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Density/Buoyancy question

Discussion in 'MCAT Study Question Q&A' started by aint settling, Aug 11, 2011.

  1. aint settling

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    The true mass of a rock is 10g. When submerged in water (D=1000kg/m3) the apparent mass of the rock is 5g.

    What is the density of the rock?


    Please help me with this question? Thanks!!!
     
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  3. geeyouknit

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    Let's say force = F; b= buoyancy, w=weight

    On the object, the overall force is Fnet;
    Fnet = Fb - Fw;
    Fw= mass * g = 0.01 kg * g (pointing down)
    Fnet is related to the apparent mass = 0.005 kg* gravity (pointing down)
    Fb = the force of buoyancy is the weight of the water displaced =
    = mass of water * g (pointing up)
    Cancel out gravity in all terms ==> -0.005 kg = Masswater - 0.01 kg
    Mass of water is 0.005 kg; Use density to get vol of water, 5e-6
    which gives volume of rock; 0.01 kg / vol of rock = density

    Nevermind: I did most of it right, I just forgot that the net force is pointing down, so Fnet should be negative.
     
    #2 geeyouknit, Aug 11, 2011
    Last edited: Aug 11, 2011
  4. inaccensa

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    is the answer 5*10^-6 m3? oh yeah, sorry I wrote the volume...oopss hope not to do that on the test, density =10^10-3/5*10^-6 = 2000kg/m3
     
    #3 inaccensa, Aug 11, 2011
    Last edited: Aug 11, 2011
  5. costales

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    This object-in-water problem is common. You can just take a ratio and multiply by the water density:
    specific gravity
    = true mass/(true mass-apparent mass)
    So it's 10/(10-5) = 2, and the answer is 2000 kg/m3.
     
    #4 costales, Aug 11, 2011
    Last edited: Aug 11, 2011
  6. docntrainin

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    I'm confused as well, I thought the problem would be solved like this:

    Fb = pfluid x g x Vsubmerged
    (10 g - 5 g/ 1000) (10) = 1000 x 10 x Vsubmerged (aka volume of rock)
    5/1000 = 1000 x V submerged
    V submerged (Volume of Rock) = 5e-6

    Density = 2000 kg/m3 (AHHHH STUPID MISTAKES LIKE THIS!!!!)
     
    #5 docntrainin, Aug 11, 2011
    Last edited: Aug 11, 2011
  7. inaccensa

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    Well, I did
    Weight air - Apparent weight = weight of the displaced fluid, although costales's formula is the easiest way, which i was not aware of.

    There is something similar density object/ density fluid = fraction submerged ---> floating liquid
     
  8. Majik

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    Fnet = Normal Force + Bouyant Force - Weight

    Note: Normal Force & Bouyant Force both act upwards; Weight acts down.
    Provided the object is touching the sea floor, the net force is zero.
    The apparent weight is equal to the normal weight.

    Therefore, rearranging we get:

    Weight - Bouyant Force = Normal Force

    Weight = 0.1N
    Normal Force (Apparent Weight) = 0.05N

    We can solve for Bouyant Force now:
    Weight - Normal Force = Bouyant Force
    0.05N = Bouyant Force = pVg (p=density of water;V=volume of object)
    0.05N / (1000kg/m^3)(10m/s^2) = Volume
    5x10^-6 m^3 = Volume

    Density of the object equals: mass / volume
    Density: 10x10^-3 kg / 5x10^-6 m^3
    Density: 2000 kg/m^3
     
  9. muhali3

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    If an object is floating, the relative density is the % submerged.

    If sunken/submerged, the relative density is the ratio of the weight to the buoyant force (W/B).
     
  10. Swagster

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    Forgive me BerkTeach for stealing your method, but I have to say it's magic.

    First, the object sinks, so we know that the weight/buoyant force ratio is equal to the relative densities (rock over water).

    Second, the weight is 10 and the buoyant force (apparent weight loss) is 5, so the W/B ratio is 2. This means that the rock is twice as dense as the water.

    For the exact breakdown of why W/B = (object density)/(water density) you need to read the page on buoyancy from the TBR book. Their approach is pure genius.
     
  11. Swagster

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    Did you take the TBR class or just use the books?
     
  12. BerkReviewTeach

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    It's not my method. That's all BR on that one. Gotta agree though, it's pretty genius the way they do certain math tricks.
     
  13. muhali3

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    relative density questions also seem to come up pretty often, so those tricks really do save a lot of time.
     

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