Archimedes' Principle Buoyancy?

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SUMMARY

The discussion centers on calculating the work done by the buoyant force on a right cylindrical flotation device during its ascent in fresh water. The cylinder has a height of 0.500 m, a face area of 4.00 m², and a density of 0.400 times that of fresh water. The key equations involved include the buoyant force equation, Fb = density * Volume * g, and the net force equation, Fnet = Fb - Fg. To find the work done, participants suggest integrating the force as a function of height above the water, leading to the conclusion that both Archimedes' principle and work equations are necessary for solving the problem.

PREREQUISITES
  • Understanding of Archimedes' principle and buoyancy
  • Knowledge of Newton's laws of motion
  • Familiarity with calculus, specifically integration
  • Basic concepts of density and volume in fluid mechanics
NEXT STEPS
  • Study the derivation and application of Archimedes' principle in fluid dynamics
  • Learn how to apply Newton's laws to fluid mechanics problems
  • Practice integration techniques for calculating work done by variable forces
  • Explore examples of buoyancy calculations involving different shapes and densities
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Students studying physics, particularly those focusing on fluid mechanics, as well as educators looking for practical examples of buoyancy and work calculations.

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Homework Statement



A flotation device is in the shape of a right cylinder, with a height of 0.500 m and a face area of 4.00 m2 on top and bottom, and its density is 0.400 times that of fresh water. It is initially held fully submerged in fresh water, with its top face at the water surface. Then it is allowed to ascend gradually until it begins to float. How much work does the buoyant force do on the device during the ascent?

Homework Equations



Fb=mf*g or Fb = density*Volume*g

The Attempt at a Solution



Fb=dwVg, where dw is the density of the fluid (water).

The net force is the difference between the buoyant force and gravity, Fg=mg=rVg, where r is the density of the cylinder’s material. Since the cylinder’s density is smaller than that of water, the upwards buoyant force has larger magnitude than the downwards gravity force, and the net force will be upwards, with magnitude Fnet=Fb-Fg=(rw-r)Vg.

Using Newton’s law, the net force is Fnet=ma, so the acceleration is upwards, and equal to a=Fnet/m=(rw-r)Vg/m=(rw-r)g/r=(rw/r-1)g=(1/0.4-1)g=1.5g.

i don't know how to answer the question.
 
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The force varies from a large value initially to zero when it reaches its equilibrium point with 60% of its volume out of the water, so you need an expression for the force as a function of h, the height that is above the surface. If you know calculus, you can integrate Fdh from h=0 to h=0.6H. If not, you can sketch the graph of F vs h and the work will be the area under the graph from 0 to 0.6H. Here H is the total height of the cylinder, which I think you can find from the given information.
 
i'm just confused.

what equation should i be using to find the answer. the Archimedes' equation or the equation for work.
 
Both! You need an expression for the force (gravitational + buoyancy) as a function of h (height above the water). Then W = integral of F*dh.
 

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