1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

2 different densitys

  1. Apr 25, 2006 #1
    50 cm of oil, on top, (obviously) and 120 cm of water down below.

    to find the density at the bottom would you find the density of both the 50cm of oil and the 120cm of water and add them together? thanks
     
  2. jcsd
  3. Apr 26, 2006 #2

    andrevdh

    User Avatar
    Homework Helper

    Assuming that you need the combined density:
    [tex]\rho_C=\frac{m_o+m_w}{V_o+V_w}[/tex]
    using the expressions for density
    [tex]=\frac{\rho_oV_o+\rho_wV_w}{V_o+V_w}[/tex]

    [tex]=\frac{\rho_o}{1+\frac{V_w}{V_o}}+\frac{\rho_w}{1+\frac{V_o}{V_w}}[/tex]
    Assuming that the oil and water is in a constant diameter, flat bottomed container you can use
    [tex]V=Al[/tex]
    for the volumes in the final equation.
     
  4. Apr 26, 2006 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    "Find the density at the bottom" makes no sense. If the oil is floating on the water then the density at the pottom is simply that of the water. Do you mean to find the total force or pressure on the bottom? In that case you would add the weight of the oil and water. While you could find an average density, as andrevdh showed, I think it would be easier just to find the weight of each separately. Since you give height but not an area, I assume you want to find the pressure on the bottom. Okay, assume a 1 sq. cm bottom, so the volume of oil is 50 cc and multiply by the density of oil to find the mass of the oil. The volume of water is 120 cc so multiply that by the density of water (1 g/cc) to find the mass of water. Add together and multiply by g= 981 cm/sec2 to find the total weight in dynes. Since that is the total weight on a 1 sq cm base, that is the pressure on the bottom in dynes/cm2.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?