Question

A cube of wood having an edge dimension of 19.5 cm and a density of 651 kg/m3 floats on water.

Required:
a. What is the distance from the horizontal top surface of the cube to the water level?
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) y ’= 0.068 m , b) m_lead = 2,599 kg

Step-by-step explanation:

a) To solve this exercise, let's use Archimedes' law, which establishes that the pressure of the water is equal to the weight of the dislodged liquid.

Let's write the forces in the equilibrium condition

B - W = 0

B = W

Thrust is

B = ρ_liquid g V_water

Body weight is

W = m g

the definition of density

ρ_body = m / V

we substitute

W = ρ_body g V_body

The volume of the body is

V_body = L³

The volume of water is

V_agua = y L²

we substitute

ρ_liquid g and L2 = rho_body g L3

ρ_liquid y = ρ_body L

y = L ρ_body /ρ_liquid

let's calculate

y = 651/100 0.195

y = 0.1269 m

this is the distance from the bottom of the cube, the distance from the top of the cube

y ’= 0.195 -y

y '= 0.195 - 0.1269

y ’= 0.068 m

b) In this case the equilibrium equation is

B - W_body - W_lead = 0

B = W_body + W_lead

ρ_liquido g V_liquido = ρ_body g V_body + m_lead g = 0

ρ_liquid L3 = ρ_body L3 + m_lead = 0

m_lead = L3 (ρ_liquid - ρ_body)

let's calculate

m_lead = 0.195 3 (1000 - 651)

m_lead = 2,599 kg