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Submerged sphere in a beaker

  1. Oct 29, 2007 #1
    http://session.masteringphysics.com/problemAsset/1011221/19/SFL_ap_6a.jpg

    A cylindrical beaker of height 0.100 {m} and negligible weight is filled to the brim with a fluid of density rho = 890 {kg/m}^3. When the beaker is placed on a scale, its weight is measured to be 1.00 {N}. View Figure

    A ball of density rho_b = 5000 {kg/m}^3 and volume V = 60.0 {cm}^3 is then submerged in the fluid, so that some of the fluid spills over the side of the beaker. The ball is held in place by a stiff rod of negligible volume and weight. Throughout the problem, assume the acceleration due to gravity is g = 9.81

    ------> What is the reading W_2 of the scale when the ball is held in this submerged position? Assume that none of the water that spills over stays on the scale.
    ________________________________________________________________
    I know that the weight of the ball is 2.94 N, and that the beaker is 1 N... And that pressure= F/A and pressure = p_0 + density of liquid(gh) ... Help??
     
  2. jcsd
  3. Oct 29, 2007 #2

    Doc Al

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    How does the water pressure change when the ball is submerged?

    Another way to analyze this setup is to examine all the forces acting on the beaker and its contents. (Start by analyzing the forces on the ball. What must they add to?)
     
  4. Oct 30, 2007 #3
    I don't understand ... Pressure force on the bottom is greater in order to balance gravity...
     
  5. Oct 30, 2007 #4

    learningphysics

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    What is the pressure at the bottom of the beaker before the ball is submerged?

    What is the pressure at the bottom after the ball is submerged?
     
  6. Oct 30, 2007 #5
    Ok, so pressure at the bottom of the beaker before the ball is submerged ... well P=F/A and A of the beaker I don't know, but I know that the weight of the beaker+fluid = 1N. A=density(V) = 890(V) ... I'm stuck
     
  7. Oct 30, 2007 #6
    after the ball is submerged, P=F/A and A of the ball is just (5000(60/(100^3)))=0.3m^2 ?? and force is just the weight which equals 2.94? so pressure is then 98 pa?? That sounds .. wrong
     
  8. Oct 30, 2007 #7

    learningphysics

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    Another way of putting it... does the pressure change at the bottom of the beaker, from before and after?

    The force on the bottom of the beaker is the pressure due to the water*area...

    The normal force upward on the beaker = force due to pressure + weight of beaker...

    this normal force is what the scale measures...

    how do either of these 2 things change... "force due to pressure" or "weight of the beaker"

    do they change at all?
     
  9. Oct 30, 2007 #8
    Ah, ok, no, replacing the fluid volume with the object does not change the fluid ... so pressure force stays the same... By the way, how do you calculate the pressure?
     
    Last edited: Oct 30, 2007
  10. Oct 30, 2007 #9

    learningphysics

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    exactly. everything's the same. it's a trick question. :wink:
     
  11. Oct 30, 2007 #10

    learningphysics

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    you don't need it... you know the pressure force is the same... the weight of the beaker is the same... so the scale measures the same force... 1N.
     
  12. Oct 30, 2007 #11
    I just looked it up and the answer key said 873 pa and i was wondering where the heck that came from. But what you're saying makes sense...
     
  13. Oct 30, 2007 #12

    learningphysics

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    yeah, that's just rho*g*h = 890*9.81*0.1 = 873 pa
     
  14. Oct 30, 2007 #13
    question: when is this p_0 101.3 kpa and when is it 0 ...
     
  15. Oct 30, 2007 #14

    learningphysics

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    it depends on what exactly the question was asking for... did the question ask for the pressure at the bottom of the beaker, from inside...

    was it asking for the difference in pressure between inside and outside?
     
  16. Oct 30, 2007 #15
    Well when you were calculating the pressure at the bottom of the beaker before the ball was submerged and after the ball was submerged, you used p_0 = 0 in the equation p=p_0+rho(g)(h) for both cases
     
  17. Oct 30, 2007 #16

    learningphysics

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    yeah... I should use p_0 = 101kpa both times. we also need to remember that there's a pressure outside the beaker acting upwards on the bottom of the beaker of 101kpa.

    The 101kpa inside and outisde cancel... so that's why I didn't include it...

    When you consider the force on the beaker...

    Fnormal - (pressure inside)*area + (pressure outside)*area - (weight of beaker) = 0.

    Fnormal - (101kpa + rho*g*h)*area + (101kpa)*area - weight of beaker = 0

    see now that the 2 101kpa's cancel

    Fnormal - rho*g*h*area - (weight of beaker) = 0

    giving the same equation as before...
     
  18. Jun 15, 2008 #17
    I am confused about this question. So what is the answer?
    Do I use the formula, W= rho * V* g?
    I know everything except for V... how can I find V when i only have the height?

    Thank you
     
  19. Jun 15, 2008 #18

    Doc Al

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    What do you think the answer is?
    You don't need that to answer the question.
     
  20. Jun 15, 2008 #19
    ohh the answer is 2.94N because nothing changes.
     
  21. Jun 15, 2008 #20
    nevermind. that is wrong. i just tried it.

    hmm looks like i'm really not understanding buoyancy! =(
     
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