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Archimedes Principle homework

  1. Nov 12, 2011 #1
    1. The problem statement, all variables and given/known data

    A piece of sealing-wax weighs 0.27 N in air and 0.12 N when immersed in water. Calculate :
    a) It's relative density
    b) It's apparent weight in a liquid of density 800 kg/m^3




    The attempt at a solution
    I did part a) I got 1.8 I have no idea how to attempt part b) this part of physics is hard my teacher at school didn't teach it well, all he did was give notes. :(
     
  2. jcsd
  3. Nov 13, 2011 #2
    Next time try a little harder with posting relevant equations. Homework's often confusing and I make post for help that are sparse too but it helps show your earnest

    Nonetheless, I believe that you must first find the volume of the wax and then the mass of the liquid that is being moved,


    Remember that mass=density*Volume
    or m=d*V

    Then find the apparent mass of the block

    and then finally find the apparent weight of the block

    weight is a "type of force" which can be given with the equation F=m*g where g is gravity's constant of 9.8ms^2

    There may be a shortcut to all this but I can't think of it of the top of my head
     
  4. Nov 13, 2011 #3
    But how would you find the volume here?
     
  5. Nov 13, 2011 #4
    Well, if mass=Volume*density and Volume=mass/density and you said you already have d=1.8

    --and f=m*g

    Then can you say how you would find the Volume?
     
  6. Nov 13, 2011 #5
    Hmm Can I calculate the mass by using f=ma and m= f/a? and use that mass in m/d?
     
  7. Nov 13, 2011 #6
    Yes

    You already have the force F newtons stated in the original problem


    solve f=m*a for the mass because you know that a in this case is equal to 9.8ms^2 and the force F newtons is given

    then you have the mass to use in the m/d
     
  8. Nov 13, 2011 #7
    Alright here's what I did, f=ma, m= f/a, .27/10= .027kg.

    volume = md = .027kg x 1800kgm-3 = 48.6 m^3

    Apparent mass = 800 x 48.6 = 38,880kg?

    :S This seems wrong.
     
  9. Nov 13, 2011 #8
    volume doesn't = m*d

    volume=m/d

    you might need to review solving for a variable in linear equations

    Ex.) V = m/d = 32 / 2700 = 0.0119 m^3
     
  10. Nov 13, 2011 #9
    Oh sorry!!! I'm very tired it's 2am here.
     
  11. Nov 13, 2011 #10
    I ended up getting 0.12 N but in the back of my book it says 0.15N.
     
  12. Nov 13, 2011 #11
    That's ok,

    Remember to use a calculator to avoid silly mistakes.

    Maybe you just need some sleep :wink: It might make sense in the morning
     
  13. Nov 13, 2011 #12
    Did you use the correct value of the Newtons?
     
  14. Nov 13, 2011 #13
    yes the weight in air is 0.27 N , so I divided that by 10 to get the mass.
     
  15. Nov 13, 2011 #14
    What is "apparent weight"?

    This seems to be some vague reference to what the weight 'appears' to be, without clarification as to how the peering is to be done.
     
  16. Nov 13, 2011 #15
    The weight a body appears to have in a fluid?
     
  17. Nov 13, 2011 #16
    In this context, this seems reasonable. From a) you should have the mass and weight of the object in air, so you have enough information to calculate the weight in the 800 grams/liter fluid to solve part b).
     
  18. Nov 13, 2011 #17
    The .12 N I got should I subtract it from the actual weight? Sorry for my ignorance, I don't know this part of physics well at all!
     
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