Pressure Force on Submerged Bodies MCQ Quiz in தமிழ் - Objective Question with Answer for Pressure Force on Submerged Bodies - இலவச PDF ஐப் பதிவிறக்கவும்

Explanation: 

Whenever a static mass of fluid comes into contact with a surface, the fluid exerts a force upon that surface. The magnitude of these forces exerted on the surface is known as hydrostatic resultant force or pressure force or total pressure.

Let us consider a vertical plane surface submerged in liquid,

Let A = Total area of the surface, h̅ = Distance of centre of gravity from the free surface of the liquid, G = Centre of gravity of plane surface, P = Centre of Pressure, h* = Distance of centre 0of pressure from the free surface of liquid

Total Force F: It is determined by dividing the entire surface into small strips, force on small strip is determined and then it is integrated get the force on the entire surface.

Let us consider a small strip of thickness dh and width b at a depth of h,

Pressure intensity on the strip p = ρgh

Area of the strip = b × dh

∴ Total pressure force on the strip, dF = p × Area = ρgh × b × dh

∴ Total pressure force on the whole surface

\(F = \;\smallint dF = \;\smallint {\rm{\rho gh}} × {\rm{b}} × {\rm{dh}} = \rho g\smallint b × h × dh\)

\(\smallint b × h × dh = \;\smallint h × dA\) = Moment of the free surface of the area about the free surface of the liquid

which is equal to Area of the surface × Distance of the Centre of Gravity from the free surface of the liquid = A × h̅

∴ F = ρgh̅ × A = Pressure at the centroid × Area of the surface

So, Total pressure = Pc  A

Where Pc is the pressure at the centroid.

Concept:

• According to Pascal’s Law, the pressure or intensity of pressure at a point in a static fluid is equal in all directions.
• For a plane surface of arbitrary shape immersed in a liquid in such a way that the plane of the surface makes an angle θ with the free surface of the liquid:

A = Total area of an inclined surface, h̅ = Depth of centre of gravity of inclined area from a free surface, h* = Distance of centre of pressure from the free surface of a liquid

\({h^*} = \frac{{{I_G}{{\sin }^2}\theta }}{{A̅ h}} + \bar h\)

For vertical plane surface: θ = 90°

\({h^*} = \frac{{{I_G}}}{{A \bar h}} + \bar h\)

Centre of the pressure of rectangular plate:

Consider a rectangular plate of width b and height h is immersed in water.

h̅ = Depth of centre of gravity of rectangular plate from free surface = h/2

Area = bh2 Moment of inertia through centre of gravity. IG = bh3/12

h* = Distance of centre of pressure from the free surface of a liquid = \({h^*} = \frac{{{I_G}{{\sin }^2}\theta }}{{A̅ h}} + \bar h\)

For vertical plane surface: θ = 90°

\({h^*} = \frac{{{I_G}{{\sin }^2}\theta }}{{A̅ h}} + \bar h=\frac{{\frac{{bh^3}}{{12}}}}{{bh~\times~\frac{{h}}{{2}}}}+\frac{{h}}{{2}}=\frac{{2h}}{{3}}\)

Additional Information

The following facts can be concluded from the above equations:

• Centre of pressure lies below the centroid because for any plane surface, the factor \(\frac{I_G}{A \bar h}\) is always positive
• Deeper the surface is lowered into the liquid (i.e. greater is the value of h̅), closer comes the centre of pressure to the centroid of the area
• Depth of the centre of pressure is independent of the specific weight of the liquid and is consequently same for all liquids

Concept: 

Total Hydrostatic Force on Inclined Plane Surface:

Let a plane surface of area A is submerged wholly in the fluid of specific weight γ. The surface is held inclined and making an angle θ with horizontal. The intersection of the plane in which the surface is placed and the free surface of the liquid is represented by axis OO.

Let h̅  and y̅  are the distance of the centroid of the area from the free surface of the liquid and axis OO along the inclined plane respectively.

Total hydrostatic force is given by;

F = γAh̅  = ρgAh̅ 

Calculation:

Given:

ρ = 850 \(\frac {kg}{m^3}\), g = 9.8 \(\frac {m}{s^2}\), A = 0.75 × 2.4 = 1.8 m²,

0.75 m side horizontal and just at the water surface ⇒ h_cg = 1.2 m;

Now Inclined at 60° ⇒ h_cg = 1.2 sin 60⁰ = 1.039 m = ̅h  

Now the force will be

F_pressure = ρg h̅ × A  = 850 × 9.8 × 1.039 × 1.8 = 15616.7 N = 15.6 kN

Note: The centre of pressure is where the resultant force will act and it is not involved in the calculation of the hydrostatic Force.

Explanation:

Hydrostatic pressure, P = ρgh

Hydrostatic force, F = ρgh̅A

Where h̅ is the depth of the centroid of the area A of the surface

\({h^*} = \frac{{{I_G}{{\sin }^2}\theta }}{{A\bar h}} + \bar h\)

Since the pressure increases with the depth, the centre of pressure P must lie below the centroid of the area of the surface.

For vertical plane surface: θ = 90°

\({h^*} = \frac{{{I_G}}}{{A\bar h}} + \bar h\)

Consider a surface is rectangular, with dimension b × d, and its one end is placed immediately beneath the free level of the liquid.

The centroid is located at d/2 below the free level of the liquid.

h̅ = d/2, I_G = bd³/12

\({h^*} = \frac{{\frac{{b{d^3}}}{{12}}}}{{bd \times \frac{d}{2}}} + \frac{d}{2} = \frac{2}{3}d\)

The centre of pressure for a plane vertical rectangular surface lies at a depth of two-third the height of the immersed surface.

Concept:

Pascal’s Law:

According to Pascal’s Law, the pressure or intensity of pressure at a point in a static fluid is equal in all directions.

For a plane surface of arbitrary shape immersed in a liquid in such a way that the plane of the surface makes an angle θ with the free surface of the liquid

where

A = Total area of an inclined surface,

h̅ = Depth of centre of gravity of inclined area from a free surface,

h* = Distance of centre of pressure from the free surface of a liquid.

\({h^*} = \frac{{{I_G}{{\sin }^2}\theta }}{{A̅ h}} + \bar h\)

For vertical plane surface: θ = 90°

\({h^*} = \frac{{{I_G}}}{{A \bar h}} + \bar h\)

The following facts can be concluded from the above equations:

• Centre of pressure lies below the centroid because for any plane surface, the factor IG/Ah̅ is always positive.
• The deeper the surface is lowered into the liquid (i.e. greater is the value of h̅), closer comes the centre of pressure to the centroid of the area.
• Depth of the center of pressure is independent of the specific weight of the liquid and is consequently the same for all liquids.

Additional Information

Total Pressure:

\(F_h=\omega ∫ydA \cos \theta\)

\(F_y=\omega ∫ydA \sin \theta\)

The total pressure on a surface immersed in a liquid depends on its inclination to the liquid surface.

Explanation:

Given:

Mass density of oil: 800 kg/m³

Acceleration due to gravity: 9.81 m/s²

Height of oil: 30 m

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Using Pascal's Law: 

Pressure due to oil height = Pressure due to water height

\((\rho g h)_{\text{oil}} = (\rho g h)_{\text{water}} \)

\(800 \times 9.81 \times 30 = 1000 \times 9.81 \times h \\ h = 24 \, \text{m} \)

Concept:

For a vertical plane surface, the depth of centre of pressure is given by

\({h_p} = \bar h + \frac{{{I_g}}}{{A\bar h}}\)

Where  is depth of centroid;

For a square plate, the moment of inertia about its diagonal axis is given by

\({I_g} = \frac{{{a^4}}}{{12}}\)   

Calculation:

Given a = 1 m

From the figure, the depth of centroid is given by 

\(\bar h = 1 + \frac{{\sqrt 2 a}}{2} = 1 + \frac{a}{{\sqrt 2 }} = 1 + \frac{1}{{\sqrt 2 }} = 1.707\)

The Depth of centre of pressure will be

\({h_p} = 1.707 + \frac{{\frac{{{a^4}}}{{12}}}}{{{a^2}\; \times\; 1.707}} = 1.707 + \frac{{{a^2}}}{{12\; \times\; 1.707}}\)

\({h_p} = 1.707 + \frac{1}{{12\; \times \;1.707}} = 1.756\;m\)

Explanation:

Centre of pressure:

The point of intersection of the line of action of the resultant hydrostatic force and the corresponding surface is called the center of pressure. The Centre of pressure is also defined as the point of application of the total pressure on the corresponding surface.

• According to Pascal’s Law, the pressure or intensity of pressure at a point in a static fluid is equal in all directions.
• For a plane surface of arbitrary shape immersed in a liquid in such a way that the plane of the surface makes an angle θ with the free surface of the liquid:

A = Total area of an inclined surface, h̅ = Depth of centre of gravity of inclined area from a free surface, h* = Distance of centre of pressure from the free surface of a liquid

\({h^*} = \frac{{{I_G}{{\sin }^2}\theta }}{{A̅ h}} + \bar h\)

For vertical plane surface: θ = 90°

\({h^*} = \frac{{{I_G}}}{{A \bar h}} + \bar h\)

The following facts can be concluded from the above equations:

• The Center of pressure lies below the centroid because for any plane surface, the factor \(\frac{I_G}{A \bar h}\) is always positive
• The deeper the surface is lowered into the liquid (i.e. greater is the value of h̅), the closer comes the center of pressure to the centroid of the area
• The depth of the center of pressure is independent of the specific weight of the liquid and is consequently the same for all liquids

Total Pressure:

\(F_h=\omega ∫ydA \cos \theta\)

\(F_y=\omega ∫ydA \sin \theta\)

The total pressure on a surface immersed in a liquid depends on its inclination to the liquid surface.

Concept:

Hydrostatic force or total pressure

The net force acting on one side of a gate due to the hydrostatic pressure intensity when the gate is supporting the fluid is given by,

F = γ × A × h̅ 

Where,

γ = Unit weight of the fluid which the gate supports = ρ × g

A = Cross-section area of gate

h̅ = Height of fluid above the C.G of gate

Calculation:

Given,

S.G of oil = 0.8

∴ ρ_oil = S.G of oil × ρ_water = 0.8 × 1000 = 800 kg/m³

F = ρgAh̅ 

\(\begin{array}{*{20}{l}} {{\rm{\bar h = 1}}{\rm{.5 + }}\frac{{\rm{2}}}{{\rm{3}}}{\rm{ × 1}}{\rm{.5 = 2}}{\rm{.5}}\;{\rm{m,}}\\{\rm{A = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ × 1}}{\rm{.5 × 2 = 1}}{\rm{.5}}{{\rm{m}}^{\rm{2}}}\;}\\ {{\rm{F = 800 × 9}}{\rm{.81 × 1}}{\rm{.5 × 2}}{\rm{.5 × 1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{ = 29}}{\rm{.43}}\;{\rm{kN}}} \end{array}\)

Calculation:

D = 1m;
y = 8 m;

F = \(\rho .g.\mathop h\limits^ - .A = \rho .g.\left( {y + {D \over 2}} \right).{\pi \over 4}{D^2}\)

\( = 1000 \times 9.81\left( {8 + {1 \over 2}} \right) \times {\pi \over 4}{1^2} = 65.4kN\)