neurosis Posted January 28, 2006 Share Posted January 28, 2006 Here's a problem I recently came across in a very old calculus book. Unfortunately, it was an even-number, and I can't quite figure out for sure how to solve it. A bag of sand originally weighing 144 lbs is lifted at a constant rate of 3ft/min. The sand leaks out uniformly at such a rate that half the sand is lost when the bag has been lifted 18 ft. FInd the work done lifting the bag this distance. The thing about it is this would be easy to solve, except that we don't know the weight of the bag *alone* and can not just assume that it's negligable. I know that the work is force times distances, and that you use infinite sums to find the work over that particular distance (since it's changing) so we have an integral from 0 to 18 of the f(X)...but what is the function of x? THe sand is all gone by 36 ft (in 12 minutes) but we don't know the weight of the bag, so therefore we don't know the weight of the sand that was lost. This is frustrating! Does anyone have any idea? (I originally thought Integral from 0-18 of 144-4x, but I don't feel right about that) Link to comment Share on other sites More sharing options...

Connor Posted January 28, 2006 Share Posted January 28, 2006 I think you have to assume the wieght of the bag is negligable, and the bag weighs 72 pounds at 18 feet. Otherwise you don't have enough information to solve the problem. Link to comment Share on other sites More sharing options...

Tartaglia Posted January 28, 2006 Share Posted January 28, 2006 Assume bag is negligible weight Work given by /6 | m(t)*g*(dh/dt)*dt /0 where t = time (minutes) m(t) = 144-12t (lb) dh/dt = 3ft/min horrendous units though Link to comment Share on other sites More sharing options...

neurosis Posted January 29, 2006 Author Share Posted January 29, 2006 I believe you guys are correct; the bag has to be assumed negligable. There really isn't any other way of solving for the bag. I just was worried maybe I missed something! But I do see now that I was most likely making it more complicated than it is. Link to comment Share on other sites More sharing options...

s pepperchin Posted February 15, 2006 Share Posted February 15, 2006 the key to this problem is to write what you know. [math]Fg(0)=144 lbs[/math] [math]dh/dt = 3 ft/min[/math] this is constant so the eqtn for height is: [math]h={3 ft/min} * t[/math] if half the weight is gone when the bag has been lifted 18 ft then [math]t={18 ft}*{1 min/3 ft}[/math] [math]t=6 min[/math] so we know know that: [math]Fg(6 min) = 1/2*144 lbs = 72 lbs[/math] so we can determine from this that the eqtn for Fg is: [math]Fg(t) = 144 lbs - (12 lbs/min)t[/math] recall that [math]W=\int_a^b F dh[/math] our particular problem has a hitch in that Fg is a function of time and we need to integrate with respect to height, h. We can fix this though because we have h as a function of time. [math]h={3 ft/min} * t[/math] we can solve for t [math]t=h*{1min/3 ft}[/math] we can plug this into Fg to make it a function of h [math]Fg(h) = 144 lbs - (12 lbs/min)(h*1min/3 ft)[/math] [math]Fg(h) = 144 lbs - (4 lbs/ft)h[/math] now we can integrate this [math]W=\int_a^b F dh[/math] [math]W=\int_0^18 144 lbs - (4 lbs/ft)*h dh[/math] [math]W=144lbs* h -(2 lbs/ft)h^2 |_0^18[/math] [math]W=2592 ft lbs - 648 ft lbs = 1944 ft lbs[/math] or 2.5 BTUs Link to comment Share on other sites More sharing options...

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