4. Alloy $A$ of two metals with a mass of 6 kg, in which the first metal is twice as much as the second, placed in a container with water, creates a pressure force on the bottom of $30 \mathrm{N}$. Alloy $B$ of the same metals with a mass of 3 kg, in which the first metal is five times less than the second, placed in a container with water, creates a pressure force on the bottom of $10 \mathrm{N}$. What pressure force (in newtons) will a third alloy, obtained by melting the initial alloys, create?

$\{40\}$
Solution. Due to the law of conservation of mass, in the resulting alloy, the mass of each metal is equal to the sum of the masses of these metals in the initial alloys. Thus, both the gravitational forces and the forces of Archimedes also add up. From this, it follows that the reaction force will be the sum of the reaction forces in the first two cases. Answer: $30+10=40$.