Would the level of the water be rising, falling, or staying the same?

Question

one day you were standing on the shore of a small lake. you looked out onto the water and saw a boat. in the boat were two men with five huge steel beams. the men were sidling the beams out of the boat and into the lake. if you could actually see it or measure it, would the level of the water in the lake be rising, falling, or staying the same?

Answer 1

The mass of rolled steel is about 7,850 Kg per cubic meter.

The mass of water is about 1,000 Kg per cubic meter.

Let's say the steel beams have total mass of 7,850 Kg. The mass of the boat plus the beams is the mass of water displaced by the loaded boat. While the steel beams are on the boat, they displace 7.850 cubic meters of water. But when the steel beams are on the bottom of the lake, they displace only one cubic meter.

You can figure out what happens to the volume of the lake when the beams are no longer in a boat borne up by, and displacing, their weight of water in the lake, but rather are on the bottom of the lake displacing their volume of only 1 cubic meter.

Answer 2

With Archimedes:

$(m_{boat}+m_{beams})g=\rho_w Vg$, with $\rho_w$ density of the water, $V$ the volume of water displaced by the boat.

So $(m_{boat}+m_{beams})=\rho_w V$.

And $\frac{m_{boat}+m_{beams}}{\rho_w}=V$.

Now we take the beams off the boat but don't submerse them in the lake yet:

$\frac{m_{boat}}{\rho_w}=V'$, with $V'$ the new, smaller displaced volume of water.

So:

$\frac{m_{beams}}{\rho_w}=V-V'$.

Now we submerse the beams in the water, their density is:

$\rho_{beams}=\frac{m_beams}{V_beams}$.

So that $V_{beams}=\frac{m_{beams}}{\rho_{beams}}$.

Assuming $\rho_{beams}>\rho_w$.

Then $\frac{m_{beams}}{\rho_w}<\frac{m_{beams}}{\rho_{beams}}$

Since as the difference $V-V'$ is smaller than the volume displaced by the beams $\frac{m_{beams}}{\rho_{beams}}$, the water level in the lake falls a little.