1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Density of a Bar of Soap/ buoyancy

  1. Sep 30, 2010 #1
    1. The problem statement, all variables and given/known data

    A rectangular bar of soap floats with 3.5 cm extending below the water surface and 1.5 cm above. what is its density?

    2. Relevant equations
    p(rho)=m/v



    3. The attempt at a solution
    I dont know how to compute the volume since we are given only two lengths.
    Athough, i think a better way to approach this problem will be with buoyancy.

    To start,can we consider this block to be in equilibrium since it is floating? I am confused because it is also partially submerged.
     
  2. jcsd
  3. Sep 30, 2010 #2
    Yes. This question is solved using the idea of buoyancy and Archimedes Principle.
    Here's a way of starting...
    Let the area of the top and bottom surface be A m²
    The volume of the soap is therefore 0.05A m³ [0.05 is the height/thickness of the soap]

    Now because the soap is floating, its weight is exactly balanced by the upthrust due to the weight of the volume of water it is displacing.
    Weight = mass x g
    Mass = volume x density

    Does this help you to get started?
     
  4. Sep 30, 2010 #3
    how did you get .05A?

    So if it is floating, mg=Buoyancy.
    mg=density x volume x g

    how do i get the mass if i dont know the density?
    Sorry, there are two unknowns still.. density and mass. i dont know how to get either with just the volume given
     
  5. Oct 1, 2010 #4
    If there is 3.5cm of soap below water and 1.5 above, then the thickness of the soap is 3.5 + 1.5 = 5cm
    5cm = 0.05m

    Volume is area of cross section times length/thickness
    If the area of cross section (of the top and bottom of the soap) is A then the volume of the soap is 0.05A

    When the soap is floating, only 3.5cm (0.035m) is below the water.
    This means that the volume under water is 0.035A
    The volume of water displaced is V=0.035A
    The mass of water is volume V times density of water ρw
    The weight of water is mass times g
    w x 0.035A x g
    This equals the buoyancy force.

    This force must be equal to the weight of the soap.
    The weight of the soap is its volume (0.05A) times its density.
    If you equate the buoyancy force to the weight of the soap you will get a value for the density of the soap.
    A and g cancel out. You need a value for the density of water.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook