Hydrostatic Force MCQ Quiz - Objective Question with Answer for Hydrostatic Force - Download Free PDF

Explanation:

Hydrostatics:

Hydrostatics deals with fluids at rest. This branch studies the conditions of fluids that are not in motion, particularly focusing on the forces exerted by fluids at rest. The main concepts include pressure distribution, buoyancy, and the effects of gravity on fluids.

• Key Concept: In a fluid at rest, the pressure at any given point is due to the weight of the fluid above that point. The pressure increases with depth because the weight of the fluid above adds to the pressure experienced at lower points.

• Applications:

  • Calculation of pressure in a fluid at a certain depth.

  • Understanding buoyancy (the force that allows objects to float).

  • Determining the forces acting on submerged bodies.

Hydrodynamics:

Hydrodynamics, on the other hand, is concerned with fluids in motion. It studies the behavior of fluids when they are moving, whether it’s the flow of water through pipes, the airflow over an aircraft wing, or ocean currents. This branch examines how forces, velocity, pressure, and energy interact in a moving fluid.

• Key Concept: In hydrodynamics, the flow of the fluid is analyzed. It may be steady or unsteady, laminar (smooth and orderly) or turbulent (chaotic and irregular), and it often involves the study of velocity distributions, forces acting on boundaries (like pipe walls), and how pressure varies with speed.

• Applications:

  • Analyzing fluid flow in pipes, ducts, and open channels.

  • Study of aerodynamics and hydrodynamics of vehicles (airplanes, ships).

  • Designing hydraulic systems and turbines.

Additional Information Key Differences between Hydrostatics and Hydrodynamics:

1. Fluid Motion:

   • Hydrostatics: Studies fluids at rest (no motion).

   • Hydrodynamics: Studies fluids in motion (flowing fluids).

2. Pressure Considerations:

   • Hydrostatics: Focuses on the pressure due to the weight of the fluid. Pressure at any point depends on the depth of the fluid.

   • Hydrodynamics: Focuses on how pressure changes with the velocity of the fluid and how forces act due to the fluid's motion.

3. Flow Conditions:

   • Hydrostatics: Since the fluid is not moving, concepts like flow type (turbulent or laminar) don't apply.

   • Hydrodynamics: It considers both laminar flow (where the fluid flows in smooth layers) and turbulent flow (where the fluid undergoes chaotic fluctuations).

In fluid mechanics, the pressure within a submerged body in a fluid at rest increases with depth due to the weight of the fluid above it. This is described by the hydrostatic pressure formula:

Hydrostatic pressure formula:

P = P₀ + ρgh

Where:

• P = Pressure at a certain depth

• P₀ = Atmospheric pressure at the surface

• ρ = Density of the fluid

• g = Acceleration due to gravity

• h = Depth below the surface

From the formula, it is clear that pressure increases with depth (h). Therefore, the pressure will be greatest at the bottommost point of the submerged body.

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°

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

The following facts can be concluded from the above equations:

• Center 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 center of pressure is independent of the specific weight of the liquid and is consequently 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.

Explanation:

Considering PQR

P_R + 4 × (0.8γw) = 10γw

⇒ PR = (10 3.2)γw = 6.8γw

Now, 

Considering force prism QRST 

\(\rm F_{1}=\frac{(6.8+10)}{2} \gamma_{w} \times 4 \times 5=168 \gamma_{w}=1680 \mathrm{kN}\)

Considering force prism PQTU 

\(\rm F_{2}=\frac{(10+16)}{2} \gamma_{w} \times 6 \times 5=3900 \mathrm{kN}\)

So, F = F₁ + F₂ = 1680 + 3900 = 5580 kN

Vertical Hydrostatic Force: \( F_v = \gamma \cdot V \), Volume of Water above the Gate: \( V = \frac{\pi r^2}{4} \cdot 1 \, \text{(per unit width)} \), \( r = 1 \, \text{m} \), Substitute: \( F_v = 9810 \cdot \frac{\pi \cdot 1^2}{4} = 7704.75 \, \text{N/m} \)

Centre of pressure

\(\begin{array}{l} {{\rm{h}}^{\rm{*}}} = {\rm{\bar X}} + \frac{{{{\rm{I}}_{\rm{G}}}}}{{{\rm{A\bar X}}}} = \frac{{\rm{h}}}{3} + \frac{{\frac{{{\rm{b}}{{\rm{h}}^3}}}{{36}}}}{{\frac{{{\rm{bh}}}}{2}.\frac{{\rm{h}}}{3}}}\\ = \frac{{\rm{h}}}{3} + \frac{{\rm{h}}}{6} = \frac{{\left( {2 + 1} \right){\rm{h}}}}{6} = \frac{{\rm{h}}}{2} \end{array}\)

Important point:

Concept:
We know that the centre of pressure is the point at which resultant pressure force due to fluid acts and it is given by-

\({\rm{h}} = {\rm{\bar x}} + \frac{{{{\rm{I}}_{\rm{G}}}}}{{{\rm{A\bar x}}}}{\left( {\sin {\rm{θ }}} \right)^2}\)

Where, h represents the centre of pressure, I_G = second moment of area, A = area, and x̅ = centre of gravity.

Note:

In case of vertical surface, θ = 90° and Sin θ = 1

\(\frac{{{{\rm{I}}_{\rm{G}}}}}{{{\rm{A\bar x}}}}{\left( {\sin {\rm{θ }}} \right)^2}\) = \(\frac{{{{\rm{I}}_{\rm{G}}}}}{{{\rm{A\bar x}}}}\)

\({\rm{h}} = {\rm{\bar x}} + \frac{{{{\rm{I}}_{\rm{G}}}}}{{{\rm{A\bar x}}}}\)

Now, as the depth of immersion increases, x̅ increases which results in an increase in h but the rate of increment would be decreased because the factor \(\frac{{{{\rm{I}}_{\rm{G}}}}}{{{\rm{A\bar x}}}}\) has x̅ in the denominator.

Hence h will come closer to x̅.

Important Points

For θ = 90°, which means vertical, the centre of pressure is farthest to the centre of gravity.

For θ = 0°, which means horizontal, the centre of pressure is coinciding with the centre of gravity.

Explanation:

In a real fluid, the following forces are present

Gravity force (Fg) due to gravity

Pressure force (Fp) due to the pressure of the fluid

Viscous force (Fν) due to viscosity

Tension force (Fs) due to surface tension

Turbulent force (Ft) due to turbulence.

and (Fc) due to compressibility.

∴ Fnet = Fg + Fp + Fν + Fs + Ft + Fc

• If Fnet = Fg + Fp + Fν + Ft this is known as Reynold’s equations of motion.
• If Fnet = Fg + Fp + Fν this is known as the Navier-Stokes equation of motion.
• If Fnet = Fg + Fp this is known as Euler’s equation of motion.

∵ the fluid is in rest i.e. the velocity gradient is zero, therefore no shear stress/force will be acting and no external force is acting on it, the only force present is due to gravity (Fg) and pressure of the fluid (Fp) i.e. normal force only.

Concept:

Total pressure is defined as the force exerted by a static fluid on a surface either on curved when the fluid comes in contact with the surface.

Centre of pressure is defined as the point of application of the total pressure on the surface.

The point of action of total hydrostatic force on the submerged surface is called the Centre of Pressure.

G = Centre of gravity of plane surface

P = Centre of pressure

F = ρgAx̅

Calculations:

Force on the dam wall (F) = Pressure at the center (P) × Projected area of the dam (A) = ρgAx̅

The pressure at the center (P) = ρ × g × x̅

Given:

Height of the dam (H) = 20 m

Height of the centroid from the water surface (x̅) = H/2 = 10 m

Density of water (ρ) = 1000 kg/m³

Width of dam (w) = 25 m

Area of dam (A) = 20 × 25 = 500 m²

Concept:

Water pressure at bottom of the tank (P) =ρgh

Where ρ = Density of water,

g = Acceleration due to gravity,

h = Height of the water in the tank.

Total water force at bottom of the tank (F) = Water pressures at bottom of the tank (P) × Area of the tank bottom (A)

Total water force at bottom of the tank (F) = ρghA

Calculation:

Given data:

Height of tank (h) = 3 m

Width of the tank (b) = 2 m and length of tank = 2 m

Total water force at bottom of the tank (F) =?

g = 9.81 m/s

ρ = 1000 kg/m³

Area of tank bottom (A) = 2 × 2 = 4 m²

Total water force at bottom of the tank (F) = 1000 × 9.81 × 3 × 4 = 117720 N

Total water force at bottom of the tank (F) = 117.720 kN

Concept:

Whenever a static mass of fluid comes into contact with a surface the fluid exerts force upon that surface. The magnitude of this force is known as the hydrostatic force or total pressure force.

The magnitude of hydrostatic force is given by F = ρgh̅A 

Where ρ = Density of the fluid, h̅ = Depth of center of gravity of the surface from free liquid surface, A = Area of the surface

Calculation:

Given, h = 10 m

h̅ = 10/2 = 5 m

Area per metre length = 10 × 1 = 10 m²                  

Concept:

When any plate is immersed vertically in water having any arbitrary shape,

total horizontal force is given as 

F_H = ρgAx̅

Where,

ρ = density of the fluid (kg/m3),

g = acceleration due to gravity (m/s2)

x̅ = depth of the center of gravity of the submerged body from the free surface (m)

A = total area of the surface (m2)

Calculation:

Given: Diameter of circular plate d = 2 m, x̅ = 2.5 m, ρ = 1000 kg/m3

\(\begin{aligned} & A=\frac{\pi}{4} \times d^2=\frac{\pi}{4} \times 2^2=3.141 \\ & F_H=\rho g A \bar{x}=1000 \times 9.81 \times 3.141 \times 2.5 \\ & F_H=77047.55 \mathrm{~N} \approx 77048 \mathrm{~N} \end{aligned}\)

 Additional Information

Note: When nothing is given about g, take g = 9.81 m/s2.

Don’t get confused between the centre of gravity and the centre of pressure while calculating F_H.

While calculating F_H centre of gravity is considered but the horizontal force acts at the centre of pressure.

Position of centre of pressure,
\(\bar{h}=\frac{I_{x x} \sin ^2 \theta}{A \cdot \bar{x}}+\bar{x}\)

Concept:

Hydrostatic force per unit width on the vertical side of beaker is given as

F = \(\frac 1 2\)ρgh2 

Calculation:

Given:

h' = h/4

The hydrostatic force will be

F' = \(\frac 1 2\)ρg(h/4)2 

F' = \(\frac{1}{16}\) ×\(\frac 1 2\)ρgh2 

F' = \(\frac{1}{16}\)F

Concept:

The Centre of Pressure is the average location of all of the pressure acting upon a body moving through a fluid.

The depth of the centre of pressure

\({h_{cp}} = \bar h + \frac{{{I_G}}}{{A\bar h}}\)

Calculation:

\({h_{cp}} = \frac{h}{2} + \frac{{b{h^3}/12}}{{\left( {b \times h} \right) \times \frac{h}{2}}}\)

Concept:
We know that the centre of pressure is the point at which resultant pressure force due to fluid acts and it is given by-

\({\rm{h}} = {\rm{\bar x}} + \frac{{{{\rm{I}}_{\rm{G}}}}}{{{\rm{A\bar x}}}}{\left( {\sin {\rm{\theta }}} \right)^2}\)

Where, h represents the centre of pressure, IG = second moment of area, A = area, and x̅ = centre of gravity.

Now, as the depth of immersion increases, x̅ increases which results into increase in h but the rate of increment would be decreased because the factor \(\frac{{{{\rm{I}}_{\rm{G}}}}}{{{\rm{A\bar x}}}}{\left( {\sin {\rm{\theta }}} \right)^2}\) has x̅ in the denominator.

Hence h will come closer to x̅.

Important Points

For θ = 90°, means vertical, the centre of pressure is farthest to the centre of gravity.

For θ = 0°, means horizontal, the centre of pressure is coinciding with the centre of gravity.