You have a beam balance, the kind of scales that tips from one side to the other, depending on the weight on each side. On the left side lies a kilogram of ice. On the right side lies the standard kilogram. The pans are in balance.

After a while all the ice melted.

Does the balance tip to the right, to the left, or does it remain unchanged?

What will be needed in order to restore the balance if necessary?

###### Solution.

The buoyant force acts on the objects from the ambient air. Usually, in comparison with the weight of the objects, the buoyant force is negligible and it is not taken into account. However, in our case, this isn’t so.

The volume of ice can be calculated by using the following equation

V_{ice} = m / ρ_{ice}

where ρ_{ice} is ice density, ρ_{ice} = 917 кг/м^{3}.

The volume of water can be calculated by using the following equation

V_{water} = m / ρ_{water}

where ρ_{water} is water density, ρ_{water} = 1000 кг/м^{3}.

It is clear that V_{water} < V_{ice}, thereby, the buoyant force on the left side decreases and the balance tips to the left.

We can calculate the change in buoyant force by using the following equation

ΔF = ρ_{air}*g*(m / ρ_{ice} – m / ρ_{water}),

where ρ_{air} is air density, ρ_{air} = 1.225 кг/м^{3}.

We can find the mass required to restore the equilibrium of the balance.

m = ΔF/g = ρ_{air}*(m / ρ_{ice} – m / ρ_{water}).

We substitute the values and we get

m = 1.225*(1/917-1/1000) ≈ 0.00011 kg ≈ 0.11 gr.

Finally, we need to add 0.11 gr to the right side of the balance.