Understanding the buoyant force in archimedes principle

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The buoyant force acting on an object is equal to the weight of the fluid displaced, which can occur even when the object is floating. When an object is floating, it displaces a volume of fluid that is less than its own volume. If the object is fully submerged, it displaces more fluid, resulting in a greater buoyant force. The same formula, p*V*g, can be applied to calculate buoyant force in both floating and submerged scenarios, as long as the correct volume of fluid displaced is used. Understanding this principle clarifies the relationship between buoyancy and fluid displacement.
suzukits
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I'm really confused about the buoyant force.
What I've understood is that the buoyant force of an object is equal to the volume of the displaced fluid even if it is floating (not fully immersed in the fluid).
How can this be possible?
The volume of the object is the amount of water displaced when THE WHOLE OBJECT IS IMMERSED right??

I would be really thankful if someone could clarify this
Thank you!
 
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The buoyant force equals the weight of the displaced fluid. When a body floats, the volume of displaced fluid is less than the volume of the object itself. If that same body is fully submerged, the buoyant force will be greater since it will displace more fluid.
 
Oh okay, and I can use the same formula for both cases? (p*V*g)
 
suzukits said:
Oh okay, and I can use the same formula for both cases? (p*V*g)
Sure. As long as you use the correct volume of fluid displaced.
 
Thank you so much!
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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