Question

The end plates (isosceles triangles) of the trough show below were designed to withstand a fluid force of 5800 lb. How many cubic feet can the tank hold without exceeding this limitation?

Assuming the density is 62.4 lb/ft^3, the maximum volume is ____ ft^3.

View the graph at:
postimg. org/image/4925g8vlp/

Answer 1

Let the bottom of the trough be at (0,0).
The width of the trough at height y is thus

x = 9/14 y

Consider a bunch of horizontal cross-sections of the trough end. Each is of length x and depth 9-y.

The force on the plate when the water is h feet deep, is thus

F = w∫[0,h] (h-y)(9/14 y) dy
where w = 62.f lb/ft^3
= 62.4(3/28 h^3)

So, if we want F=5800, we just solve

62.4(3/28 h^3) = 5800
h = 9.54 ft

Since the tank is 32 ft long, that means it can hold

v = 1/2 (9.54)(9/14 * 9.54)(32) = 936 ft^3

Answer 2

In my book I have diff answer. Can you double check your work?