Consider a 4-m-long, 4-m-wide, and 1.5-m-high above-the-ground swimming pool that is filled with

Consider a 4-m-long, 4-m-wide, and 1.5-m-high...

Key Concepts

Hydrostatic Pressure

Hydrostatic pressure is the pressure exerted by a fluid at equilibrium at any given point within the fluid, due to the force of gravity. This pressure increases linearly with depth because it depends on the weight of the fluid above the point, following the relationship p = ?gh, where ? is the fluid density, g is acceleration due to gravity, and h is the depth below the free surface.

Pressure Distribution on Submerged Surfaces

For a submerged, vertical surface, the pressure varies linearly from zero at the surface (if open to the atmosphere) to a maximum at the bottom. This linear distribution means that the pressure distribution is not uniform, which is important when calculating the total force acting on the surface. The result is that pressure increases proportionally with depth, leading to a non-uniform load over the surface area.

Integration over a Submerged Surface

Calculating the hydrostatic force on a submerged surface involves integrating the pressure over the entire area of that surface. This integration takes into account the continuously varying pressure with depth, resulting in an expression for the force that is proportional to the area of the surface and the depth squared in the case of a vertical wall. This concept is essential for determining both the magnitude of the hydrostatic force and its distribution along the surface.

Center of Pressure

The center of pressure is the point on a submerged surface where the resultant hydrostatic force is considered to act. Due to the linear increase in pressure with depth, the center of pressure is located below the centroid of the surface. Its calculation involves taking the moments of the pressure distribution about an axis and is crucial for understanding how forces produce torque on structures in contact with fluids.

Scaling Effects on Hydrostatic Force

The effect of scaling the dimensions of a submerged surface, such as doubling the wall height, is significant in hydrostatics. Since the hydrostatic force on a vertical wall depends on the square of the depth (due to the integration of pressure that itself increases linearly with depth), increasing the wall height leads to a more than proportional increase in force. This means that doubling the wall height results in the force increasing by a factor of four, rather than merely doubling, making the geometry of the container a critical factor in structural design.

Transcript

00:03 We're told that a swimming pool is 20 feet wide and 40 feet long, with an inclined plane for the bottom, and the shallow end of the plane has a depth of 3 feet, deep end has a depth of 9 feet.

00:19 We're told that the pool is also full of water, and we're asked to estimate the hydrostatic force on different parts of the pool.

00:28 So in part a, we're asked to estimate the force on the shallow end of the pool.

00:40 Well, if we're looking at the shallow end of the pool, the area of a strip is approximately the width of the pool 20 feet times delta x, and the pressure on the strip is, because we're using feet, i'll use the density unit delta times the depth, which is a very much.

01:31 X i and therefore the force is the integral from well the depth could be as little as zero and as large in the shallow end as three of the pressure on the strips so delta times x i star which becomes x in the limit times twenty the area which is twenty and then delta x becomes d x so we get taking into derivatives 20 delta times 1 half x squared, evaluated from 0 to 3, or 20 delta times 9 halves, which is 90 delta, which is equal to approximately 90 times 62 .5, which is equal to 5 ,625 pounds.

02:53 And to round this using scientific notation, this is approximately 5 .63 times 10 to the third pounds.

03:19 In part b, we're asked to find the hydrostatic force on the deep end of the pool.

03:28 So this is similar to part a, except for now we're integrating from 0 to 9.

03:35 So integral from 0 to 9 of delta x, 20 dx.

03:45 This is equal to 20 delta times 1 half x squared, evaluated from 0 to 9, which is equal to 20 delta times 81 halves, or 810 delta.

04:10 And subbing in delta equals 62 .5, this is approximately equal to 5 ,6002 ,000.

04:22 25, or sorry, 50 ,625.

04:27 And rounding this to three places and using scientific notation, this is about equal to 5 .06 times 10 to the fourth.

04:41 And the unit is pounds.

04:49 Part c, press to find the force on one of the sides of the pool.

05:04 Well, think of this, as an as a cross section, we have this rectangle here, and this is the shallow end over here, and then we have the deep end over here.

05:25 So we have a right triangle under this rectangle, and the right triangle makes an angle theta here from the rectangle.

05:40 Now, consider, oh sorry, it's actually part d, my mistake.

06:09 So instead, the first three feet, the length of the side is constant, add 40 feet.

06:25 So for the first three feet below the surface, this is true.

06:42 Now once we go below three feet, in particular if we go from three to nine, so if our depth is between three and nine, then, by similar triangles, we have that the length a over the length 40, this is going to be equal to the height of the smaller triangle, 9 minus x over the height of the larger triangle, which is 9 minus 3 or 6.

07:35 And so we have the length a is equal to 40 times 9 minus x over 6.

07:58 Therefore, the force on the side on one of the sides is going to be the integral, first from 0 to 3, of we have the density, delta, times the depth, which is x, times the area, which is 40 times dx.

08:31 Plus integral from 3 to 9 of the density, which is delta times x times 40, well the area which is 40 times 9 minus x over 6, dx.

09:00 And so simplifying and taking a derivative, we get 40 delta times 1 half x squared from 3 to 9.

09:15 Sorry, from 0 to 3, plus and then 20 thirds delta integral from 3 to 9, 9x minus x squared, dx.

09:53 So we get 180 delta plus 20 over 3 delta times 9 halves x squared minus 1 3x cubed from 3 to 9...

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