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?. What Force must be applied to the original cube of wood, so that it is 15% above the surface?

Explanation / Answer

a) at equilibrium,

buoyant force = weight force

and buoyant force = Volume of fluid displaced * density of fluid * g

and weight = volume of block * density og block * g

Volume of fluid displaced * density of fluid * g - volume of block * density og block * g = 0

0.70V * 1 = V * rho

rho_wood = 0.7 g/cm^3

B) extra buoyant generated by wood block;

Fex = 0.30 V * rho_water * g

     = (0.30 * 0.20^3 m^3) (1000 kg/m^3) (g m/s^2) = 2.4g

so weight of lead will be equal to that extra force.

hence m = 2.4 kg

c) density = mass/ volume

11.3 g /cm^3 = (2.4 x 10^3 g ) / V

V = 212.4 cm^3

d) weight of water equal to that volume.

Fex = 0.15V * rho_water * g

= (0.15 * 0.20^3) ( 1000 ) (9.8) = 11.76 N

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