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Homework Help: Work integration application problem, answer check

  1. Oct 21, 2011 #1
    1. The problem statement, all variables and given/known data
    A leaky 6lb bucket is lifted from the ground to a height of 30ft at a constant speed with a rope that weighs .5lb/ft. Initially, the bucket contains 60lbs of water, but the water leaks out at a constant rate and finishes draining just as the bucket reaches the 30ft level. How much total work is done lifting the bucket?

    2. Relevant equations
    Function I found for the weight of the rope: (15-.5x)

    Function I found for the weight of the bucket: (60-2x)

    so adding all together I got = 6 + (15-.5x) + (60-2x) (Δx) (x)

    then simplified to get the work function = 81x-2.5x2dx

    3. The attempt at a solution

    integrated work function 81x-2.5x2 over 0 to 30 and got 13950 ft/lbs total work.
    Last edited: Oct 21, 2011
  2. jcsd
  3. Oct 21, 2011 #2
    Always look at your answer to see if it is realistic to you. Suppose the bucket were not leaking and the rope was coiled up inside the bucket. Total weight of rope laden bucket is 60+6+15 = 81 lbm. To raise 81 pounds to a height of 30 feet is only 2430 ft-lbm. How can your answer be a larger number?
  4. Oct 21, 2011 #3
    Another hint when you are deriving equations. Make sure the units you end up with jive with what you seek.
  5. Oct 21, 2011 #4
    Thanks for your help. Obviously my answer is way high. I will keep working at it.

    Concerning the units, what does lbm stand for?
  6. Oct 21, 2011 #5
    Same as lbf in this case. I use them interchangeably when I'm on earth. : )
  7. Oct 21, 2011 #6
    You are going about it in the correct manner. You just have a slight mistake. How is work defined?
  8. Oct 21, 2011 #7
    Got to hurry on this. My happy hour begins at 4 PM EDT.
  9. Oct 21, 2011 #8
    W=force x distance


    I'm not getting 81 pounds as 2430 ft/lbs

    (81 pounds)(distance) = W

    so integrate 81x over 0 to 30, and I'm getting 36450... what the heck.
  10. Oct 21, 2011 #9
    ok, 81 x 30 = 2430, but shouldn't I be integrating the work function to get the total work (area)?
  11. Oct 21, 2011 #10
    The work function is force times distance. The force is 81 pounds. The distance is dx. remember, your answer must have the units of ft-lbs. The integrand must have the units of ft-lbs. You are putting an additionl space dimension in it.

    W = integral (F * dx) from 0 to 30. Integral of F * dx is F*X evaluated from 0 to 30. Answer is 81 * 30 = 2430.

    Do you see where you have gone awry?
    Last edited: Oct 21, 2011
  12. Oct 21, 2011 #11
    I guess I'm not understanding how distance is Δx. I thought that was the width of the "slice" of work done at xi.
  13. Oct 22, 2011 #12
    dx is the width of the slice. Integration is simply a method that 'adds' up all the little slices letting the function take on its different values as the summing progresses over the limits of integration. Delta x in calculus terminology (when it approaches 0 in the limit) becomes dx.

    Work is the area under the force-distance curve. Suppose you have a constant force and you graphed it with force on y-axis. The work would be represented by the area of the rectangle because that area is force multiplied by distance. Rectangle's height would be the force (y-axis). dx is a slice which is a very small rectangle whose height is F and whose width is dx. dW, the area of the slice, is F * dx. Note that the units of F*dx are ft-lbs. Since I indicated above that integration is basically a summing procedure, the units do not change. (3 apples plus 4 apples is 7 apples).

    You have inserted an extra space dimension in your integration. Delta W is W multiplied by delta x. dW is W multiplied by dx. In the limit as one likes to say in 'calculese', the delta x becomes dx.

    In your problem the force is not constant. The function you specified for it is correct. What is wrong is that you have inserted an extra x in the integrand.
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