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Finding density of item in a liquid

  1. Oct 20, 2006 #1
    I am suppose to find the density of a thick syrup-like liquid (does not disperse).
    The only information I have is that when a small portion is poured into a cup of unknown volume, it takes 5 seconds for it to reach the bottom. The cup has a height of 12 cm.

    I am not sure how to approach this problem. When I was doing a FBD, the only thing I can see to put on it is weight, and buoyancy, but I don't have a mass, volume, or density of the unknown heavy liquid.
    The only equation I have that has enough information to fill out is:

    pressure = (initial pressure) + (density)(gravity)(height)

    That would give me the pressure on the unknown liquid at the bottom, which I can't see how that helps me.

    With kinematics, I can find the deceleration necessary, but I can't find the buoyancy force because I don't have the mass.
    How should I start this problem?
  2. jcsd
  3. Oct 21, 2006 #2


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    I assume the cup is full of water before you pour in the unknown liquid, and I assume also you are ignoring any drag effects. So what you have is a constant buoyant force and gravity acting on the blob of liquid. The bouyant force is related to pressure, but by Archemedes principle it is equal to the weight of a volume of water equal to the volume of the unknown liquid.

    Write a Newton's second law equation for the blob

    ma = F

    where m is the mass of the blob, a is the constant acceleration that you calculate from the information given in the problem (5 seconds to fall 12cm starting from rest) and F is the difference between the weight of the object and the weight of the water it displaces.

    Your equation will involve the mass of the blob and the mass of an equal volume of water. You do not need to know the mass of the blob or its volume numerically. If you divide each mass by the volume of the blob, the equation will become an equation relating the densities (m/V). You can solve for the density of the unknown liquid in terms of the density of water. Then use the known density of water to complete the calculation.
  4. Oct 22, 2006 #3
    I'm still trying to get my head around this.

    So it sounds like:
    F = ma = F
    (mass of unknown)*(acceleration) = ((mass of unknown)-(mass of displaced water))*g

    a = 0.0096 m/s^2 (1/2at^2 = .12 m)

    (density of unknown) = (mass unknown)/(volume of unknown)
    (density of water) = (mass of displaced water)/(volume of unknown)
    (volume of unknown) = (mass of displaced water)/(density of water)

    d(water) = 1.0g/cm^3
    I seem to be stuck at this point, since I don't have the mass of the displaced water. I think I did something wrong somewhere.

    I tried just putting in numbers to see what would happen:
    1.0g/cm^3 = 1.0g/1 cm^3
    volume of unknown = 1 cm^3
    mass of displaced water = 1.0g

    (mass of unknown)*(acceleration) = ((mass of unknown)-(mass of displaced water))*g
    (mass of unknown)(0.96 cm/s^2) = ((mass of unknown)-(1.0g))9.8
    (0.0096 m/s^2) = (1 - (1.0g/mass of unknown))(9.8m/s^2)
    (0.0096 m/s^2)/(9.8m/s^2) = (1 - (1.0g/mass of unknown))
    9.8E-4 = 1 - (1.0g/(mass of unknown))
    -.999 = -1.0g/(mass of unknown)
    (mass of unknown) = 1.0g
    Which means I just ended up with basically the exact same density as water. Which wouldn't make sense at all.
  5. Oct 22, 2006 #4


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    The volume of the unkown is the same as the volume of the water it displaces. Call that simply volume. Divide your equation

    (mass of unknown)*(acceleration) = ((mass of unknown)-(mass of displaced water))*g

    by that common volume and replace the ratios (mass/volume) by densities. Using the acceleration you calculated and the density of water you will have an equation for the density of the unknown.
  6. Oct 22, 2006 #5
    So when I'm dividing by the volume, this isn't a normal math operation where I have to apply it to all terms right? It only applies to the mass? I am kind of just substituting it looks like.

    p(u)*a = (p(u)-p(w))*g

    So I guess if that was the case, the answer is real near the same density of water. I got 1.001 kg/m^3 vs. 1.000kg/m^3 for water. Actually I think the way where I just started with the density of water to fill in mass and volume yields the exact same answer I just need to keep a lot of significant figures.
  7. Oct 22, 2006 #6


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    It absolutely is a normal math operation. When you divide a product of factors by something, you divide the whole product once and only once. Equivalently, you can divide any one of the factors in the product before you multiply.

    The blob is sinking quite slowly, so the density is very nearly the same as that of water.
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