Question

A block of wood has a mass of 3.57 kg and a density of 588 kg/m³. It is to be loaded with lead so that it will float in water with 0.82 of its volume immersed. The density of lead is 1.13 ? 10⁴ kg/m³.

(a) What mass of lead is needed if the lead is on top of the wood?

(b) What mass of lead is needed if the lead is attached below the wood?

Answer 1

V = volume = 3.57/588 = 6.07*10^-3 m^3

v1 =volume inside water = 0.82*V = 4.97*10^-3 m^3

P = density of water =1000 kg/M^3

a) m1 = V1*P - 3.57 = V1*1000 -3.57 = 4.97 -3.57 = 1.4 kg

b) Let V' = volume of lead below the wood

   m' = mass of lead having volume V'

   m' = 1.13*10⁴ *V'

then    3.57 +m'   = (V' + V1)*1000

            3.57 + 1.13*10⁴ *V'   = V' *1000 + 4.97

                                   V' = 1.4/(10300) =1.359*10^-4 m^3

m' = mass of lead= 1.13*10⁴ *V'    = 1.53 kg