Question

A block of wood of mass 10.8 kg and density 4.42 x 102 kg/m3, floats on water. (a) What is the minimum amount of aluminum (density 2.70 x 103 kg/m3) that can be placed on top of the wood block and cause the block of wood and the aluminum to sink? (b) If instead it was a block of ice of mass 10.8 kg and density 8.90 x 102 kg/m3. What is the minimum amount of aluminum that can be placed on top of the ice block and cause the block of ice and the aluminum to sink? This question has been posted several times but they all have a different answer with little feedback.

Answer 1

a)
given
m_wood = 10.8 kg
rho_wood = 442 kg/m^3

volume of the block, V = m_wood/rho_wood

= 10.8/442

= 0.02443 m^3

let V_aluiminum is the minimum volume of the aluminum required.

in the equilibrium,

buoyant force = weight of woodedn block + weight of aluminum

rho_water*(V + V_aluminum)*g = m_wood*g + m_aluminum*g

rho_water*(V + V_aluminum) = m_wood + m_aluminum

rho_water*(V + V_aluminum) = m_wood + rho_aluminum*V_aluminum

1000*(0.02443 + V_aluminum) = 10.8 + 2700*V_aluminum

==> V_aluminum = 0.008018 m^3

so, m_aluminum = rho_aluminum*V_aluminum

= 2700*0.008018

= 21.6 kg <<<<<<<<<<<----------------------------Answer

b)

m_Ice = 10.8 kg
rho_Ice = 890 kg/m^3

volume of the Ice, V = m_Ice/rho_Ice

= 10.8/892

= 0.0121 m^3

let V_aluiminum is the minimum volume of the aluminum required.

in the equilibrium,

buoyant force = weight of Ice block + weight of aluminum

rho_water*(V + V_aluminum)*g = m_Ice*g + m_aluminum*g

rho_water*(V + V_aluminum) = m_Ice + m_aluminum

rho_water*(V + V_aluminum) = m_Ice + rho_aluminum*V_aluminum

1000*(0.0121 + V_aluminum) = 10.8 + 2700*V_aluminum

==> V_aluminum = 0.0007647 m^3

so, m_aluminum = rho_aluminum*V_aluminum

= 2700*0.0007647

= 2.06 kg <<<<<<<<<<<----------------------------Answer