Chapter 7 Exercise 5. Fluid Forces
We consider the fluid forces exerted on a triangle submerged vertically into water. We assume the weight density of water is 62.4 lb per cubic foot.
The diagram below shows the arrangement of the body under water.
Triangle
Clearly, the deepest point of the triangle is at a depth of 11 feet. The relevant integral to calculate the fluid force is shown below and uses key idea 7.6.1 for fluid force o a vertically oriented plate.
\[ F = \int_{a}^{b} w \cdot d(y) \cdot L(y) dy \] where \(d(y)\) denotes the distance between the surface of the fluid and the plate at coordinate \(y\) and \(L(y)\) represents the length of the plate at \(y\).
Solution
Our fluid force can be presented as the integral:
\[ F = \int_{0}^{6} w (5 + h) L(h) dh \] where \(w = 62.4\), \(L(h) = \frac{2}{3}h\) and \(h\) represents the height from the top of the triangle to the depth of \(5+h\) below the surface of the water.
This gives the fluid force integral:
\[\begin{align} F & = & \int_{0}^{6} w ( 5+h)L(h) dh \\ & = & \frac{2}{3}w \int_{0}^{6} (5 + h) h dh \\ & = & \frac{2}{3}w \left[ \frac{5}{2}h^2 + \frac{ h^3}{3} \right]_{h=0}^{6} \\ & = & \frac{2}{3}w \left[ \frac{5}{2}36 + \frac{6}{3} 36 \right] \\ & = & \frac{2}{3} 36 \cdot w \left[ \frac{5}{2} + 2 \right] \\ & = & 108 \cdot w \\ & = & 108 \cdot 62.4 \\ & = & 6739.2 \\ \end{align} \]
We conclude that the fluid force on the submerged triangle is 6739.2 ft per cubic inch.