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Integration buoyancy problem

  1. Apr 28, 2008 #1
    1. The problem statement, all variables and given/known data
    A 4 cm diameter cylinder floats in the water. How much work must be done to push the cylinder 10 cm deeper into the water?


    2. Relevant equations



    3. The attempt at a solution

    I did the integral from 0 to .1 m of (.02m)^2 * pi * (1000kg/m^3) * (9.80) * x dx

    = (12.3 x^2)/2 evaluated from 0 to .1 = .0615 J. However, the back of my book says .615 J. Somehow I ended up a decimal place off. Could someone help me?
     
  2. jcsd
  3. Apr 29, 2008 #2
    Anyone have any thoughts on this one?
     
  4. Apr 29, 2008 #3
    Isn't the radius=0.2 instead of 0.02 ?
     
  5. Apr 29, 2008 #4
    no, I'm pretty sure the radius is correct

    4 cm/2 = radius of 2 cm 2cm * 1m/100cm = .02 m
     
  6. Apr 29, 2008 #5
    Any thoughts would be greatly appreciated
     
  7. Apr 29, 2008 #6
    I think that given the values you are working with that your answer is correct.
     
  8. Apr 29, 2008 #7
    I am concerned because the back of my book says it's .615 J and I can't figure out what's wrong with mine
     
  9. Apr 29, 2008 #8
    Can somebody find my mistake?
     
  10. Apr 30, 2008 #9
    I need some help with this one pretty soon because my homework is due tomorrow. Thanks!
     
  11. Apr 30, 2008 #10
    Sometimes the answer in the book is wrong...or the problem data has been misread. In either case it looks like to me that your answer is correct, If not, I would also like to know why and maybe another reader can set us both strait.
     
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