NameRoll no. Arjun Vekariya Brijesh Ashlaliya06 Dhaval Gohel02(D2d) Prashant Manavar01(D2d) Fultariya Arpit05

The atmospheric air exerts a normal pressure upon all surfaces with which it is in contact, and it is known as atmospheric pressure.

A gauge pressure can be given using height of any fluid.

Manometers use the relationship between pressure and head to measure pressure 1- Simple manometer

Solution

Vertical Single manometer The volume that move from x-x to z-z datum = the volume that rise in right tube above x-x datum 1 2

Inclined single column manometers p

Solution

1 2

Example 1 The figure Solution: w gasoline = 6.65 kN/m2, w air = kN/m2

Solution:

Total pressure and centre of pressure Total Pressure: Total pressure is defined as the force exerted by a static fluid on a surface either plane or curved when the fluid comes in contact with the surface. This force always acts normal to the surface. Consider a vertically immersed surface as shown in figure. P= P1A1+P2A2+P3A3… Where, P1, P2, P3….= Intensity of pressure on different strips of the surface. A1, A2, A3….= Area of corresponding strips. centre of pressure The point at which, the total pressure acts on the surface immersed in liquid, is known as centre of pressure.

● PRESSURE DIAGRAME Total pressure and centre of pressure for a plane surface wholly submerged in a static liquid either vertically or inclined may also be determined by drawing a pressure diagramed. A Pressure diagram is graphical representation of the variation of the pressure intensity of over a surface a pressure diagram is prepared by plotting the pressure intensities at various points on the surface since pressure at any point acts in the direction normal to the surface the pressure intensities at various point on the surface are plotted normal to the surface. Show typical pressure diagram for horizontal, vertical and inclined plane surface.

Pressure Diagram (Cont.)

Horizontal plane surface If the submerged plane surface is horizontal, the pressure at all points lying on it is same and the pressure distribution is rectangular. Pressure intensity, p = w.h1 Where, w= specific weight of liquid. h1 = vertical depth of surface below free liquid surface. Hydrostatics force on the surface is F = pressure x area of the surface. F= wAh1

Vertical surfaces: In case of vertical plane surface immersed in water, pressure at any point varies with the depth of point below the free surface of the liquid. Let the top bottom edge of the plane surface be at vertical depth of h1 and h2 respectively below free surface of the liquid. Pressure intensity at the top edge, p1 = wh1 Pressure intensity at the bottom edge, p2 = wh2 Let, b = height of plate L= Length of plate Average pressure intensity p= wh1 + wh2 / 2

Hydrostatic force on vertical plane F= average pressure x area. Total pressure on curved surface Consider a curved surface AB wholly Submerged in a static fluid as shown in fig. Let DA is the Area of a small strip at a depth of h from water surface. Then intensity of pressure on the area dap = qgh pressure force on area da df = p x da This force df acts normal to the surface total force on the curved surface, F =∫ ∞g h. da But, here the direction of the forces on the small elements are not in the same direction, but changes from point. Hence, integration for curved surface is not possible.

The total force on the curved surface is, F= √ fx2+fy2 and inclination of resultant with horizontal, tanф= fy/fx Resolving the force df, Dfx = df sin =Q gh. da sin Dfy= dfy cos =Q gh. da. Cos as shown in fig. EF = dA FG= da sin = vertical projection of area da EG= da cos = horizontal projection of area da Hence, = Qg ∫ h. da.sin ф represents the total pressure force on the projected area of the curved surface on the vertical plane.

∫ h.da. cos is the total volume contained between the curved surface extended up to free surface.

Total pressure and centre of pressure consider a vertical surface immersed in a liquid as shown in fig. G= C.G. Of immersed surface P= centre of pressure h= Depth of cg. of immersed surface from the liquid.

h* = Depth of centre of pressure from the liquid surface. Let US Consider a small strip of thickness dh width b at depth h from of liquid. Total pressure on one strip, df= p. d. dh = wh. b. dh F= WA h ~ ….. Total pressure To find centre of pressure (h * ) : Total pressure (f) is acts at p, at a depth h * from the free surface of liquid. Moment of force df acting on a strip about free surface of liquid, = dF x h =wh x b x dh x h

Sum of moments of all such forces about fee surface of liquid. = ∫ wh x b x dh x h =w ∫ b. h². Dh = w ∫. h² dh =w ∫ h². dA B. dh =dA But, ∫ h². dA = ∫ b. h². Dh = M.I. of surface about free liquid surface = I ˛ Sum of moments about free surface = w. I ˛. Equating (ii) and (iii) F x h* = w. I ˛

wAhˉ x h* = w. I˛ h* = I˛ / Ahˉ Now, By the ‘parallel axis theorem’ of M.I. I˛ = Ig + ah² Where, Ig = M.I. of area about an axis passing through the C.G. of the area and parallel to the free surface of liquid. Substituting Ig in eq. (iv) h* = Ig + ah² / Ahˉ = Ig / Ahˉ + hˉ …. …… Center of pressure.

Archimedes ( BC), pre-eminent Greek mathematician and inventor, who wrote important works on plane and solid geometry, arithmetic, and mechanics. "Archimedes",Microsoft« Encarta« Encyclopedia ⌐ Microsoft Corporation. All rights reserved. Who is Archimedes?

Archimedes' Principle, law of physics that states that when an object is totally or partially immersed in a fluid, it experiences an upthrust equal to the weight of the fluid displaced. The principle is most frequently applied to the behaviour of objects in water, and helps to explain floating and sinking, and why objects seem lighter in water. It also applies to balloons in the air.

The key word in the principle is “upthrust” (or buoyant force), which refers to the force acting upward to reduce the actual weight of the object when it is under water. for example, a metal block with a volume of 100 cm 3 is dipped in water, it displaces an equal volume of water, which has a weight of approximately 1 N. The block therefore seems to weigh about 1 N less. UPTHRUST AND BUOYANT FORCE

SINKING AND FLOATING OBJECTS The reading of spring balance is 2.7 N The reading of spring balance is 1.7 N

What is the reading of spring balance if the wood is attached to it ? ZERO

From Archimedes’s Principle : Buoyant Force= Weight of fluid displaced = mg (note : F = ma) =  Vg (note :  = m ) V Thus F B =  V g Where …… F B = Buoyant Force or Upthrust  = Density of fluid V = Volume of fluid displaced or the volume of the object that immersed in the fluid.

Buoyant force = weight  the object floats and stationary Buoyant force > weight  the object moves up Buoyant force < weight  the object moves down

A floating object displaces its own weight of fluid in which it floats.

warm fresh water cold fresh water warm sea water cold sea water THINK!!!!! 1. Why the depth of ship immersed in the water different?

Fresh water less dense than sea water and warm water less dense than coldwater so warm fresh water need to be displaced more to keep the uptrust force equal with weight of the boat so it still can float.

2. If the plasticine is formed into a ball, it will sink. But when it is formed into a hull it will float. Why? -

BECAUSE…..

If ballast tanks empty  Upthrust > weight  submarine rises to surface If ballast tanks full  Upthrust < weight  submarine sinks to bottom SUBMARINE

1. The weight of the rock in air is 0.85N. When it is completely submerged in water, its weight is 0.45N. What is the buoyant force acting on the rock when it is completely submerged in the water ? Solution : Buoyant force = Actual weight – Apparent weight = 0.85 – 0.45 = 0.4N Example

2. A concrete slab weight 180N. When it is fully submerged under the sea its apparent weight is 105N. Calculate the density of the sea water if the volume of the sea water displaced by the concrete slab is 4800 cm 3. [ g = 9.8 Nkg -1 ] Solution : Buoyant force = actual weight – apparent weight = 180 – 102 = 72N

According to Archimedes’s principle Buoyant force = weight of sea water displaced Therefore, F = pVg so…. p = F / Vg = 72 / (4800 x x 9.8 ) = kg m -3

BUOYANCY When a body is immersed in a fluid, an upward force is exerted by the fluid on the body. This upward force is equal to the weight of the fluid displaced by the body and is called the force of buoyancy or simply buoyancy 4.3 CENTRE OF BUOYANCY It is defined as the point, through which the force of buoyancy is supposed to act. As the force of buoyancy is a vertical force and is equal to the weight of the fluid displaced by the body, the centre of buoyancy will be the centre of gravity of the fluid displaced.

META-CENTRE It is defined as the point about which a body starts oscillating when the body is tilted by a small angle. The meta-centre may also be defined as the point at which the line of action of the force of buoyancy will meet the normal axis of the body when the body is given a small angular displacement. Consider a body floating in a liquid as shown in Fig. 4.5 (a). Let the body is in equilibrium and G is the centre of gravity and B the centre of buoyancy. For equilibrium, both the points lie on the normal axis, which is vertical.

META-CENTRIC HEIGHT The distance MG, i.e., the distance between the meta-centre of a floating body and the centre of gravity of the body is called meta-centric height. 4.6 ANALYTICAL METHOD FOR META-CENTRE HEIGHT Fig. 4.6 (a) shows the position of a floating body in equilibrium. The location of centre of gravity and centre of buoyancy in this position is at G and B. The floating body is given a small angular displacement in the clockwise direction. This is shown in Fig. 4.6 (b). The new centre of buoyancy is at 5). The vertical line through B^ cuts the normal axis at M. Hence M is the meta-centre and CM is meta-centric height.

ANALYTICAL METHOD FOR META-CENTRE HEIGHT

65 Example 4.7 A rectangular pontoon is 5 m long, 3 m wide and 1.20 m high. The depth of immersion of the pontoon is 0.80 m in sea water. If the centre of gravity is 0.6 m above the bottom of the pontoon, determine the meta-centric height. The density/or sea water = 7025 kg/m3.

66 Example solution

CONDITIONS OF EQUILIBRIUM OF A FLOATING AND SUB-MERGED BODIES Stability of a Sub-merged Body

CONDITIONS OF EQUILIBRIUM OF A FLOATING AND SUB-MERGED BODIES Stability of a Sub-merged Body (a) Stable Equilibrium. When W = Fg and point B is above G, the body is said to be in stable equilibrium. (b) Unstable Equilibrium. If W = Fg, but the centre of buoyancy (B) is below centre of gravity (G), the body is in unstable equilibrium as shown in Fig (b). A slight displacement to the body, in the clockwise direction, gives the couple due to W and Fg also in the clockwise direction. Thus the body does not return to its original position and hence the body is in unstable equilibrium. (c) Neutral Equilibrium. If Fg = W and B and G are at the same point, as shown in Fig (c), the body is said to be in Neutral Equilibrium.

CONDITIONS OF EQUILIBRIUM OF A FLOATING AND SUB-MERGED BODIES Stability of Floating Body.

CONDITIONS OF EQUILIBRIUM OF A FLOATING AND SUB-MERGED BODIES Stability of a Sub-merged Body (a) Stable Equilibrium. If the point M is above G, (b) Unstable Equilibrium.. If the point M is below G,. (c) Neutral Equilibrium. If. If the point M is at the centre of gravity of the body, the floating body will be in neutral equilibrium.

71 Example A solid cylinder of diameter 4.0 m has a height of 4.0 m. Find the meta- centric height of the cylinder if the specific gravity of the material of cylinder = 0.6 and it is floating in water with its axis vertical. State whether the equilibrium is stable or unstable. D=4m h = 4 m =0.6 = Sp. gr. x h = 0.6 x 4.0 = 2.4 m

72 Example solution

PHYSICS IS SIMPLY FUN