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A trough is 5 feet long and 1 foot high. The vertical cross-section of the trough parallel to an end is shaped like the graph of y=x^4 from x=−1 to x=1 . The trough is full of water. Find the amount of work in foot-pounds required to empty the trough by pumping the water over the top. Note: The weight of water is 62 pounds per cubic foot.

2007-10-10 10:54:50 · 1 answers · asked by mjs3382 1 in Science & Mathematics Physics

1 answers

But in reality, you can set up a siphon and empty the trough with negligible work.

But, it appears that what they want you to do is to compute the work required to bring each drop of water to the edge where it can drop on the ground.

The trough is one foot high. We put the origin at the bottom of the trough with the length along the z axis, the width along the x axis, and the height along the y axis.

Doing it the hard way, the total work is the integral of the work done for each dy, for y = 0 to y = 1. (Think of this as emptying the trough one thin sheet of water at a time instead of drop by drop.)

The work in foot-pounds for each dy is the weight times distance.

The weight is the volume times the density; the distance is the vertical height to the edge.

Volume is the area times the thickness. The density is given as 62 pounds per cubic foot.

The area is that of a rectangle 5 feet long and 2x wide. That is, if the cross-section is defined by:

y = x^4 for x in [-1,+1]

It is also defined by:

x = +/- y^(1/4) for y in [0, 1]

So the area is 2y^(1/4) and the volume is 2y^(1/4)dy

The distance is the vertical distance from y to the edge, so it is (1-y).

So we can now compute the incremental weight and distance, set up the integral, and integrate.

2007-10-12 18:50:19 · answer #1 · answered by simplicitus 7 · 0 0

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