52–54. Force on a window A diving pool that is 4 m deep and fu...

Question

52–54. Force on a window A diving pool that is 4 m deep and full of water has a viewing window on one of its vertical walls. Find the force on the following windows.

The window is circular, with a radius of 0.5 m, tangent to the bottom of the pool.

Accepted Answer

Hello. In this video we are told that a swimming pool is filled with water to a depth of 4 m and has a circular vertical viewing window of radius 0.5 m that is tangent to the bottom of the pool. Using calculus, we want to find the hydrostatic force on the window and take the water density as 1000 kg per meter cubed and gravity g as 9.8 m per second squared. So, let's go ahead and think about the force of the water. Now, the force of the water P of Y is going to equal to row multiplied by G multiplied by a function of Y. Now here, row is going to be the force of the water, which is 1000 kg per meter cubed, and G is going to be our gravitational constant, which is 9.8 m per second squared. So these are going to be the constant values that the problem has provided for us. But now what we want to do is we want to go ahead and determine the value of Y. Now Y is going to be the value of the geome of the geometry of the circular window. So we're going to have Y_C which is equal to 4. -0.5, which is going to give us 3.5 m. So, that tells us that the value of y has to be between 3 and 4. But more importantly, if we express. The horizontal strip of the depth of y that is going to be xy, which is equal to the square root of the radius square minus y minus yc 2. This is going to simplify to be the square root of 25, 0 sorry, 0.25. Y minus 3.5 quantity squared. And so now, at the full width, the full width of this function is going to be 2 multiplied by X of Y. And so that is going to be 2 multiplied by the square root of 0.25 minus Y minus 3.5 quantity squared. Now, what we want to go ahead and do is we want to go ahead and integrate in order to find the hydrostatic force. Now, the hydrostatic force is going to be defined as F is equal to the integral from A to B. Of 2 row G. Multiplied by Y multiplied by W of Y. DY and so here we know 2 row and G are constants to the integral. So we will have 2 multiplied by row multiplied by g, and we know that our boundaries are going to be from 3 to 4. Of why Multiplied By the WY function we just came up with, which is going to be the square root. Of 0.25 minus Y minus 3.5 quantity squared DY. Now, here, we can go ahead and simplify the integral by using a U substitution. If we allow U to equal to Y minus 3.5, then DU is equal to DY. So, we can rewrite the integral to be to row G, multiplied by the new boundaries, which is from -0.5 to 0.5. Now, we need a substitution for Y. If we solve for Y in our U sub, we get Y is equal to U + 3.5. So we get the quantity U + 3.5 multiplied by the square root of 0.25 minus U 2 DU. Now, here's the thing about simplifying this function. We can distribute the square root to both quantities in the given U + 3.5 quantity, but if we take the integral of U multiplied by the square root of everything else, this ends up being an odd integral, and that odd integral gives us an output of zero. So, if we were to distribute and then separate the integrals, we can further simplify the integral to be 2 multiplied by row, multiplied by G integrated from -0.5 to 0.5 of 3.5 multiplied by the square root of 0.25. Minus U 2 DU Now here, if we were to integrate, this is going to simplify to be 2 row g multiplied by 3.5 multiplied. By 1/8 pi. When we evaluate from -0.5 to 0.5, and by plugging in or multiplying the constants together, we get row multiplied by G, which is 0.875 pi. And finally, by plugging in the constants of row and G, we get 1000 multiplied by 9.8, multiplied by 0.875 pi. And that is going to give us a hydrostatic force of 8,575 pi Newtons, which is going to be the solution to the problem, and the overall solution is going to be option B. So I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

52–54. Force on a window A diving pool that is 4 m deep and full of water has a viewing window on one of its vertical walls. Find the force on the following windows. The window is circular, with a radius of 0.5 m, tangent to the bottom of the pool.

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52–54. Force on a window A diving pool that is 4 m deep and full of water has a viewing window on one of its vertical walls. Find the force on the following windows.

The window is circular, with a radius of 0.5 m, tangent to the bottom of the pool.