# Hydrostatic force on a semicircular wall

• Ryker
In summary, the conversation discusses the computation of hydrostatic force on a semicircular wall. The width of the strip is 2x and not πx as it is a planar wall and not a spherical one. The conversation also clarifies the misunderstanding about the wall being viewed in 3D.
Ryker

## Homework Statement

This isn't really a homework problem, it's a passage in the lecture notes that I don't quite get wholly. So here's the passage:

"To compute the hydrostatic force on a semicircular wall of radius a which forms the end of a swimming pool filled with water, between D(y) = 0 and D(y) = a, we compute
$$F = \rho g \int_0^a yw(y)dy.$$If (x, y) is a point on the semicircular edge of the wall, then x2 + y2 = a2. Hence w(y) = 2x = ..."

## The Attempt at a Solution

The part I don't get is the bolded part. Why is width of the strip 2x and not πx? It seems I'm not picturing the semicircular wall correctly, because I can't wrap my head around this. But from the description it seems as if the semicircular wall is a quarter of a sphere with radius a. So the x component would denote the radius of the horizontal strip on that sphere, and since that horizontal strip represents half of the perimeter of the whole circle, its width would be πx.

I don't know, any help here would be greatly appreciated.

edit: Or is this semicircular wall meant in a way that the going from the middle of the pool to this wall, that wall in your view would represent the bottom half of the circle as viewed in, say, a circle in the coordinate system sketched in your notes? Then if y is the down component, and x is the lateral one, 2x would make sense, of course, but is this how you picture the wall from the wording, as well?

Last edited:
Because this this is a semi-circle that is symmetric about the y-axis. The distance from (0, y) to (x, y) on any curve is x. The distance from (-x, y) to (x, y) is 2x. That has nothing to do with the curve being a circle.

So you interpreted the wall as being like what I posted under "edit" then (ie. a vertical wall with the shape of a semicircle), and not the spherical thing I was talking about in the part of the post prior to the edit?

Hi Ryker!
Ryker said:
"To compute the hydrostatic force on a semicircular wall of radius a which forms the end of a swimming pool filled with water, …"

Yes, "semicircular" must mean planar, so the wall is flat.

tiny-tim said:
Hi Ryker!

Yes, "semicircular" must mean planar, so the wall is flat.
Ah, thanks, it's easy to see 2x in this case, and it seems I shouldn't have gone 3D on the problem

## 1. What is hydrostatic force on a semicircular wall?

Hydrostatic force on a semicircular wall is the pressure exerted by a fluid on a semicircular wall due to the weight of the fluid above it. It is a result of the fluid's weight and density.

## 2. How is hydrostatic force calculated on a semicircular wall?

Hydrostatic force on a semicircular wall can be calculated using the formula F = ρghA, where F is the hydrostatic force, ρ is the density of the fluid, g is the acceleration due to gravity, h is the depth of the fluid, and A is the area of the semicircular wall.

## 3. What factors affect hydrostatic force on a semicircular wall?

The hydrostatic force on a semicircular wall is affected by the density of the fluid, the depth of the fluid, and the acceleration due to gravity. It is also influenced by the shape and size of the semicircular wall.

## 4. How is hydrostatic force on a semicircular wall used in real-life applications?

Hydrostatic force on a semicircular wall is used in various real-life applications, such as designing dams, calculating the pressure on a ship's hull, and understanding the stability of structures in water.

## 5. Can the hydrostatic force on a semicircular wall be negative?

No, the hydrostatic force on a semicircular wall cannot be negative. It is always positive and acts outward from the surface of the wall, perpendicular to the fluid's surface.

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