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Buoyant Force and FBD

  1. Oct 22, 2007 #1
    I am not sure why this is not clicking...

    An object hangs from a spring balance. The balance registers 30 N in the air and when it is immersed in water it reads 20 N. What is the buoyant force on this object? Draw a FBD to solve this.


    [tex]\uparrow F_b[/tex]
    [tex]\downarrow F_W[/tex]

    Do I need to take into account any other forces?

    It is multiple choice a) 20 N up
    b) 10 N up
    c) 10 N down
    d) 20 N down

    I am pretty sure it is not c or d......as that wouldn't seem like a "buoyant" force:rolleyes: But I am not sure how to set up my Newton's 2nd equation?

    Is it just [tex]\sum F=F_{weight}-F_{buoyant}=20[/tex]
    so F_buoyant=10N?

    It just seems strange that we spent all day taking atmospheric pressure into account, but now we do not...
    Last edited: Oct 22, 2007
  2. jcsd
  3. Oct 22, 2007 #2


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    Well, that's correct, but your free body diagram technically has 3 forces acting on the object: its weight down (30N), The buoyant force up (F_b), and tension in the scale's cord acting up (20N). So your FBD equation using Newtyon 1 is [tex] F_{net} = T + F_b -W = 0 [/tex] from which 20 + F_b -30 = 0, that is, F_b = 10N up, whuch is what you got, but don't take shortcuts. Atmospheric pressure for all practical purposes cancels out of the equation, because the difference is rather small between the top and bottom of the object.
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