16 of 100: What Happens?

A Solution Using Limits

The rock sinks, therefore it is denser than water, but few people have good intuition for using that knowledge to guess at what will happen when the stone is cast out of the boat.

So, instead of dealing with small forces and slight changes that we don't have strong intuition for, consider the limiting case where the stone is much, much denser than water: imagine it as a tiny pebble that weighs several tons! When it's in the boat, the boat would be pulled deeply into the water, pushing the shoreline up as the water is displaced by the boat.

Then, when the stone is thrown overboard, the boat rises in the water because the weight of the stone is no longer pulling it down. Meanwhile, the stone sinks to the bottom of the lake and displaces only a small volume of water.

In summary, the stone displaces far more water when it's in the boat, pulling it down, compared to the amount of water it displaces when resting on the bottom of the lake. Therefore, when the stone is thrown out of the boat, the shoreline lowers. Relaxing the limit: this logic holds so long as the stone is denser than water. We know it is denser than water because it sinks.

It does not matter how big the stone in this scenario is, so long as it's denser than water. When the stone is in the boat, the extra amount of water displaced by the boat pushing down is equal in mass to the mass of the stone. When the stone is thrown into the water and sinks, the amount of water displaced is equal in volume to the volume of the stone. How large and dense the rock is will determine the magnitude of the effect, but so long as the rock is denser than water, the shoreline will lower some amount when the rock is tossed out.

Correct me if I'm wrong, but this is the simplest way I could frame the problem to myself: when the stone is in the boat, it displaces a volume of water having a mass equal to the mass of the stone. When the stone is on the bottom of the lake, it displaces a volume of water equal to the volume of the stone. Since the stone is denser than water, the volume of water displaced is greater in the first case than it is in the second. Therefore, the water level drops (ever so slightly) when the stone is thrown into the lake.

Volume/Mass of Boat with stone < Volume/Mass of Boat and Stone separately!

So, water level is lower!

When the stone in the boat, the boat would be pulled deeply into the water, pushing the water level up as the water is displaced by the boat.

When the stone is thrown, the boat rises in the water because the weight of the stone is no longer pulling it down.

The stone displaces more water when it's in the boat, pulling it down, compared to the amount of water it displaces when resting on the bottom of the lake.

In the initial state displaced volume of water is $V_i=(m+M)/ \rho$ , where $m$ and $M$ are masses of stone and boat respectively. $\rho$ is the water density. In the final state displaced volume of water is $V_f=m/ \rho_s+M/ \rho$ , where $\rho_s$ is stone density. Since $\rho_s>\rho$ , then $V_f=m/ \rho_s+M/ \rho<m/ \rho+M/ \rho= (m+M)/ \rho=V_i \ \ \ \Longrightarrow \ \ \ V_f<V_i$ So in the final state the water level is lower.

Here's an intuitive way of thinking about this problem:

You have a boat. You tie a rock to a rope and hang it over the side of the boat into the water. The weight of the rock pulls the boat downward, causing the boat to displace more water than it would otherwise.

Now the rock slips out of the rope. It's still in the water, so it's effect on the water level remains unchanged. But the boat, free of the extra mass, floats slightly higher, displacing less water than it had before.

With less of the boat's volume in the water, the overall water level is lower.

Imagine an extremely light paper boat with a huge volume. Normally it would barely be in the water, but a rock on it makes the boat sink much farther than normal. When you take the rock out, the boat jumps way back up to the surface. The amount of boat volume lost can easily be intuitively imagined to be far greater than the rock volume added to the water, so the water level decreases.

Think of an extreme case where the rock is a black hole XD or is just very tiny and massive. Then when it's in the bost, the entire volume of the boat is submerged raising the water level, but when you throw it off board, only its tiny volume displaces water.

The rock in the boat displaces by its mass. In the water only by its volume. If the rock were a very small piece of a black hole, yet the boat could still remain afloat, it would displace the near maximum amount of water. Once tossed overboard it would displace a negligible amount of water, the boat would float higher and the water level would drop commensurately.

Archimedes principle states that the bouyancy force on a fully or partially submerged object equals the weight of the displaced fluid.

When an object is floating, we are at force equilibrium. Therefore,

$mg=V\rho_w g$

Thus, the volume of the displaced fluid is

$V=m/\rho_w$

When the stone is in the boat, everything floats, and the displaced volume is

$V_1=(m_b+m_s)/\rho_w$

When the stone is at the bottom of the lake, only the boat is floating, so the displaced volume is

$V_2=m_b/\rho_w + V_s$

Now,

$V_1-V_2 = m_s/\rho_w - V_s = V_s(\rho_s/\rho_w-1)$

Since $\rho_s > \rho_w$ (because the stone sinks!), $V_1>V_2$ , so the water level is lower after the stone has been thrown overboard.

One way to see it is that when the stone is at the bottom, its contribution to the displaced volume is just its own volume. Any density above the water's density is in this case "wasted", whereas when the stone is in the boat, the displaced volume just keeps increasing with the density of the stone.

This seemed simple but. You can scale this down to a beaker full of water (to the rim). then you place a rock in the beaker the rock displaces the water which means there is less water in the beaker. Scale this up to the size of a lake, even though it would be minor there would still be a displacement.

When you go into a bathtub full of water, the water goes up. When you put something that sinks into the bathtub, the water does not go up.

Mass of Boat with stone < Mass of Boat and Stone separately!

So, water level is lower!

Volume/Mass of Boat with stone < Volume/Mass of Boat and Stone separately!

So, water level is lower!

When the stone in the boat, the boat would be pulled deeply into the water, pushing the water level up as the water is displaced by the boat.

When the stone is thrown, the boat rises in the water because the weight of the stone is no longer pulling it down.

The stone displaces more water when it's in the boat, pulling it down, compared to the amount of water it displaces when resting on the bottom of the lake.

We can use extreme cases, the rock has no volume but a total mass of 1gram, the boat is has volume and no mass. When the rock is on the boat, the water level is higher than when there is no boat.

The stone makes the boat heavy and holds it lower In the water. When he got it out the boat raised and the rock went to the bottom.

Perhaps, the most basic thing to note is that once you throw the stone outside, the net weight of the ship decreases( as a part of it has been thrown out), and thus the upthrust immediately pushes it upward. Thus, relative to the ship, the water level has LOWERED( as it has itself risen). However, the catch is that the stone too displaces a certain volume of water. As others have pointed out, since the volume of the stone is negligible compared to that of the ship, ultimately it is the rise of the ship that dominates. Done!

All the things in the lake in total provides constant force to the bottom. At first, the bottom only receives the force from water. Since the rock drops, the bottom directly receives some force from the rock, occupies the force through water. In other words, the force from the water reduces. With the equation for liquid: Force(N)/area(m^2)=pressure(N/m^2)= density(kg/m^3) gravity(N/kg) depth(m), the bottom area, water density, gravity are constant, the Force reduce must be followed by the pressure and the depth reduce.

This problem can be easily solved by considering extreme case. Take the stone to be extremely dense. Then when the stone is on the 🚣, the thing that matters is its mass. But when it is thrown then the main role is played by its volume(which is very less than its density numerically).

Hence, the stone displaces more volume in the first case.

Imagine taking a huge volume of water from the lake and replacing it with a tiny rock of equal mass. The total volume of the lake will decrease, thus lowering the water level. This is analogous to throwing the rock overboard.

A floating body displaces its own weight . A sinking body does not diplace its own weight of water ,it diplaces less than before (that is why it sinks). The stone in the boat was floating . Now the stone is not floating therefore the contribution of the stone is lesser than before .

Hence the water level will go down. (Floating man and floating boat make no difference)

Solution #1:

Before: While the stone was floating along with the boat, according to Archimedes' Principle,

weight of water displaced = weight of the stone

and because the stone is denser than water,

volume of water displaced > volume of the stone .

After: Now clearly

volume of water displaced = volume of the stone .

Therefore the stone now displaces less water than it did before; the water level goes down.

Solution #2:

Before: The buoyant force was sufficient to keep the stone afloat.

After: The buoyant force cannot keep the stone afloat; the stone sinks until some other force (normal force from floor of lake) stops it.

Thus the buoyant force decreased. Since the buoyant force equals the weight of the displaced water, this means that the volume of displaced water decreased; the water level went down.

Treat the overall change in water level as a superposition of two changes ( $v$ denotes volume).

Change #1: The stone becomes fully submerged in water $\large{\Delta v_1 = +v_{stone}}$

Change #2: From Archimedes' Principle, in order to float the stone in the boat, the weight of the water displaced (just to float the stone) must equal the weight of the stone (here, $\rho$ is density). Since the stone no longer floats, this water is no longer displaced.

$\large{W_{stone} = v_{stone} \, \rho_{stone} \, g = v_{w,disp} \, \rho_{w} \, g \\ v_{w,disp} = v_{stone} \frac{\rho_{stone}}{\rho_{w}} \\ \Delta v_2 = -v_{stone} \frac{\rho_{stone}}{\rho_{w}} }$

The overall displaced volume change is:

$\large{\Delta v = \Delta v_1 + \Delta v_2 = v_{stone} -v_{stone} \frac{\rho_{stone}}{\rho_{w}} \\ \\ \frac{\rho_{stone}}{\rho_{w}} > 1 \implies \Delta v < 0}$

Thus, the net effect is that the displaced volume decreases, and the water level lowers.