Consider a horizontal layer of the dam wall of thickness dx located a distance x above the reservoir floor. what is the magnitude df of the force on this layer that results from adding the water to the reservoir?

Respuesta :

The force on the layer will be equivalent to the weight of water on it. This is given by:
F = mg; m is the mass of water and g is the acceleration due to gravity.

Answer:

[tex]F = [P_a + \rho g(H-x)]Ldx[/tex]

here

[tex]P_a[/tex] = atmospheric pressure

L = length of dam wall

H = height of water in the dam

[tex]\rho[/tex] = density of water

Explanation:

Let the length of the wall is "L"

Now here we know that pressure due to liquid filled in the dam at given position is given as

Let say the height of water in the dam is "H"

now the wall height where we took a small element is at height "x" from bottom

So pressure at that position is given due to water is

[tex]P = \rho g (H - x) + P_a[/tex]

here we know

[tex]\rho[/tex] = density of water

H - x = height measured from the top surface

[tex]P_a [/tex] = atmospheric pressure

now we have to find the Force on the wall

so here area of the element is given as

[tex]A = Ldx[/tex]

now force is given as

[tex]F = [P_a + \rho g(H-x)]Ldx[/tex]