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

2.
volume of wood = 0.20^3 = 8 x 10^-3 m^3

weight of wood = rho_wood * 8 *10^-3 * g

buoyant force = weight of water dispalced

= 1000 * (0.7 * 8 *10^-3) g

and for equilibrium weight = buoyatnt force

rho_wood * 8 *10^-3 * g = 1000 * (0.7 * 8 *10^-3) g

rho_wood = 700 kg/m^3 Or 0.7 g/cm^3

b & c ) maximum buoyant force = (8*10^-3 + Vlead) *1000 * g

and weigt force = (700 * 8 *10^-3 * g ) + (Vlead * 11300 * g)

(8*10^-3 + Vlead) *1000 * g = (700 * 8 *10^-3 * g ) + (Vlead * 11300 * g)

8 + 1000 Vlead = 5.6 + 11300Vlead

Vlead = 2.33 x 10^-4 m^3 ...........Ans(c)

mass = Vlead x 11300 = 2.63 kg ......Ans(b)

d) total volume = (8 + 0.233) * 10^-3 =8.233 *10^-3 m^3

15% is out of water then

buoyant force equal to that 15% have to be applied.

F = 0.15 * 8.233 * 10^-3 *1000 * 9.8 = 12.1 N

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