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B About buoyancy and 2 equations

  1. Jan 26, 2017 #1
    1) My teacher says that the apparant weight of an object in water (floating,sunken,submerged etc) is equal to it's actual weight-buoyant force acting on it.

    That is, wt (ap)=wt (ac)- F(b)

    Where wt (ap) = apparant weight,
    weight (ac)= actual weight.
    And f (b) = buoyant force

    2) But..., if an object is floating or is submerged under water, it's weight is completely supported by the buoyant force acting on it. So shouldn't the buoyant force be equal to the apparant weight of the floating object.?
    i.e., f (b)=wt (ap)

    The equations in 1) and 2) contradict each other.

    Please tell me where I am wrong.
  2. jcsd
  3. Jan 26, 2017 #2


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    Staff: Mentor

    What about a stone submerged under water? Is it fully supported by buoyancy?
  4. Jan 26, 2017 #3
    A submerged object (resting on the bottom) is not completely supported by the buoyancy force , some support comes from the sea bottom , and this support from the sea bottom is the 'apparent weight' ...
    If floating there is no apparent weight ...
  5. Jan 26, 2017 #4


    Staff: Mentor

    An object submerged in water will sink to the bottom if its actual weight is more than the buoyant force.

    If it floats below the surface but doesn't sink, then the weight and buoyant forces exactly balance.

    If the weight is less than the buoyant force, it floats to the surface and only part of the object will be submerged, like a boat.
  6. Jan 26, 2017 #5


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    Gold Member

    Just to complete the answers with specific reference to where you went wrong:


    No, in this case the buoyant force is equal to the actual weight. That is why the object is in equilibrium and its apparent weight is zero. But apparent.. for whom? For you, for example: you don't need to make any effort to avoid that the object sinks and if you put a balance scale under the object it will read zero, just as if it were supported by a rope from a hook. But for the water itself, things are different: the water has suffered the actual weight (well, to be accurate, the water has suffered contact force equal to the object's actual weight and has reacted by applying another contact force of the same magnitude, i.e. the buoyant force).
  7. Jan 27, 2017 #6
    I knew there was something off about the submerged part.
  8. Jan 28, 2017 #7
    Restating a comment from above, if an object is completely submerged and is not moving vertically up or down then it is in equilibrium, the buoyant force equals the actual weight. The apparent weight is zero.
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