Thread: Fluid Pressure Fluid Force

1. Fluid Pressure Fluid Force

A gate in the vertical face of a dam has the shape of a semicircle of radius 4 ft whose top edge is its diameter (also parallel to the water surface). Find the total force of water (62.4 pounds per cubic foot) on this gate if its top is 10 ft beneath the surface of the water.

Ugh, I feel like I'm drowning!!! (Pardon the pun) Any help at all would be appreciated... I think I just have trouble "visualizing" what's going on, and getting started. Thanks in advance.

2. I am also having a difficult time picturing the problem given the description.

I am thinking it is a semicricular gate with radius 4 whose top is 10 feet beneath the top of the water of the dam. I hope that's correct.

If so, we have the integral:

$\displaystyle 124.8\int_{0}^{4}(10+y)\sqrt{16-y^{2}}dy$

3. Originally Posted by UDaytonFlyer
A gate in the vertical face of a dam has the shape of a semicircle of radius 4 ft whose top edge is its diameter (also parallel to the water surface). Find the total force of water (62.4 pounds per cubic foot) on this gate if its top is 10 ft beneath the surface of the water.

Ugh, I feel like I'm drowning!!! (Pardon the pun) Any help at all would be appreciated... I think I just have trouble "visualizing" what's going on, and getting started. Thanks in advance.

So from the above immage we see that the area of each sliver of water is

$\displaystyle 2ydx$ we can solve for y using the pythagorean theorem and

$\displaystyle y=\sqrt{16-x^2}$

also from the above diagram we see that the depth at any point is (10+x)

so the change in volume $\displaystyle dV=2(10+x)\sqrt{16-x^2}dx$

the total force will be $\displaystyle F=Volume \times 62.4$

so we end up with the integral for the force

$\displaystyle 62.4 \int_{0}^{4}2(10+x)\sqrt{16-x^2}dx=124.8\left(\int_0^4 10\sqrt{16-x^2}dx+\int_0^4x\sqrt{16-x^2}dx\right)$

$\displaystyle 1248\int_0^4\sqrt{16-x^2}dx+124.8\int_0^4x\sqrt{16-x^2}dx$

The first integral is just the area of a quarter circle with a radius or 4

$\displaystyle \int_0^4\sqrt{16-x^2}dx=\frac{1}{4} \pi 4^2=4\pi$

and the second can be solved with a u sub

good luck.