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Applications of integration: work problems

  1. Dec 2, 2009 #1
    1. The problem statement, all variables and given/known data

    1. A cylindrical gasoline tank 3 feet in diameter and 4 feet long is carried on hte back of a truck and is used to fuel tractors. The axis of the tank is horizontal. Find the work done in pumping the entire contents of the fuel tank into a tract if the opening on the tractor tank is 5 feet above the top of the tank in the truck. Gasoline weighs 42 pounds per cubic foot.

    2. Consider a 15-foot chain hanging from a winch 15 feet above ground level. Find the work done by the winch in winding up the specified amount of chain, if the chain weighs 3 pounds per foot.


    2. Relevant equations



    3. The attempt at a solution

    1. I visualized it as being chopped into little chunks of rectangles:
    dW = dF * x
    dF = (weight)
    weight = 42*4*1.5*dh
    dF = 42*4*1.5*dy * (5-y)
    [tex]\int^{3}_{0}42*4*1.5*dy * (5-y)[/tex] = 2646
    Book Answer: 2457pi

    2. dW = dF * x
    dF = 3 dy
    d = 10-y
    [tex]\int^{10}_{0}3*(10-y) dy[/tex] = 150
    Book answer: 300


    Can anyone help me see my mistakes?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 2, 2009 #2
    For number 1), It doesn't look like you're correctly taking the geometry of the tank into account. The axis of the tank is horizontal, so the volume of a tiny horizontal slice of dy thickness would change depending on y. I'd expect the expression inside the integral to be a much more complicated function than just (3-y). Although, I really don't think calculus is the way to go for this problem. There's a MUCH simpler way, if your prof. will let you take the easy way out.

    I'll look at 2 tomorrow.
     
  4. Dec 3, 2009 #3

    Mark44

    Staff: Mentor

    For #1, you are assuming that all of the "layers" of gasoline are 1.5 ft wide, which isn't true. You need to use the equation of the circular cross-section of the tank. Since you are integrating from y = 0 to y = 3, you are assuming that the origin is at the bottom of the tank, which has an impact on the equation of the circular cross-section of the tank. This assumption also impacts the expression that represents the height each layer of gas is lifted. For example, the slab at a level y has to be lifted to a height of 8 ft (the top of the tank is at (0, 3) according to your coordinate system, and has to be lifted to (0, 8)). This means that this typical layer has to be lifted (8 - y) ft.
     
  5. Dec 3, 2009 #4
    I'd also like to add that your notation seems really odd to me. First off, usually work is expressed as F(x)*dx, not dF*x. It makes a difference--you're dividing the work up into increments of x (thin layers of gasoline, tiny links of chain), not increments of force. You also write this 'd', and I can't tell if you're thinking of it as distance, or misusing the d in dF or dx.

    I'll check this forum tomorrow, so please don't hesitate to ask more questions if you're confused!
     
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