mth174
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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A fuel tank is an upright cylinder, buried so that its circular top is 10 feet beneath ground level. The tank hasa radius of 5 feet and is 15 feet high, although the current level of oil is only 6 feet deep. Calculate the work required to pump all of the oil to the surface. Oil weighs 50lb/ft^3
-We first need to find the volume of a slice: pi * r^2 d'x. Can we draw a circle that, in it, we have the triangle and know that our cross sectional radius is 5ft. the depth at this point is 6ft so r^2+ (6-h)^2= 5^2, then solve for r giving the sqrt(5^2-(h-6)^2 ). Then you need to find the force on a slice so its 50*g*pi* (5^2-(h-6)^2 ) d'x * distance. Would the distance just be 15 since that's where its pumping up to? or would it be (15-H). I know the limits of integration are from 0 to 15. My main concern is whether or not I have the correct volume of slice and distance, since they are the key to correctly doing the problem.
Your setup is very good, however it does not correspond to the given orientation of the tank. If the circular top is at a constant depth below the surface, then the horizontal cross-sections of the cylinder are circular, and since the tank is cylindrical they all have the same area pi * (5 ft)^2.
Your expression in terms of r and h would hold if the tank was lying on its side, but in that case the oil surface in the tank would be rectangular, with one dimension equal to the 15 ft altitude of the cylinder (in this orientation the altitude would run in the horizontal direction) and the other dimension being 2 *sqrt(5^2-(h-6)^2)
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In any case the orientation is as in my first paragraph so the expression is much simpler.
The height to which oil at depth h, relative to the bottom of the tank, is raised will been 10 ft + (6 ft -h) = 16 ft - h. The limits on the integral will be the limits on h.
A trough is full of water. Find the force of the water on a triangular end. Find the work to pump all the water over the top. The trough has a length of 15ft, the two triangles on the side have a width of 2 ft and a heigh of 3 ft. The volume of a slice will be l*w*h.
Since the the thing is full of water would the volume of a slice just be 15 w d'x . Then can we use similiar triangles to find w?
Right, assuming your x-axis is vertical so that it is the x-axis that is being partitioned. However it isn't the volume that exerts the pressure; this increment would be fine for determining the volume, but instead you need to determine the area are and pressure (see below)
But since the tank is full and we aren't submerging any, so we aren't going down any distance of x to find. But the force of water will just be 62.5lb/ft^2 * 3ft= 187.5 lb/ft^3.
The force of the water on an area at the end of the tank is pressure * area. The pressure changes with depth.
So it is actually the trianglular region that needs to be partitioned into horizontal strips, over each of which the pressure is nearly constant.
I'm using rectangles to find the volume of the slice, but would it make more sense to use triangles? Using similiar triangles you can have s/2=3-x / 3 and get s= 2/3(3-x) then have s^2 d'h. then multiply that by 187.5lb/ft^3*g* volume of slice, then the work done on the slice can equal force* distance( since the tank is full, can we say that the distance is 3ft?) then the limits of integration on this function will be from 3 to 0.
The pressure on a triangular region would not be constant. You need to think in terms of horizontal strips along the triangle.