Archimedes principle boat displacement

AI Thread Summary
The discussion revolves around understanding Archimedes' Principle and its application to a boat submerged in water. The submerged volume of the boat, 35m^3, is equal to the volume of water displaced, which allows for the calculation of the boat's mass using the equation pV = m. Participants clarify that the mass calculated from the displaced water directly corresponds to the mass of the boat when in equilibrium. The conversation also touches on the concept of neutral buoyancy, where an object can be fully submerged without touching the bottom if its density matches that of water. Overall, the thread emphasizes the relationship between the mass of the displaced fluid and the mass of the floating object, clarifying common misconceptions.
Drizzy
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Homework Statement


35m^3 of a boat's volume is submerged under water. What does the boat weigh? The boat is 14 metres long.

Homework Equations


pVg=mg

The Attempt at a Solution


pVg=mg
pV=m
m=pV=1000*35

This is what my teacher said, but I don't understand it. The submerged volume of the boat is equal to the volume of the displaced water. So by calculating m we are getting the water's mass, which is not equal to the boats mass. or is it?
 
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The "m" that appears in your force balance equation is the mass of the boat. So, of course whatever you calculate for "m" has to be the boat's mass. You are probably a little confused because \rho_{w} V is also the mass of the water that has been displaced.
But if you trust your mathematical manipulations, that result is exactly the implication of Archimedes Principle: that for an object partially submerged (floating) in a fluid, the mass of fluid displaced is equivalent to the mass of the object itself.
 
this principle is confusing me so much :(

"Archimedes' principle indicates that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces."

Doesn't that mean that m is the waters mass?
 
Archimedes' Principle is essentially
F_{up} = m_{w}g = \rho_{w} V g

The first equation you have up there is actually not so much a statement of Archimedes' Principle but rather a force balance equation, that the upthrust acting on the boat is equal to the weight of the boat:
F_{up} = F_{g} = m_{boat}g
 
So the boat's mass is equal to the waters mass and the waters volume is equal to the boats volume that is under water?
 
Drizzy said:
the waters volume is equal to the boats volume that is under water?
Yes, it would be strange if it wasn't so, wouldn't it?
Drizzy said:
So the boat's mass is equal to the waters mass
In equilibrium, yes. That is what applying Archimedes' Principle tells us.
 
oh now I get it! what if the object is kinda floating but still not. it is fully under water but it is not touching the ground/bottom. The boyant force is still equal to mg if the ibject is at rest rght?
 
Well, technically yes. Although such a system is not a stable equilibrium.
 
This situation, an object floating so that it is completely underwater but not touching the bottom, can happen only if the object has exactly the same density as water.
 
  • #10
HallsofIvy said:
This situation, an object floating so that it is completely underwater but not touching the bottom, can happen only if the object has exactly the same density as water.
This is only true if the object has uniform density. If the object is hollow, for example, then no. The situation that Drizzy is describing is called "neutral buoyancy," and is attained when the weight of the displaced water is exactly equal to the weight of the fully submerged object.

Chet
 
  • #11
If the density of the object is NOT uniform, we have to talk about its average density and what I said is still true.
 
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