 m³ / 0.00764 m³ = 65.6% and therefore 34.4% exposed.

since the cross-section does not change, that means the part exposed is 34.4% of 19.7 cm

b) displaced water if the wood is totally submerged is
0.00764 m³ x 998 kg/m³ = 7.62 kg
subtract the weight of the wood to get weight of lead, 7.62 kg – 5.007 kg = 2.62 kg