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Really lost, pulling ball through water

  1. Nov 8, 2004 #1
    I need to calculate thr force required to pull a copper ball of radius 2cm upwards through a fluid at the costant speed of 9 cm/s. The drag force is porportional to the speed with porportionality constant 0.950 kg/s. Ignoring the buoyant force, I need to caluate the force.

    I fould out the mass=2 .67E-4 kg, after finding copper's density then m=d/v

    I think in the end, I will go back to the equation F=ma. So, I need to find acceleration:

    a=g-bv/m
    b=porportionality constant 0.950 kg/s
    v=costant speed of 9 cm/s
    m=2 .67E-4 kg

    what in the world would g equal? Any help is appreciated
     
  2. jcsd
  3. Nov 8, 2004 #2

    Doc Al

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    Staff: Mentor

    Since the speed is constant, the acceleration is zero.

    You need to figure out the force you need to exactly balance the other forces on the ball (its weight plus the drag force). So:
    F = weight + drag
     
  4. Nov 8, 2004 #3
    F=mg+bv/m
    =(2 .67E-4 kg)(9.8m/s^2)+(0.950 kg/s)(9m/s)/(2 .67E-4 kg)

    like that?
     
  5. Nov 8, 2004 #4
    by dimensional analysis, bv/m is in units of acceleration. F=ma, leading me to beleive you may want to try bv.
     
  6. Nov 8, 2004 #5
    F==(2 .67E-4 kg)(9.8m/s^2)+(0.950 kg/s)(9m/s)

    The answer I get is 8.8N

    However, the book says that it's 3N

    What am I doing wrong?
     
  7. Nov 8, 2004 #6
    you stated the velocity was 9 cm/s not m/s -- thats the only thing that sticks out to me
     
  8. Nov 8, 2004 #7
    F=(2 .67E-4 kg)(9.8m/s^2)+(0.950 kg/s)(.09m/s)
    F=.09N

    ackkk where am I wrong?
     
  9. Nov 8, 2004 #8
    Didn't do the calculation myself but the formula you stated there is wrong.

    Remember d=m/v so m=dv
     
  10. Nov 8, 2004 #9
    thanks! that did the trick!
     
  11. Nov 8, 2004 #10
    I've learned to catch those kind of things because I do them so much myself. Spent forever one night trying to solve a buoyancy problem because I thought it was necessary to solve without the density of an object. When I read the problem again I saw the object was aluminum with a density clearly listed in a table several pages before :grumpy:
     
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