Chapter -5(Section-1) Friction in Solids and Liquids

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1 Chapter -5(Section-1) Friction in Solids and Liquids Que 1: Define friction. What are its causes? Ans : Friction:- When two bodies are in contact with each other and if one body is made to move then the property due to which a tangential force is exerted between them, at their surfaces in contact, tending to oppose relative motion between them is called friction. Force of friction:- The force which opposes relative motion of two bodies in contact is called force of friction. There are two main causes of friction 1. Roughness of surface in contact 2. Strong atomic and molecular attraction between the two surfaces in contact. Que 2: Explain the origin of friction. Ans 1. Roughness of surface:- An opposing force always acts between the two bodies in contact, whenever one body tends to move or move over the body. This opposition force depends upon the nature of surfaces in contact. Surface of moving body appears smooth, but when observed through powerful microscope it is found that there are hills and dales. In rough as well as smooth surface irregularities are found. When two rough solid surfaces are in contact, the elevation of one surface fit into depressions of the other surface. As a result of such interlocking of the irregularities, the relative motion between the two surfaces is opposed, and which gives rise to the force of friction.. 1

2. Intermolecular force The molecules at the point of contact between the irregularities of the surface exert force of attraction on each other.

2 2. Intermolecular force The molecules at the point of contact between the irregularities of the surface exert force of attraction on each other. Similar types of molecule exert cohesive force on each other and different types of molecules exerts adhesive forces on each other, thus the point of contact are Cold welded together. The surface adhesion gives rise to frictional force which opposes relative motion. Hence there is opposition to the relative motion between two surfaces in contact. We can conclude that the causes of friction are. 1) Interlocking between the irregularities in the contact surface. 2) Intermolecular forces of attraction between the two surfaces. Note:- Adhesive forces between molecules of different substance are smaller as compacted to cohesive force between molecules of same substance. To reduce friction ball bearings used in a machine are of different material Que 3: Explain the nature of frictional force. Ans: Consider a rectangular block at rest on a horizontal surface having pulley at one end of surface. As shown as in figure.the forces acting on the block are. 1) The weight (w) of the block acts vertically downward direction which presses the surface and 2) the force exerted by the surface on the block in 3) vertically upward direction is called as normal reaction (N) 4) If applied force (F) on the block is increased by increasing weight in the pan the block initially remains at rest. 5) At particular value of applied force. Block is just starts moving on the surface. This value of force is called limiting force of static friction. 6) If an applied force is further increased then block is sliding continuously with uniform speed over the surface, such minimum required force is called force of kinetic friction. 2

3 Conclusion:- i. The forces of friction exist between two surfaces in contact only when one body tends to move or slide over the other surface. ii. In equilibrium, limiting force of friction action between two surfaces in contact is equal and opposite of applied force. iii. When a body is sliding on a surface in a particular direction, then force of friction is in opposite direction. If the direction of applied force is changed then the direction of limiting force of friction also changes with constant magnitude. Therefore force of friction is called as self adjusting force with magnitude and direction. Que 4 why bearings are made up of different materials from that of the material of machine? Ans : 1. The bearings are used in machine for reducing friction. 2. The force of attraction between the molecules of the same substance (cohesive force) is greater than the molecules of different substances (adhesive force). 3. So, if the two surfaces of same material are in contact, friction between them is greater than the surface of two different materials. Due to this bearing are used in machine are made up of different material from that of the material of machine. Que 5: Define static friction, Kinetic friction, rolling friction. Ans : A. Static Friction: Friction between two bodies in contact when one body just moves or tends to move over the other is called as static friction. B. Kinetic Friction: Friction between two bodies in contact when one body just move is sliding over the other is called as kinetic friction. (Dynamic or sliding friction) C. Rolling Friction: Friction between two bodies in contact, when one body is rolling over the other is called is called as rolling friction. Hence, Static friction > Kinetic fraction > Rolling friction. Que 6: Why is the kinetic friction less than static friction? 3

4 Ans: A large force is required to set a heavy box moving along a plane surface. If this box starts moving, smaller force is needed to keep it in motion. This observation shows that the kinetic friction is less than static friction. Que 7: state law of static friction. Ans : law as of Static friction : i) The limiting force of static friction (Fs) is directly proportional to normal reaction (N) between the two surfaces in contact. Fs N Fs = s N Where s is constant of proportionality and is called as coefficient of static friction. Thus the coefficient of static friction is defined as the ratio of liming force of static friction to the normal reaction. ii) The limiting force of static friction for any two given surfaces the surfaces in contact so long as the normal reaction remains the same. iii) The limiting force of static friction depends upon the nature and materials of the surfaces. Que 8: Describe the experiment to verify that limiting force to static friction depends upon the normal reaction between two surfaces in contact. Ans : Verification of first law : A wooden block of weight (W1) is kept on horizontal surface. One end of string is tied to the block passes over the pulley and of the light pan is attached at the other end string. Weights are added in the pan so as to increase the driving force (F1), till the block just begins to move, w1 and f1 are noted. 4

5 Repeat the above procedure by changing the weight on the block as shown in Fig. (b). W2 and F2 are noted. When block just moves then applied force is equal limiting force of static friction, F1 = Fs1 andf2 = FS2 and also W1 = N1 and W2 = N2 It is found that This shows that Fs N, Thus first law of static friction is verified. Que 9: Describe the experiment to verify that limiting force to static friction is independent of the apparent area of contact. Ans : Experimental verification of second law of Static Friction : Two identical blocks of same material and same polished surfaces are arranged one above the other on horizontal plane surface as shown in Figure (a). Weights are added in the pan. Till the block just tends to move. W and F1 are noted. Normal reaction = N1 + N2 = W Then keep the two block one after another, on the horizontal plane surface (as shown in Fig. (b). in this arrangement, area of surfaces in contact increases but W=N1 + N2 remains the same, weights are added in the pan till the blocks just begins to move. F2 is noted. It is found that F1 = F2 5

Thus shows that the limiting force of static friction is independent of apparent area of surfaces in contact as long as normal reaction remains same.

6 Thus shows that the limiting force of static friction is independent of apparent area of surfaces in contact as long as normal reaction remains same. Que 10: Describe the experiment to verify that limiting force to static friction depends upon nature and material of surface in contact. Ans : Experimental verification of third law of Static Friction : a) The Wooden block is placed on a horizontal surface and find limiting force static friction as explained earlier, F1 is noted. Another wooden block of same material and same normal reaction but surface is polished. Find the limiting force of static friction, F 2 is note. It is found that F1 F2 This shows that limiting force of static friction. Depends upon nature of surface in contact. b) When blocks of different materials (such as glass, wood, iron, aluminum etc.) having same normal reaction and same polished surface are used. The values of limiting force of static friction in each case is noted it is found that F1 F2 i.e. limiting force of static friction depend upon material of surface in contact. Que 11: state laws of kinetic friction Ans : Laws of Kinetic Friction : The friction that exists when one body is sliding over the other is called kinetic friction. The laws of kinetic friction are stated as follows: i).the force of kinetic friction (Fk) is directly proportional to the normal reaction between two surfaces in contact. Fk N Fk = k N 6

7 Where k is constant of proportionality and is called as coefficient of kinetic friction. The ratio of force of kinetic friction to the normal reaction is called as the coefficient of kinetic friction. ii) Force of kinetic friction is independent of shape and apparent area of the surfaces in contact. iii) Force of kinetic friction depends upon the nature and material of the surface in contact. iv) The Magnitude of the force of kinetic friction is approximately independent of the relative velocity, provided the relative velocity is neither too large nor too small. NOTE:- The laws of kinetic friction can be verified in the same manner as that of the laws of static friction. The only difference in this experiment is that applied force is adjusted in such manner that the body is set sliding (in uniform motion) on horizontal surface. Que 12 Explain the term a) Coefficient of static friction, b)coefficient of kinetic friction Ans : a) Coefficient of static friction ( s): The coefficient of static friction is defined as the ratio of liming force of static friction to the normal reaction b) Coefficient of kinetic friction ( k): The ratio of force of kinetic friction to the normal reaction is called as the coefficient of kinetic friction Que 13 : Define pressure, state its SI, CGS unit and dimensions. Ans : Pressure : The thrust exerted by a liquid at rest per unit area normal to the surface in contact with the liquid is called pressure. P = F/A SI unit of pressure = N/m² CGS unit of pressure = dyne/cm² 1 N/m² = 1 pascal (P a ) Dimension of pressure = [M¹ L ¹T ²] 7

8 Que 14: Derive an expression for pressure exerted by liquid column. Ans: Suppose a homogeneous liquid of density is contained in a cylinder of cross sectional area A Liquid is partly filled in a cylinder, let h be the height of liquid column as shown in figure. Weight of that liquid column exerts a downward thrust. Weight of liquid column = volume of liquid x density of liquid x g Therefore = Ah g weight of liquid column Pressure = cross sectional area = = h g This is an expression for pressure. The pressure due to liquid column is directly proportional to i) Depth of the point below the free surface of the liquid ii) The density of the liquid Let P0 be atmospheric pressure, which acts on the free surface of liquid, therefore the total pressure at the depth h from the free surface is equal to the atmospheric pressure and pressure due to liquid column. Total pressure = P0 + h p g = P0 + Gauge Pressure 8

9 Hydrostatic pressure PHYSICS-MANIA Que 15 : Explain the hydrostatic paradox? Ans : Hydrostatic paradox:- The Shape of the vessel containing the liquid does not affect the pressure. This is known as Hydrostatic paradox. Explanation:- 1. Consider three different shapes of vessel A,B and C of same base areas. 2. Let all the vessel are filled with liquid up to same height but volume of liquid are different in different vessels. 3. Liquid exerts force on the base as well as on the wall of the vessels. At the same time the base as well as of the vessel exert equal and opposite force on liquid. 4. Vessel A is in vertical position, force of reaction exerted on water by the wall of container is horizontal and there is no vertical component; the force exerted by water at the bottom is exactly equal to weight of water in the vessel. 5. Vessel B. its open end is wider, the incline surface exerts force of reaction on water and this force split up in to two components, upward vertical components add together & balance the part of weight of the water on the bottom. This reduces the effective force. 6. vessel C, its open end is smaller as compare to vessel B, the incline surface exerts force of reaction on water and this force split up in to two components, downward component are add together. This is added to the weight of the water in the vessel. This increases the effective force. Condition 1 st, 2 nd 3 rd shows that the downward force exerted by water on the bases of all the vessels are equal. 7. Therefore, the force exerted by liquid on the bases of all the vessels is equal. Que 16: State pascal s law. State application of pascal s law. Ans: Pascal s law: The pressure applied to an enclosed fluid is transmitted equally to all other point of the liquid and to the walls of the container. 9

10 Application of Pascal s law: i) Hydraulic press: Hydraulic press is based on Pascal s law. It consists of two cylinders C and D filled with water. A large platform M is fitted to the piston P2 & strong metal frame (N) fixed at its top. Cotton is kept between M and N.A2 is chose to be greater than A1. Suppose piston p1 is pressed downwards with a force FP this pressure is transmitted undiminished to all points in water. It is also acts on P2 P1 =, P2= P1 = P2 = F2 = If A1 > A2 Then > Hence large force acts on the cotton kept between M and N. In hydraulic press. A smaller force is applied on column of liquid and it is converted to the very large force in upward direction on the cotton bale. It is useful for. a) Compressing cotton bales or bundles of waste paper. b) Extraction of oil by crushing the oil seeds. 10

11 c) Pressing metal sheets to a desire shape etc. ii) Hydraulic lift: PHYSICS-MANIA Hydraulic lift is used to lift or support heavy objects such as cars. Trucks etc. It is an important application of Pascal s law. In hydraulic lift a small force pushes down a smaller piston, this small force produce greater thrust on larger piston and raises the load placed on it. iii) Hydraulic brakes: It is a system based on Pascal s law, thus small force applied to brakes pedal is immediately transmitted equally by the brake. Fluid in the cylinders produces a large thrust on the wheels and vehicle stops. Que 17: Explain the effect of gravity on fluid pressure. Ans: Effect of gravity on fluid pressure Consider liquid of density in vessel. Let us find pressure between two point P and Q as shown in fig. h be distance between point P and Q, consider imaginary cylinder of liquid of cross sectional area A, Point P and Q lie on upper and lower surface respectively. Weight of imaginary cylinder = mg = vg= Ah g It acts vertically downwards. Let P1, P2 are pressure at point P and Q respectively. Force on face = F1 = P1 A Total downward force Q = F1 = P1 A + mg Vertically upward force on face Q = F2 = P2 A Since imaginary cylinder is in equilibrium the total downward force = Total upward force F1=F2 P1A + mg = P2A (P2 P1) A = mg (P2-P1) = = 11

12 Pressure difference = h g This equation gives value of pressure difference In absence of gravity, P1 = P2 and the pressure at every point inside the liquid in equilibrium would be the same. In presence of gravity, P2 is not equal to P1 and the pressure cannot remains same at all points in the liquid. The pressure at all points at the same horizontal level is the same but increases with depth. Que 18: Define Viscosity Ans : The property of fluid by virtue of which it opposes the relative motion between its layers is called viscosity. Que 19 : Explain the term (a)stream line flow.(b) turbulent flow. Ans : (a) stream line flow :- Flow of liquid over plane surface or through a tube, so long as the velocity of fluid is less than a certain limiting value (called as critical velocity) is called the streamline flow or laminar flow. (b) Turbulent flow: - If the velocity of fluid in streamlines flow is gradually increased so that it is greater than critical velocity of the fluid then the flow becomes suddenly irregular and unsteady. This disorderly motion of fluid is called turbulent flow. Que 20: What do you mean by viscous drag? Ans : The tangential force acts on adjacent layer of fluid, which opposes the relative motion between the layers of the liquid or gases, is called viscous force or viscous drag. Que 21 : State and Explain Newton s law of viscosity. Ans : when a liquid flow is streamline,the viscous force acting on any layer is directly proportional to (i) area of layer and (ii) velocity gradient. Let A be the area of layer and be the velocity gradient, the viscous force f is given by F (A) ( ) 12

13 F = (A) = Where is a constant, is called coefficient of viscosity of the liquid, which is depends upon nature of the liquid. If A = 1 and = 1 then then = F (numerically) Thus coefficient of viscosity is defined as the viscous force per unit area per unit velocity gradient SI unit of is N. s/m2 CGS unit of is dyne s/cm2 or poise Relation between SI and CGS unit of is 1 Ns/M 2 = 10 poise Dimension of are = [L -1 M 1 T -1 ] Poise: if a force of one dyne per square centimeter exists between two layers of fluid with unit velocity gradient, then its coefficient of viscosity is one poise. The viscosity of liquid decreases with temperature, while the viscosity of gases increases with temperature. Que 22 :Define (a) velocity gradient Ans : (b) Coefficient of viscosity i. Consider streamline flow of the liquid on horizontal surface. Let us imagine that the liquid is divided into thin layers which are parallel to each other and to the horizontal surface as shown in fig. ii) Velocity of the liquid in given layer is assumed to be constant but differ from layer to layer. The velocity of bottom layer in contact with surface is zero and increases in the upward direction. The velocity of top surface is a maximum. 13

14 The rate of change of velocity of with distance measured form stationary layer is called velocity gradient. Let v and v+ dv be the velocities of layer of liquid which are at distance of x and x + dx respectively from the bottom. Velocity gradient = Unit of velocity gradient is per second and its dimensions are [L 0 M 0 T -1 ] (b) Coefficient of viscosity: - coefficient of viscosity is defined as the viscous force per unit area per unit velocity gradient. Que 23: state SI and CGS unit of coefficient of viscosity and find conversion factor between them. and also state their dimension Ans : SI unit of h is N. s/m2 CGS unit of h is dyne s/cm2 or poise Relation between SI and CGS unit of h is 1 Ns/M2 = 10 poise Dimension of h are = [L -1 M 1 T -1 ] Que 24: State stoke s law and derives it by dimensional analysis. Ans : Statement : - The viscous force acting on a small sphere falling through a medium is directly proportional to the radius r of the sphere, its velocity v through fluid and coefficient of viscosity viscosity of the fluid. F r v 14

15 F = 6 r v Stokes law can be partially proved by the method of dimensional analysis. We can write F = k 2 r b v c (1) Where k is numerical constant and a, b & c are the power to be determined [F] = k [ ] a [r] b [v] c [M 1 L 1 T -2 ] = k [M1L-1T-1]a [M 0 L 1 T 0 ] b [M 0 L 1 T -1 ] c [M 1 L 1 T -2 ] = k [M a L -a+b+c T -a-c ] Value of constant k = 6 I determined experimentally. It is dimensionless quantity. Equating indices of M,L and T on both sides, we get a = 1 - a + b + c = 1 - a c = -2 Solving this equation we get - a - c = -2 -c = -1 c = 1 1 = -a + b + c 1 = -1 + b + 1 b = 1 Substituting in equation (1) we get F = k 1 r 1 v 1 F = 6 r v This is Stokes law Que 25: obtain an expression for the terminal velocity of small spherical body falling under gravity through viscous fluid. Ans : Equation for Terminal Velocity : When a spherical body falls through a viscous fluid, it experiences a viscous force. The motion of body is initially accelerated and its velocity begins to increase. 15

16 The spherical body attains a constant velocity called as terminal velocity. coefficient of viscosity is ( ). i) Total downward force acting on the sphere = weight of the sphere = mass x gravitational acceleration = volume x density x g = ( ) r3 g ii) The total upward force acting on the sphere is given by The total upward force = viscous force + up thrust due to the medium =6 rv + ( ) r3 g iii) Since the sphere is moving with its terminal velocity, then Total downward force = total upward force ( ) r3 g = 6 rv + ( ) r3 g At terminal velocity of body, total downward force is balanced by resultant upward force due to up thrust and viscosity of the medium. Suppose that a sphere of radius r and density ( ), is falling 6 rv = r3 ( - )g r3 ( - )g v = r 2 r² ( - )g v=

17 9 This is the expression for terminal velocity of the sphere; coefficient of viscosity is given by 2 r² ( - )g = v Que 26: what is Reynold s number. Ans : Flow of liquid inside a tube remains streamline flow, if is velocity is less than a critical value. If velocity is greater than the critical value, laminar flow con vets into turbulent. According to Reynolds is critical velocity Vc = N / D Where D = Decimeter of tube = coefficient of viscosity = density of liquid N = Reynolds number Thus Reynolds number N = Vc D/ Reynolds number is pure number, so is it has no unit and dimensions. Note:- Conditions 1) If N lies between 0 to 2000 then flow of liquid is laminar. 2) If N lies between 2000 to 3000, the laminar flow may change to turbulent flow. If N is greater than 3000 then flow of liquid is turbulent Que 27: state Bernoulli s principle. Ans : Statement : In case of the streamline flow of no viscous incompressible fluid, the sum of (i) pressure energy, ii) K.E. per unit mass and ii) P.E. per unit mass always remains constant at every point. 17

18 Let. = density of liquid h = height of fluid v = velocity of the fluid K.E. = mv2 = ( V)v 2 KE/V = v 2 Potential Energy = P.E. = mgh = V gh PE/V = gh According to Bernoulli s principle P + ( v²)+g h= constant P/ + v2 + gh OR =Pressure energy +KE/mass + PE/mass = constant Bernoulli s principle stated above is true under the following assumptions i) The fluid is non viscous. ii) The flow is stream line. Alternate Statement of Bernoulli s principle In the case of stream line flow, the total energy per unit mass of a fluid always remains constant. Que 28 : state limitation and application of Bernoulli s principle. Ans :Limitation of Bernoulli s Principle We have considered that there is no loss of energy. But in reality there is loss of energy in the experiment. Applications of Bernoulli s Principle 1) Vacuum Brakes it works on the principle of Bernoulli to stop railway train. 2) Bunsen s Burner the gas escape through the fine nozzle with high velocity. Due to which pressure decrease in near nozzle, sine pressure inside burner is lowered, air enters through holes. The air and gas mixture rise up and produce a flame, when burned. 3) Venturimeter- This is device used to determine the velocity of liquid flowing through a pipe and its rate of 18

19 flow across any part of the pipe. 4) Air Purifier This is a device used for spraying fragrance in air, it is based on Bernoulli s principle.important FORMULAE 1. Coefficient of static friction ( s): 2. Coefficient of kinetic friction ( k): 3. Pressure = h g 4. Newton s formula for the force due to viscosity of liquid is given by F = (A) Velocity gradient = 5. Stokes law F = 6 r v 6. Terminal Velocity 2 r² ( - )g = v 2 r² ( - )g v= Reynolds number N = Vc D/ 8. Bernoulli s principle=p + ( v²)+g h= constant 19

MECHANICAL PROPERTIES OF FLUIDS

CHAPTER-10 MECHANICAL PROPERTIES OF FLUIDS QUESTIONS 1 marks questions 1. What are fluids? 2. How are fluids different from solids? 3. Define thrust of a liquid. 4. Define liquid pressure. 5. Is pressure