A cube of wood, 20 cm on each side, floats in water so that 30% is above the sur

Question

A cube of wood, 20 cm on each side, floats in water so that 30% is above the surface of the water, and 70% is below. What is the density of the wood? (You must justify your answer) What mass of lead mass has to be placed on top of the block of wood, so that it will just be totally submerged? What volume of lead mass has to be placed on top of the block of wood, so that it will just be totally submerged? Instead of placing weight on top of the wood, how much how much force would have to be exerted on the wood to submerge it? Densities in gm / cm3: Water Iron Lead 1.0 7.8 11.3

Explanation / Answer

a) If the wood block is 70% below the water, that means it's 70% as dense as water, or 0.7g/mL (or 700 kg/m^3).

b) The buoyancy force acting on the wood if it were totally submerged would be

B =Water*V*g = (1000kg/m^3)*(0.2m)^3*(9.8m/s^2) = 78.4N

Whereas the block of wood would only weigh

W = Block*V*g = (700kg/m^3)*(0.2m)^3*(9.8m/s^2) = 54.9N

To completely submerge the block, it needs a chunck of lead that weighs 23.5N, so

m = 23.5N/9.8m/s^2 = 2.40 kg

c) The density of lead is 11.3 g/mL = 11,300 kg/m^3, so the volume is

V = m/ = (2.4kg)/(11,300 kg/m^3) = 0.00021 m^3

d) You could just push down with 23.5N to submerge it.

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