Question

A cube of wood having an edge dimension of 18.7 cm and a density of 645 kg/m3 floats on water. (a) What is the distance from the horizontal top surface of the cube to the water level? 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?

Answer #1

**a) let y is the distance from the horizontal top surface
of the cube to the water level**

**In the equilibrium**

**Buoyant fore = weight of the cube**

**weight of the dispalced water = m*g**

**rho_water*V_displaced*g = rho_cube*V_cube*g**

**1000*18.7^2*y*g = 645*18.7^3*g**

**==> y = (645/1000)*18.7**

**= 12.06 cm
<<<<<<<<<<<------------------Answer**

**b) let M is the mass of lead is required**

**In the equilibrium**

**Buoyant fore = weight of the cube + lead**

**weight of the dispalced water = m*g + M*g**

**rho_water*V_cube*g = m*g + M*g**

**rho_water*V_cube = m + M**

**M = rho_water*V_cube - rho_cube*V_cube**

**= 1000*0.187^3 - 645*0.187^3**

**= 2.32 kg
<<<<<<<<<<<------------------Answer**

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