Solving a work problem using integrals? please help!
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Solving a work problem using integrals? please help!

[From: ] [author: ] [Date: 12-07-17] [Hit: ]
If you want to covert it to BTUs or Joules, you can do so yourself.......

We need to form an equation for the volume of the pyramid based on the height

Cross sectional area changes linearly with cross section of (756 ft)^2 on the bottom and 0 at 481 ft

h1 = 0, A1 = (756 ft) ^2 571536 ft^2
h2 = 481, A2 = 0

A = m*h + b

m = (0 - 571536) / (481 - 0 ) = -1188.2 ft^2 / ft

0 = -1188.2 * 481 + b

b = 571536

A = -1188.2 h + 571536
V = A*h = -1188.2 h^2 + 571536*h

Work = integral from h to hf (W * dh)
Work = integral from h to hf (density * g * V * dh)
Work = int from h to hf ( density * g * (-1188.2 h^2 + 571536*h) dh )
Work = density * g * (-369 h^3 + 285768 h^2) from h to hf

Now here's the tricky part with english units, mass is a slug, not a pound... weight = mass* g = slug * g = pound, since the denisty is already given in lbs/ft^3 you can ignore the g term... The correct term for g in english units is 32.2 ft/s^2.

Work = 3.75 * 10^12 ft*lb

FT*LB is a correct unit for work in the english system. If you want to covert it to BTU's or Joules, you can do so yourself.
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