A boat is loaded with heavy statues on a small lagoon. The boat overturns and the statues sink to the bottom of the lagoon. The boat, now empty, displaces less water than when it was full of cargo. The question is: Will the lagoon's water level rise or fall because of the statues on its bottom?

Question

anonymous · Answer

I think it will rise.

jamesj · Answer

No.  Think this through.  While a statue is in the boat, it displaces water equal to its weight.  The density of water is less than the density of the statue.

Now ... what happens when a statue is dropped in the water?  Compare the volume of the water displaced when the statue is in the boat vs. when the statue is in the water.

anonymous · Answer

i think ARSHMIDES PRINCIPLE apply on it!!!!!!!

jamesj · Answer

It is called Archimedes Principle and that is the theory you need to understand the water displaced when the statue is in the boat.  But that is not the complete answer.

anonymous · Answer

a submerine works on this principle of arshmides

anonymous · Answer

when water filled on submerine tank then it move down and when we want to take it on the surface then we empty the tank,this idea can be useful for the above question

jamesj · Answer

@josh, making sense?

anonymous · Answer

Yes, I think so. So while they are still in the boat, the statues displace water equal to their  weight, as you said, and when the statues are submerged they displace water equal to their volume.. So their volume must displace less water, right?, because they are affected by the gravity of water when they are submerged??

anonymous · Answer

which means the water level would fall

jamesj · Answer

Exactly

anonymous · Answer

Thank you :)

anonymous · Answer

@jamesJ 
why should we take buoyant force into account when we have water displaced by statues (having same volume)
the volume of water getting displaced (by statue+boat) is going to remain same before and after sub=merging
i am confused as to why the water level would fall because of decrease in apparent weight whereas the volume of boat+statue(determining the dispplaced water) is going to remain the same....

jamesj · Answer

While a statue is in the boat, its weight is added to the weight of the boat and additional water must be displaced vs. if the statue were on the ground.  

Now, the volume of the water displaced by that statue is greater than the volume of statue itself, because we're assuming the density of the statues is greater than that of water; that is why the statues sink.

Hence when the statues are in the water, the volume of water displaced by the statues is less than the volume of water displaced when the statues were in the boat.

anonymous · Answer

but according to archimedes principle or normally how can the volume of statues increase or decrese while in water(volume only depends upon the dimensions of statue which do not change.....)
while being above in the boat the statues alone displace= weight of water displaced=density of water*g*volume of statue
the same thing is applied when they are immersed in water right?

jamesj · Answer

When the statue is on the bottom of the lagoon, there are three forces acting: gravity, buoyancy **and the normal force from the bottom of the lagoon**.  It is the last force that is new vs. the statue being in the boat, and that is the reason why the buoyant force of the water is not equal to the entire weight of the statue itself.

jamesj · Answer

Further, when the statue is dropped in the lagoon, the very reason it accelerates down is because the net force acting on it is down: the buoyant force is not sufficient to counter-act the force of gravity.

jamesj · Answer

If you're not convinced by this, do the experiment yourself with a glass/beaker of water, a small plastic cup as a boat, and pebbles for statue weights.

anonymous · Answer

but there is buoyant force acting on the boat to allow it to float too?
that will have an effect on the statues as well? buoyant force====weight of water displaced by them?

anonymous · Answer

yeah will certainly try that!

jamesj · Answer

Let W be the weight of the boat and w the weight of a statue of mass m, i.e., w = mg.  When the statue is in the boat, the weight of water displaced is
W + w

and hence the volume of water, V, displaced satisfies
$$ W + w = mg = (\rho_{water} V)g $$
and hence
$$ V = \frac{W + w}{\rho_{water} g} $$

Now, when the statue is in the water, the volume of water displaced by the boat is
$$ V_1 = \frac{W}{\rho_{water} g} $$
but the volume of water displaced by the statue is
$$ V_2 = \frac{ m}{\rho_{statue}} = \frac{mg}{\rho_{statue}g} = \frac{w}{\rho_{statue}g} $$

You now show that
$$ V > V_1 + V_2 $$

anonymous · Answer

everything looks pretty but how did V2=mg/density of statue*g
instead of V2=mg/densitry of water*g ?
taht is disobeyal of archimedes right?

jamesj · Answer

Because the statue has mass $ m $, volume $ V_2 $ and density $ \rho_{statue} $.  By definition of density
$$ \rho_{statue} = \frac{m}{V_2} $$
this volume MUST be the volume of water displaced by the statue.  It has nothing to do with Archimedes principle.

anonymous · Answer

oh my god!
i was totally confused thinking mg(weight)=weight of water displaced while actually the body isnt floating for the two to get cancelled!
thank u for ur patient responses
i shud concentrate more in the future !