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Questions on Archimedes principle

  1. Oct 26, 2011 #1
    1-A hollow ball is suspended in water,has a larger density than the density of the water..how?

    Fb=Fg (okay)
    ρ(of water)* (Vol of ball substance+Vol.of the cavity)=ρ(of the ball) * vol of the ball substance)
    (not okay)
    I don't understand how we calculate the buoyant force
    The buoyant force should be calculated according to the equation: Fb=ρ(of water)*Vol of the substance*g
    not Fb=ρ(of water)* (Vol of ball substance+Vol.of the cavity)*g



    2-when a body is suspended in water,does the level of water rise up when we try to move this body to the bottom?and does the buoyant force increases as we go down?

    Thanks in advance..
     
  2. jcsd
  3. Oct 26, 2011 #2

    Doc Al

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    Staff: Mentor

    What do you mean by 'suspended'? Is it being held by a string or other support?

    The buoyant force equals the weight of the displaced fluid. In this case: ρ(of water)*(Vol of ball)*g
     
  4. Oct 26, 2011 #3
    No, it should be "Volume of water displaced by submerging the object". See Archimedes' law.
    If the object is homogenous and completely submerged, then your formula works too. But is a quite special case. Here you do not have a homogenous object.
     
  5. Oct 30, 2011 #4
    Nope, but Fg=Fb
    see this image

    [PLAIN]http://img193.imageshack.us/img193/5597/unlewd.jpg [Broken]
    [/PLAIN]
    so you mean that wheather the ball is hollow or not,it displaces the same amount of water-since the external volumes is the same???

    yeah,okay so is the volume of the displaced water equal to the volume of the substance?since there's a cavity in the ball?

    don't forget my second question,i really need to know the answer

    Thanks in advance
     
    Last edited by a moderator: May 5, 2017
  6. Oct 30, 2011 #5
    The volume of the displaced water in your case is the volume (external) of the ball, when the ball is completely submerged. If by substance you mean only the walls of the ball, then no. The volume of "substance" is much less than the total volume of the ball. If the ball has a hole in it so that the water can enter and the air can exit, then the volume displaced will be equal to the volume of "substance".

    2. The level will rise until the ball is completely submerged. After that it does not change.
     
  7. Oct 31, 2011 #6

    Doc Al

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    Staff: Mentor

    So you have something of neutral buoyancy, that remains in equilibrium when submerged.

    Right.

    The fact that the ball contains a cavity is irrelevant to the amount of fluid it displaces.

    The buoyant force equals the weight of the displaced fluid. Does that change as the object is lowered? Once the object is completely submerged, why would the water level change?
     
  8. Oct 31, 2011 #7

    sophiecentaur

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    Gold Member

    If the ball cannot deform (which would introduce another level of complication) then only its volume would count towards the buoyancy force. However, an uneven distribution of mass or irregular shape can cause the floating / suspended object to rotate, if it is free, so that its centre of mass is at a point which minimises the gravitational potential. This is why a stick tends to lie on its side in the water and not upright - despite experiencing exactly the same upthrust all the time.
     
  9. Nov 8, 2011 #8
     
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