The 5 -m-wide wall is in the form of a parabola. Determine the magnitude of the resultant force

The 5 -m-wide wall is in the form of a parabola. Determine...

Key Concepts

Hydrostatic Pressure

In fluids at rest, the pressure increases linearly with depth according to the relation p = ?gh, where ? is the fluid density, g is the acceleration due to gravity, and h is the depth. This concept is fundamental for analyzing forces on submerged surfaces because it defines how pressure varies over the area in contact with the fluid.

Differential Force Integration

The net force on a submerged surface is evaluated by summing the contributions of infinitesimal elements. Each differential area, dA, experiences a differential force given by dF = p dA. Integrating these differential forces over the entire surface provides the resultant force, which is essential in fluid statics.

Submerged Surface Geometry

The shape of a submerged structure directly influences how the pressure distribution is applied across its surface. For instance, a curved or parabolic wall requires careful parameterization of the surface area element during integration. Understanding the geometry is therefore crucial for accurately computing the resultant forces.

Resultant Force Calculation

The resultant or net force on any submerged body is obtained by integrating the pressure over the area of the object. This process involves setting up an integral that accounts for both the variation of pressure with depth and the geometry of the surface, resulting in a function describing the force as a function of depth.

Graphical Analysis of Force Variation

Plotting the computed resultant force versus depth allows for visualizing how the force increases nonlinearly with depth. Such graphical representations help in understanding the impact of varying depth on the magnitude of the force, which is important for design and analysis in engineering applications.

Transcript

00:01 In this problem, we're asked to consider a 5 meter wide well in the form of parabola.

00:07 And i've drawn the diagram here.

00:11 This is water.

00:19 And we're asked to determine the magnitude of the resultant force on the wall as a function of depth.

00:24 H or as the plot the results of the force versus h for the values given here in increments of delta h of 0 .5 meters.

00:36 Okay, so let's first find our horizontal component of our force.

00:43 This will be the density of water times g times h of a and b.

00:50 And this will equal 1 ,000 kilograms per cubic meter times 9 .81 meters per second squared times h times five meters.

01:11 This will equal 49 .05 times 10 to the third h.

01:20 And then for fh we can get this will be equal to one half times 49 .05 times 10 to the third h times h.

01:44 That'll equal 24 .525 times 10 to the third h squared.

01:54 Okay, and then the vertical component to the result in force is the weight of the column of water.

02:41 And that's above the surface ab.

02:51 Okay, so let's figure what this is going to be.

02:53 This will be equal one -third times h squared over four times h.

03:20 And then we'll get our force of v will be equal to, and that will be equal to 1 ,000.

03:49 Okay.

03:52 Then the magnitude of the resulting force is equal to a square root of f, and this will be equal to equal to plus.

04:40 So my f r where h equals m.

05:14 Okay, so here's my answer for my result and force.

05:19 Then we're asked to plot some values.

05:22 And i'm gonna put the plot values right here, and we're told to go from zero.

05:35 I'm gonna put these down.

05:59 Okay, so then my fr in kilonutons would be i'll have to do the rest on another line...

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