Question

A cube of wood having an edge dimension of 18.1 cm and a density of 655 kg/m floats on water. (a) What is the distance from the horizontal top surface of the cube to the water level? This is the length of the edge of the wood block that lies below the surface. 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? The block is at rest so the buoyant force must equal the weight of the block plus the lead on top. What is the magnitude of the b Need Help? Read it

Answer 1

here,

the dimension of cube , a = 18.1 cm

a = 0.181 m

density of wood , p1 = 655 kg/m^3

a)

let the distance of horizontal top surface of cube to the water be h

equating the forces vertically

buoyant force = weight

pw * volume submerged * g = m * g

pw * volume submerged = density of wood * total voluem

1000 * a^2 * ( a - h) = 655 * a^2 * a

(0.181 - h) = 0.655 * 0.181

h = 0.0624 m = 6.24 cm

the distance of horizontal top surface of cube to the water is 6.24 cm

b)

let the mass of lead be m2

when the wood is completely submerged

equating the forces vertically

buoyant force = weight

density of water * a^3 * g = (m + m2) * g

1000 * 0.181^3 = ( 0.181^3 * 655 + m2)

m2 = 2.05 kg

the mass of lead is 2.05 kg