Forces on Fluid Boundaries

Solids And Fluids of Class 11

Forces on Fluid Boundaries

Whenever a fluid comes in contact with solid boundaries it exerts a force on it. Consider a rectangular vessel of base size l × b filled with water to a height H as shown in the fig.(12.26). The force acting at the base of the container is given by

Fb = p × (area of the base)

because pressure is same everywhere at the base and is equal to ρgH.

Therefore, Fb = ρgH(lb) = ρglbH

Since lbH = V (volume of the liquid)

Thus,Fb = ρgV = weight of the liquid inside the vessel

Forces on Fluid Boundaries

Unlike the base, the pressure on the vertical wall of the vessel is not uniform but increases linearly with depth from the free surface. Therefore, we have to perform the integration to calculate the total force on the wall. Consider a small rectangular element of width b and thickness dh at a depth h from the free surface. The liquid pressure at this position is given by

p = ρgh

The force at the element is

dF = p(b dh) = ρgbh dh

The total force is F = ρgbForces on Fluid Boundaries

The total force acting per unit width of the vertical wall is

Forces on Fluid Boundaries(12.23)

The point of application (the centre of force) of the total force from the free surface is given by

hc = Forces on Fluid Boundaries(12.24)

where Forces on Fluid Boundaries is the moment of force about the free surface.


Forces on Fluid Boundaries

Since F = Forces on Fluid Boundaries, therefore,

hc = 2/3H (12.25)

The net resultant force acts at a depth 2/3 H  from the free surface.

Example: 12.19

Find the force acting per unit width on a plane wall inclined at an angle θ with the horizontal as shown in the figure(12.27).

Forces on Fluid Boundaries


Consider a small element of thickness dy at a distance y measured along the wall from the free surface. The pressure at the position of the element is

p = ρgh = ρgysinθ

The force is given by

dF = p(b dy) = ρgb(y dy) sin θ

The total force per unit width b is given by

Forces on Fluid Boundaries

Note that the above formula reduces to Forces on Fluid Boundaries for a vertical wall (θ = 90o)

Alternatively, the force on the inclined wall may be obtained in two parts viz. horizontal and vertical.

The horizontal force Fx acts on the vertical projection of the incline wall, i.e. Fx = 1/2 ρgbH2

The vertical force Fy acts due to weight of the liquid supported by the wall, i.e.

Forces on Fluid Boundaries

Fy = 1/2 ρgb(H)(H cot θ) = 1/2 ρgbH2 cotθ

The magnitude of the resultant force is given by

F = Forces on Fluid Boundaries 1/2 ρgbH2 cosec θ (12.27)

or F = Forces on Fluid Boundaries

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