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Solving for volume and density

  1. Dec 3, 2015 #1
    <post removed from here by mentor, but copied to a later post>
    Last edited by a moderator: Dec 4, 2015
  2. jcsd
  3. Dec 3, 2015 #2
    1. The problem statement, all variables and given/known data
    So, just took an exam, one of the questions on the exam was the following:

    An object is attached to a scale and submerged in water (p=1000 kg/m^3) and it's weight reads 34N. The same object is then submerged in oil (p=830 kg/m^3) and it's weight reads 34.7N.

    a.) What is the objects volume?
    b.) What is the objects density?

    What I know: (p= m/v) (m=weight/g) (V=Fbuoyancy/ p(g)) (Fbuoyancy= p(v))

    For the life of me I couldn't answer this question, could someone please help, I will get the chance to make corrections.
    2. Relevant equations
    What I know: (p= m/v) (m=weight/g) (V=Fbuoyancy/ p(g)) (Fbuoyancy= p(v))

    3. The attempt at a solution
    V= T1+(p*g)-mg

    This is the clue my proffesor gave us prior to exam. However when I plugged in numbers it seemed way off.
    I know how to derive density with weight in air also known, but I don't have a clue where to begin with so little data.
  4. Dec 3, 2015 #3


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    Start with Archimedes' principle. If the object has volume V, what weight of oil will it displace? What weight of water?
  5. Dec 3, 2015 #4


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    Fb = ρgV


    Fb = buoyant force, N
    ρ = density of the fluid, kg/m3
    g = acceleration due to gravity, m/s2
    V = submerged volume of the body, m3

    You should always check to make sure your formulas give consistent units.
  6. Dec 3, 2015 #5
    No. Weight is a force and its unit IS a Newton. Grams and kilograms are units of MASS, not weight. So the statement of the problem is correct. Also, the density and volume are properties of the sample, dependent on temperature, not on what you dip the sample in.
  7. Dec 3, 2015 #6
    Don't think of it as a formula into which you somehow plug in numbers. Think of the process and the principle that governs the process. If you submerge an object in a liquid, the principle of Archimedes has something to say about what happens. Start from there. State the principle, then use the data in the principle to find what you need.
  8. Dec 3, 2015 #7
    Archimedes' principle indicates that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces.

    In this case the object IS fully submerged, however I don't have a weight of fluid dispersed. I do however have weight while in water, and weight while in oil with a difference of .7N. I can find mass while in the water and oil ( m=W/g ) however I can't seem to figure out volume without Fbuoyancy (Fb= mg = (p*V*g)).
  9. Dec 3, 2015 #8
    It definetely seems my professor goes out of his way to derive problems that google or youtube can't find. lol

    I've found every other way to approach density problems except this way.
  10. Dec 3, 2015 #9
    The text book doesn't offer anything about this approach either. :eek:
  11. Dec 3, 2015 #10


    Staff: Mentor

    Take the volume of the sample as variable V, W its weight in air. Then you know the weight of water it displaces depending on V. The same with the oil. So you get two values of reduced weight and you already know the difference: 0.7N. All depend on V. 2 equations, 2 variables. Density follows from V and W plus g.
  12. Dec 3, 2015 #11
    I gotcha, I do remember my professor saying something along the lines of comparison of the separate equations. Let me give that a shot, I'll get back at ya.
  13. Dec 3, 2015 #12
    Ok so this is what I got for just the submerged in water part. I know that if I can find volume of the object I can find density (d = m/V )

    V= T + (p*g) - mg

    wouldn't tension be equal to mg and cancel, therefor leaving (p*g) this number would seem to large though

    V= (1000 kg/m^3 * 9.8 m/s^2)
    V= 9800
  14. Dec 3, 2015 #13
    forgot units: V = 9800 m^3
  15. Dec 3, 2015 #14
    Archimedes' principle indicates that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces.

    The object obviously has volume and has a weight of 34N (water) and 34.7N (oil). To answer the second and third question, what weight in oil will it displace? The object will displace the same volume as the water it displaced, however how do I determine weight of displaced fluid when all I know is weight of the object while submerged?
  16. Dec 3, 2015 #15


    Staff: Mentor

    To be honest, that confuses me. Why is there tension? What is p*g? Do the units fit together? If p should be a density, then you'll have ##m^3## as numerator on the left side and as denominator on the right.

    In the first step you could calculate with the differences. Let W denote the weight of your sample, V its volume, m its mass, g earth's acceleration and d its density.
    You know the differences in density and the difference in weight, so ##0.7 \cdot N = 170 \cdot kg \cdot m^{-3} \cdot g \cdot V##. From there on you can go ahead with V to calculate weight (using the water measurement of weight, weight of displaced water + 34N), mass and density.
    I really recommend you to carry all units explicitly within your equations. Thus you know whether it comes out right!
  17. Dec 3, 2015 #16
    • So... V=170 *kg *m^-3 * (9.8 m/s^2) all divided by .7
  18. Dec 3, 2015 #17


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    You know two important pieces of information: the mass of the object and its volume do not change when the object is submerged in oil or in water.

    You can write an equation to determine the mass of fluid displaced by using the volume of the object and the density of the fluid. In fact, you already had a good equation, but you wrote it in such a manner to obscure its usefulness:


    It's better to write this:

    Wfluid = ρgV - mg, where Wfluid is the weight of the object when it is submerged in a fluid having a density ρ. V is the volume of the object.

    Once you write this equation for one fluid, you can write it for the other fluid. The difference of the weight of the object in the two fluids depends only on the difference in the density of the fluids, since the volume of the object doesn't change.

    Once you figure out the volume of the object, then you can use Archimedes' Principle to find the mass of the object. Your Uncle Bob then drops by for a visit, and knowing both the mass of the object and its volume, he leaves you with the density.
  19. Dec 3, 2015 #18
    ... all divided by .7 * N
  20. Dec 3, 2015 #19
    = 2380 m^3
  21. Dec 3, 2015 #20


    Staff: Mentor

    No! The other way round. But with N it's better and since ##N = kg \cdot m \cdot s^{-2}## only ##m^3## will be left. (Hint: it's a little lighter than lead and heavier than iron; if I made no mistake.)
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