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Work and Pumps

  1. Jun 28, 2006 #1
    Me again :smile:

  2. jcsd
  3. Jun 28, 2006 #2
    Never fails, just after a post, I figure it out myself... haha. Well, at least part of it.

    So I get the part of the limits of integration and where they come from, that is, the total distance the fluid has to move minus the length of the pump. Right?

    So, if the tank has a 6' diameter, and the the pump is 4' tall, then the total distance the fluid has to move is 10'. That gives the upper limit. The lower limit is found by simply subtracting the hight of the container (in this case 6') from the total distance, 4'. Does this sound physically correct?

    Now I am still working on the 24... Where the heck did that come from??
  4. Jun 28, 2006 #3
    Ok, Now I confused myself more by trying to work out another problem...

    Have a look:

  5. Jun 28, 2006 #4


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    One thing I recommend in problems like this is to note specifically what the variables represent. In the first problem, x is the distance from a "layer of oil" to the pump. At the very top of the cylinder, x= 4, and at the bottom, x= 4+ 6= 10. The center of the circular cylinder would be at x= 4+ 3= 7 so we can write the equation of the circle as (x- 7)2+ y2= 9. That's where [itex]y= \sqrt{9- (x-7)^2}[/itex] comes from. You set up the integral by imagining the oil in "layers" each a fixed distance, x, from the pump. Each "layer of oil" is a rectangle having length 12 and width [itex]2y= 2\sqrt{9- (x-7)^2}[/itex]. The area of that rectangle is [itex](12)(2\sqrt{9-(x-7)^2}= 24\sqrt{9-(x-7)^2}[/itex]. That's where the "24" comes from. Taking dx as the infinitesmal thickness, the volume is [itex]24\sqrt{9- (x-7)^2}dx[/itex] and the weight of that "layer of oil" is [itex]24\rho \sqrt{9-(x-7)^2}dx[/itex]. Since it must be lifted a distance x, the work done to lift that "layer of oil" is [itex]24\rho x\sqrt{9- (x-7)^2}dx[/itex]. Since x runs from 4 to 10, the total work done is [itex]\int_4^{10}24\rho x\sqrt{9-(x-7)^2}dx= 24\rho\int_4^{10}x\sqrt{9-(x-7)^2}dx[/itex].

    In the second problem, there are two differences. First, the tank is vertical so no matter what x is the area of a "layer of oil" is [itex]\pi r^2= 4\pi[/itex] so the weight of that "layer" is [itex]4\pi \rho dx[/itex]. The second difference is that nothing is said about where the pump is in relation to the cylinder. You appear to be assuming that it is sufficient to pump the water to the top of the cylinder. In that case, water at the top, half way up the cylinder, only has to be pumped 5 ft while water at the bottom must be pumped 10 ft. Taking x to be the distance from the layer to the top of the cylinder, we integrate the weight times x: the work done is [itex]4\pi\rho\int_5^{10}xdx[/itex].

    More correctly they describe the height the fluid must be lifted- the fluid might be directed horizontally for a while but that would not require any work. In any integral, the limits of integration give the minimum and maximum values of the variable of integration. Here x represents the height any "piece of oil" or "layer of oil" must be lifted.

    Not quite. The force (weight) of each thin layer of water is actually [itex]4\pi \rho dx[/itex], the volume of a thin disk times the density.

    Again, not quite. You seem to be thinking "force times distance" as "constant in front of integral times integral". That's not the case. In both problems, x is the "distance". It just happens that the force in the second problem is a constant and so can be taken out of the integral. Of course, [itex]x\sqrt{9- (x-7)^2}[/itex], is not a "distance"- look at the units. If x is measured in feet then that is feet squared. It's actually the area of a "layer of oil". multiplying it by density and "thickness", dx, gives the weight or force needed to lift it.

    Remember how an integral could be defined in terms of a Riemann sum? Fortunately for us all, it turns out we can do an integral in terms of an "anti-derivative" instead- that's much easier. But you should still think of integrals in terms of Riemann sums because that's how we set up integrals in problems like these. In both problems the weight (force to lift) a little drop of oil or water is constant. But we can't just multiply "force times distance" because the distance to be lifted varies. What we do is look at the set of all drops of oil that will be lifted the same distance. In the first problem, with the tank horizontal, that is a rectangular layer, in the second, a disk. In the first problem the size of the rectangle depends on x, in the second it happens to be constant. But in both cases, we calculate the area, multiply by the "infinitesmal" thickness dx to get its "infinitesmal" volume, and then by density to find its weight- the force necessary to lift it. We multiply that by x to find the work done in lifting that single layer. That's exactly the same as calculating the area of those narrow rectangles that make up the "area under the curve". Adding all the work for all those "layers" gives (approximately) the total work just as adding the areas of all those rectangles gives (approximately) the total area under the curve. The last step is to take the limit so that the Riemann sum becomes an integral and we get the exact values.
    Last edited: Jun 28, 2006
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