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Fluid Mechanics?

  1. Oct 3, 2004 #1
    Block A hangs by a cord from spring balance D and is submerged in a liquid C contained in beaker B. The mass of the beaker is 1.00 kg and the mass of liquid is 1.80 kg. Balance D reads 3.50 kg and balance Balance E( supporting beaker B) reads 7.50 kg. The volume of block A is 3.80 * 10^-3 m^3. a) What is the density of the liquid? b) What will each balance read when Block A is pulled out of liquid? Do I need to know what the liquid is in order to solve this? I know that p(density) = mass/volume. So in this case is it just mass of liquid/volume of liquid?
  2. jcsd
  3. Oct 3, 2004 #2
    I suspect that you will need to find the density of the block in order to find the density of the liquid. This can be done by setting up equations for the apparent weight of the block given by balances D and E. Hope that helps.
  4. Oct 3, 2004 #3
    Can you please explain further?
  5. Oct 3, 2004 #4
    Weight loss/Apparent weight=Weight of body-weight of liquid displaced
    weight= mass*g = Volume*density*g.
    Need any more help?
    Last edited: Oct 3, 2004
  6. Oct 3, 2004 #5
    There will be two equations. The first equations will relate the apparent weight of the block given by balance D to the actual weight of the block and the product of the buoyant force (set this up using Archimedes' pricincple). The second equation will relate the weight given by balance E to the weight of beaker B, liquid C and block A. You'll have two equations and two unknowns (one being the density of the block and the other the density of the liquid).
  7. Oct 3, 2004 #6

    If you don't mind my asking, can you give me the equations?
  8. Oct 3, 2004 #7
    Why don't you derive the equations and post them here. Then I'll tell you if you're on the right track.
  9. Oct 3, 2004 #8
    w(D) = W(A) * F(B)
    W(E) = W(B) + W(liq) + W(A)
  10. Oct 3, 2004 #9
    You might want to explain what those symbols mean. Keep in mind Archimedes' principle.
  11. Oct 3, 2004 #10
    w(D) = weight read by Balance D
    w(A) = weight of Block A
    w(B) = weight of Beaker B
    w(C) = weight of liquid C
    F(B) = bouyant force
  12. Oct 3, 2004 #11
    Why in the world are you multiplying w(A) and F(B)? That makes absolutely no sense. Maybe you should go back to the basics.
  13. Oct 5, 2004 #12
    solved it. Thanks fellas
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