Friction Notes and Worksheet PDF
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Uploaded by EfficientRose
Lilavatibai Podar High School
2024
ISC
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These notes provide derivations and explanations of concepts related to friction, including angle of friction, angle of repose, and work done against friction. The document is intended for high school physics students.
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LILAVATIBAI PODAR HIGH SCHOOL, ISC Sub: PHYSICS Academic Year 202425 Grade: XI FRICTION DERIVATIONS: Q.1. Define angle of friction. Deduce its relation with coefficient of friction. Ans....
LILAVATIBAI PODAR HIGH SCHOOL, ISC Sub: PHYSICS Academic Year 202425 Grade: XI FRICTION DERIVATIONS: Q.1. Define angle of friction. Deduce its relation with coefficient of friction. Ans. Angle of friction: The angle of friction may be defined as the angle which the resultant of the limiting friction and the normal reaction makes with the normal reaction. W is the weight of the body, R is the C B max normal reaction, fs is the limiting Re m su ax an lta friction, P is the applied force and OC is R fs nt max of the resultant of fs and R. d Applied force R max BC OA fs A P tan = = = max fs O OB OB R max f But s = s = coefficient of static friction R tan = s W Thus the coefficient of static friction is equal to the tangent of the angle of friction. Q.2. Define angle of repose. Deduce its relation with coefficient of static friction. Ans. It is the minimum angle that an inclined plane makes with the horizontal when a body placed on it just begins to slide down. m : mass of the block : angle of repose max fs : limiting force of static friction R : normal reaction R balances mg cos . R = mg cos … (1) max In equilibrium, mg sin balances fs max fs = mg sin … (2) 1 LILAVATIBAI PODAR HIGH SCHOOL, ISC Sub: PHYSICS Academic Year 202425 Grade: XI FRICTION (2) (1), max fs mg sin = R mg cos max fs = tan R max fs But s = = coefficient of static friction R s = tan But s = tan tan = tan = Angle of repose is equal to angle of friction. Q.3. Find an expression for work done in sliding a body over a rough horizontal surface. Ans. m : mass of the block fk : force of kinetic friction S : displacement of the block R : normal reaction Work done = Force Displacement W = fk S … (1) But fk = kR fk = kmg … (2) From (1) and (2), W = kmgS Q.4. Find an expression for the work done against friction when a body is made to slide up an inclined plane. Ans. m : mass of block fk : force of kinetic friction : angle of friction R : normal reaction 2 LILAVATIBAI PODAR HIGH SCHOOL, ISC Sub: PHYSICS Academic Year 202425 Grade: XI FRICTION mg cos balances normal reaction R = mg cos … (1) In equilibrium of the block, F = mg sin + fk But fk = kR = k(mg cos ) … from (1) F = mg sin + k(mg cos ) F = mg(sin + k cos ) Work done in pulling the body through distance (S) W=FS W = mg(sin + k cos )S Q.5. Derive an expression for the work done against friction when a body is made to slide down on an inclined plane. Ans. m : mass of block fk : force of kinetic friction R : normal reaction mg cos balances normal reaction R = mg cos … (1) In equilibrium, F + mg sin = fk But fk = kR = k(mg cos ) … from (1) F + mg sin = k(mg cos ) F = mg[k cos sin ] Work done in sliding the body through distance (S) is given by W = FS = mg(k cos sin )S 3 LILAVATIBAI PODAR HIGH SCHOOL, ISC Sub: PHYSICS Academic Year 202425 Grade: XI FRICTION Q.6. Derive an expression for the acceleration of a body sliding down a rough inclined plane. Ans. m : mass of block fk : force of kinetic friction : angle of friction a : acceleration of the block R : normal reaction mg cos balances the normal reaction. R = mg cos … (1) Net force acting down the plane, F = mg sin = fk … (1) But fk = kR = k(mg cos ) … from (1) From (2), Net force = F = ma = mg sin = k mg cos a = g(sin k cos ) Q.7. Why is it easier to pull a lawn roller than to push it? Explain. Ans. (i) Pulling a roller: R : normal reaction F : force applied to pull a lawn 4 LILAVATIBAI PODAR HIGH SCHOOL, ISC Sub: PHYSICS Academic Year 202425 Grade: XI FRICTION R + F sin = W R = W F sin fk : force of kinetic friction fk = kR = k(W F sin ) … (1) (ii) Pushing a roller: Here, F sin acts in the downward direction. R' = W + F sin fk' = kR' fk' = k(W + F sin ) … (2) From (1) and (2), Force of kinetic friction is more in case of pushing rather than pulling so it is easier to pull a body than to push it. FRICTION: Friction: Whenever a body moves or tends to move over the surface of another body, a force comes into play which acts parallel to the surface of contact and opposes the relative motion. This opposing force is called friction. 5 LILAVATIBAI PODAR HIGH SCHOOL, ISC Sub: PHYSICS Academic Year 202425 Grade: XI FRICTION Direction of motion Force of friction Direction of motion Force of friction Friction as the component of contact force: The component of the contact force F normal to the contact surface is called normal force or normal reaction N. The component parallel to the contact surface is called friction. F = contact force N = normal force f = friction Origin of friction: The force of friction is due to the atomic or molecular forces of attraction between the two surfaces at the points of actual contacts. Apparent area of contact Actual area of contact 6 LILAVATIBAI PODAR HIGH SCHOOL, ISC Sub: PHYSICS Academic Year 202425 Grade: XI FRICTION STATIC, LIMITING AND KINETIC FRICTIONS: F fs F fs (a) (b) max F fs F fs (c) (d) v a F fk F fk (e) (f) Types of friction The force of friction which comes into play between two bodies before one body actually starts moving over the other is called static friction (fs). max The maximum force of static friction (fs ) which comes into play when a body just starts moving over the surface of another body is called limiting friction. Clearly, max fs fs. The force of friction which comes into play when a body is in a state of steady motion over the surface of another body is called kinetic or dynamic friction (fk). Variation of the force of friction 'f' with the applied force F: Obviously, as the applied force F increases, the static friction fs increases accordingly to balance it. This shows that static friction is a self adjusting force. The kinetic friction is max max always less than the limiting friction fs , i.e., fk < fs. s) (f on Friction (f) cti fri Limiting Kinetic ic friction (fsmax) at friction (fk) St Applied force (F) LAWS OF FRICTION AND COEFFICIENTS OF FRICTION: Laws of limiting fiction: (i) The limiting friction depends on the nature of the surfaces in contact and their state of polish. (ii) The limiting friction acts tangential to the two surfaces in contact and in a direction opposite to the direction of motion of the body. (iii) The value of limiting friction is independent of the area of the surface in contact so long as the normal reaction remains the same. 7 LILAVATIBAI PODAR HIGH SCHOOL, ISC Sub: PHYSICS Academic Year 202425 Grade: XI FRICTION max (iv) The limiting fraction (fs ) is directly proportional to the normal reaction R between the two surfaces. max i.e., fs R max or fs = sR max fs Limiting friction or s = = R Normal reaction The proportionality constant s is called coefficient of static friction. It is defined as the ratio of limiting friction to the normal reaction. R F f W Laws of kinetic friction: (i) The kinetic friction opposes the relative motion and has a constant value, depending on the nature of the two surfaces in contact. (ii) The value of kinetic friction fk is independent of the area of contact so long as the normal reaction remains the same. (iii) The kinetic friction does not depend on velocity, provided the velocity is neither too large nor too small. (iv) The value of kinetic friction fk is directly proportional to the normal reaction R between the two surfaces. i.e., fk R or fk = kR fk Kinetic friction or k = = R Normal reaction The proportionality constant k is called coefficient of kinetic friction. It is defined as the ratio of kinetic friction to the normal reaction. max As fk < fs or kR < 2R k < s 8 LILAVATIBAI PODAR HIGH SCHOOL, ISC Sub: PHYSICS Academic Year 202425 Grade: XI FRICTION FORMULAE USED Limiting friction 1. Coefficient of limiting friction = Normal reaction max fs or s = R max or fs = sR Kinetic friction 2. Coefficient of kinetic friction = Normal reaction fk or k = R or fk = kR 3. For a body placed on horizontal surface, R = mg max fs = smg and fk = kmg max 4. Static friction, fs fs or fs sR max 5. Kinetic friction, fk < fs 6. If is the angle of friction, then s = tan 7. If is the angle of repose, then s = tan 8. Angle of repose = Angle of friction, i.e., = 9. For a body moving on a rough horizontal surface with retardation ‘a’, f ma a = = = R mg g R 10. fr = r and r < k < s r where r is the coefficient of rolling friction, fr is the rolling friction and r is the radius of the rolling body. Units used: Force of friction f and normal reaction R are in newton (N), coefficient of friction has no units. Solved Examples R 1. A block of weight 20 N is placed on a horizontal table and a tension T, which can be increased to 8 N before the begins to slide, is applied at the block as shown T in the figure. A force of 4 N keeps the block moving at constant speed once it has been set in motion. Find the coefficient of static and kinetic friction. mg = 20 N 9 LILAVATIBAI PODAR HIGH SCHOOL, ISC Sub: PHYSICS Academic Year 202425 Grade: XI FRICTION Sol. For static friction: We have the following relations for vertical and horizontal component of forces: R mg = 0 or R = mg = 20 N max max and T fs = 0 or fs =T=8N max fs 8 Coefficient of friction, s = = = 0.40 R 20 For kinetic friction: We can write the following relations: R – mg = 0 or R = mg = 20 N and T fk = 0 or fk = T = 4 N fk 4 Coefficient of kinetic friction, k == = 0.20 R 20 2. A force of 49 N is just sufficient to pull a block of wood weighing 10 kg on a rough horizontal surface. Calculate the coefficient of friction and angle of friction. Sol. Here, P = applied force = 49 N m = 10 kg, g = 9.8 ms2 f P 49 Coefficient of friction, = = = = 0.5 R mg 10 9.8 As tan = = 0.5 = tan1(0.5) = 2634 3. A mass of 4 kg rests on a horizontal plane. The plane is gradually inclined until at an angle = 15 with the horizontal, the mass just begins to slide. What is the coefficient of static friction between the block and the surface? Sol. Here = 15 is the angle of repose. Coefficient of friction, = tan = tan 15 = 0.27 4. A body rolled on ice with a velocity of 8 ms1 comes to rest after travelling 4 m. Compute the coefficient of friction. Given g = 9.8 ms 2. Sol. Here, u = 8 ms1, v = 0, s = 4 m, a = ? 2 2 As v u = 2as 2 2 64 2 0 8 = 2a 4 or a = = 8 ms 8 Negative sign indicates retardation. f ma a 8 Now, = = = = = 0.8164 R mg g 9.8 5. The coefficient of friction between the ground and the wheels of a car moving on a horizontal road is 0.5. If the car starts from rest, what is the minimum distance in which it can acquire a speed of 72 kmh1? Take g = 10 ms2. 10 LILAVATIBAI PODAR HIGH SCHOOL, ISC Sub: PHYSICS Academic Year 202425 Grade: XI FRICTION Sol. Here, = 0.5, u = 0, 1 5 1 v = 72 kmh = 72 = 20 ms 18 f ma a As = = = R mg g 2 2 400 (20) 0 = 2 5 s or s = = 40 m 10 6. A bullet of mass 0.01 kg is fired horizontally into a 4 kg wooden block at rest on a horizontal surface. The coefficient of the kinetic friction between the block and the surface is 0.25. The bullet gets embedded in the block and the combination moves 20 m before coming to rest. With what speed did the bullet strike the block? Sol. Mass of block, M = 4 kg Mass of bullet, m = 0.01 kg After the bullet gets embedded in the block, the force of kinetic friction is, fk = kR = k(M + m)g If the kinetic friction produces retardation ‘a’ in the system, then fk = (M + m)a (M + m)a = k(M + m)a or a = k g = 0.25 9.8 = 2.45 ms2 After the bullet enters the block, suppose the system attains velocity V. Now the system comes to rest after covering a distance, s = 20 m, as shown in the figure below. R (M + m) Bullet v V vel = 0 m s (M + m)g 2 2 As v u = 2as 2 1 0 V = 2 (2.45) 20 or V = 98 = 9.8995 ms If ‘v’ is the velocity with which the bullet struck the block, then applying the law of conservation of momentum, we get mv = (M + m)V M+m 4 + 0.01 or v= V= 9.8995 = 401 9.8995 m 0.01 1 = 3969.7 ms 11 LILAVATIBAI PODAR HIGH SCHOOL, ISC Sub: PHYSICS Academic Year 202425 Grade: XI FRICTION 7. What is the acceleration of the block and the trolley system shown in the following figure, if the coefficient of kinetic friction between the trolley and the surface is 0.04? What is the tension in the string? Take g = 10 ms2. Neglect the mass of the string. Trolley a T 20 kg fk T a 3 kg 30 N Sol. As the block and the trolley are connected together by a string of fixed length, both will have same acceleration ‘a’. Applying Newton’s second law to the motion of the block, 30 – T = 3a … (i) Applying second law to the motion of the trolley, T – fk = 20a But fk = kR = kmg = 0.04 20 10 = 8 N T – 8 = 20a … (ii) Adding (i) and (ii), we get 22 2 22 = 23a or a = = 0.96 ms 23 From (i), 30 – T = 3 0.96 T = 30 – 2.88 = 27.12 N 8. A block A of mass 4 kg is placed on another block B of mass 5 kg, and the block B rests on a smooth horizontal table. For sliding the block A on B, a horizontal force of 12 N is required to be applied on it. How much maximum horizontal force can be applied on B so that both A and B move together? Also find out the acceleration produced by this force. Sol. Here, m1 = 4 kg, m2 = 5 kg Force applied on block A = 12 N This force must at least be equal to the kinetic friction applied on A by B. 12 = fk = kR = km1g or 12 = k g 12 3 or k = = 4g g 12 LILAVATIBAI PODAR HIGH SCHOOL, ISC Sub: PHYSICS Academic Year 202425 Grade: XI FRICTION The block B is on a smooth surface. Hence to move A and B together, the (maximum) force F that can be applied on B is equal to the frictional forces applied on A by B and applied on B by A. 3 F = km1g + km2g = k(m1 + m2)g = (4 + 5)g = 27 N g As this force moves both the blocks together on a smooth table, so the acceleration produced is, F 27 a= = = 3 ms2 m1 + m2 4 + 5 9. An engine of 100 H.P. draws a train of mass 200 metric ton with a velocity of 36 kmh1. Find the coefficient of friction. Sol. Power of engine, P = 100 H.P. = 100 746 = 74600 W Velocity, v = 36 kmh1 = 10 ms1 If the frictional force overcome by the engine is F, then P 74600 P = F v or F = = = 7460 N v 10 Normal reaction, R = mg = 200 1000 9.8 N F 7460 Coefficient of friction, = = = 0.0038 R 200 1000 9.8 10. A block of metal of mass 50 g when placed on an inclined plane at an angle of 15 slides down without acceleration. If the inclination is increased by 15, what would be the acceleration of the block? Sol. Here, m = 50 g = 0.05 kg Angle of repose, = 15 = tan = tan 15 = 0.2679 New angle of inclination = 15 + 15 = 30 Let ‘a’ be the downward acceleration produced in the block. Net downward force on the block is, F = mg sin f ma = mg sin mg cos … [ f = R = mg cos ] a = g(sin cos ) = 9.8(sin 30 0.2679 cos 30) = 9.8(0.5 0.2679 0.866) = 9.8 0.2680 = 2.6 ms2 13 LILAVATIBAI PODAR HIGH SCHOOL, ISC Sub: PHYSICS Academic Year 202425 Grade: XI FRICTION 11. When an automobile moving with a speed of 36 kmh 1 reaches an upward inclined road of angle 30, its engine is switched off. If the coefficient of friction involved is 0.1, how much distance will the automobile move before coming to rest? Take g = 10 ms2. Sol. As shown in the figure below, when a body moves up an inclined plane, force of friction ‘f’ acts down the plane. So the force against which work is needed to be done is, F = mg sin + f = mg sin + mg cos = mg(sin + cos ) If ‘m’ is the mass of the automobile, retardation produced in it will be, F a = = g(sin + cos ) = 10(sin 30 + 0.1 cos 30) m 2 = 10(0.5 + 0.1 0.866) = 5.866 ms 1 1 2 Now, u = 36 kmh = 10 ms , v = 0, a = 5.866 ms , s = ? 2 2 As v u = 2as 2 2 0 10 = 2 (5.866)s or s = 8.52 m 12. An engine of mass 6.5 metric ton is going upon incline of 5 in 13 at the 1 rate of 9 kmh1. Calculate the power of the engine if = 12 and 2 g = 9.8 ms. 2 Sol. Here, m = 6.5 metric ton = 6500 kg, g = 9.8 ms , 1 5 1 5 v = 9 kmh =9 = 2.5 ms , sin = , 18 13 2 25 12 cos = 1 sin = 1 = 169 13 Total force required against which the engine needs to work, F = mg sin + f = mg sin + mg cos = mg(sin + cos ) 5 1 12 = 6500 9.8 + = 29400 N 13 12 13 Power of the engine = Fv = 29400 2.5 = 73500 W = 73.5 kW 14 LILAVATIBAI PODAR HIGH SCHOOL, ISC Sub: PHYSICS Academic Year 202425 Grade: XI FRICTION 13. Two blocks of mass 2 kg and 5 kg are connected by an ideal string passing over a pulley. The block of mass 2 kg is free to slide on a surface inclined at an angle of 30 with the horizontal whereas 5 kg block hangs freely. Find the acceleration of the system and the tension in the string. Given. = 0.30. Sol. Let ‘a’ be the acceleration of the system in the direction as shown and T the tension in the string. Equations of motion for 5 kg and 2 kg blocks can respectively be written as, 5g – T = 5a … (1) and T – 2g sin f = 2a … (2) where ‘f’ is the limiting friction and is given by, f = R = mg cos = 0.3 2g cos 30 Adding (1) and (2), we get 5g – 2g sin f = 7a or 5g – 2g sin 30 0.6g cos 30 = 7a or g(5 2 0.5 0.6 0.866) = 7a 9.8 3.4804 2 or a= = 4.87 ms 7 From (1), T = 5g 5a = 5(9.8 4.87) = 5 4.93 = 24.65 N 14. A truck tows a trailer of mass 1200 kg at a speed of 10 ms1 on a level road. The tension in the coupling is 1000 N. (i) What is the power expended on the trailer? (ii) Find the tension in the coupling when the truck ascends a road having an inclination of 1 in 6. Assume that the frictional resistance of the incline is the same as that on the level road. Sol. On the level road, force applied by the truck is equal to the friction overcome. f = 1000 N Speed of truck, v = 10 ms1 Power expended on the trailor, P = fv = 1000 10 = 104 W When the truck ascends a road of 1 inclination 1 in 6 i.e. sin = , 6 15 LILAVATIBAI PODAR HIGH SCHOOL, ISC Sub: PHYSICS Academic Year 202425 Grade: XI FRICTION 1 F = f + mg sin = 1000 + 1200 9.8 6 = 1000 + 1960 = 2960 N METHODS OF CHANGING FRICTION: (i) By polishing (ii) Lubrication (iii) Streamlining (iv) By using ball-bearings WORK SHEET 1. A block of mass 1 kg lies on a horizontal surface in a truck. The coefficient of static friction between the block and the surface is 0.6. If the acceleration of the truck is 5 ms2, calculate the frictional force acting on the block. Ans. 5 N 2. A body weighing 20 kg just slides down a rough inclined plane that rises 5 m in every 13 m. What is the coefficient of friction? Ans. 0.4167 3. A scooter weighs 120 kg f. Brakes are applied so that wheels stop rolling and start skidding. Find the force of friction if the coefficient of friction is 0.4. Ans. 48 kg f 4. An automobile is moving on a horizontal road with a speed 'v'. If the coefficient of friction between the tyres and road is , show that the shortest distance in v2 which the automobile can be stopped is g. 2 5. Find the distance travelled by a body before coming to rest, if it is moving with a speed of 10 ms1 and the coefficient of friction between the ground and the body is 0.4. Ans. 12.75 m 6. A motor car running at the rate of 7 ms1 can be stopped by applying brakes in 10 m. Show that total resistance to the motion, when brakes are on, is one- fourth of the weight of the car. 7. Find the power of an engine which can maintain a speed of 50 ms1 for a train of mass 3 105 kg on a rough line. The coefficient of friction is 0.05. Take g = 10 ms2. Ans. 7500 kW 8. A train weighing 1000 quintals is running on a level road with a uniform speed of 72 km h1. If the frictional resistance amounts to 50 g wt per quintal, find power in watt. Take g = 9.8 ms2. Ans. 9800 W 9. An automobile of mass 'm' starts from rest and accelerates at a maximum rate possible without slipping on a road with s = 0.5. If only the rear wheels are driven and half the weight of the automobile is supported on these wheels, how much time is required to reach a speed of 100 kmh1? Ans. 11.3 s 16 LILAVATIBAI PODAR HIGH SCHOOL, ISC Sub: PHYSICS Academic Year 202425 Grade: XI FRICTION 10. A suitcase is gently dropped on a conveyor belt moving at 3 ms1. If the coefficient of friction between the belt and the suitcase is 0.5, how far will the suitcase move on the belt before coming to rest? Ans. 0.92 m 11. A truck moving at 72 kmh1 carries a steel girder which rests on its wooden floor. What is the minimum time in which the truck can come to stop without the girder moving forward? Coefficient of static friction between steel and wood is 0.5. Ans. 4.08 s 12. A bullet of mass 10 g is fired horizontally into a 5 kg wooden block, at rest on a horizontal surface. The coefficient of kinetic friction between the block and the surface is 0.1. Calculate the speed of the bullet striking the block, if the combination moves 20 m before coming to rest. Ans. 3136.26 ms1 13. In the figure shown below, the masses of A and B are 10 kg and 5 kg respectively. Calculate the minimum mass of C which may stop A from slipping. Coefficient of static friction between A and the table is 0.2. Ans. 15 kg 14. A block of mass 2 kg rests on a plane inclined at an angle of 30 with the horizontal. The coefficient of friction between the block and the surface is 0.7. What will be the frictional force acting on the block? Ans. 11.9 N 15. A block of mass 10 kg is sliding on a surface inclined at an angle of 30 with the horizontal. If the coefficient of friction between the block and the surface is 0.5, find the acceleration produced in the block. Ans. 0.657 ms2 16. Find the force required to move a train of mass 105 kg up an incline 1 in 50 with an acceleration of 2 ms2. Coefficient of friction between the train and the rails is 0.005. Take g = 10 ms2. Ans. 2.25 105 N 17. A block slides down an incline of 30 with an acceleration equal to g/4. Find 1 the coefficient of kinetic friction. Ans. 2 3 17 LILAVATIBAI PODAR HIGH SCHOOL, ISC Sub: PHYSICS Academic Year 202425 Grade: XI FRICTION 18. A 10 kg block slides without acceleration down a rough inclined plane making an angle of 20 with the horizontal. Calculate the acceleration when the inclination of the plane is increased to 30 and the work done over a distance of 1.2 m. Take g = 9.8 ms2. Ans. 1.8 ms2, 21.6 J 19. A railway engine weighing 40 metric tons is travelling along a level track at a speed of 54 kmh1. What additional power is required to maintain the same speed up an incline rising 1 in 49. Given = 0.1, g = 9.8 ms2. Ans. 120 kW 20. A metal block of mass 0.5 kg is placed on a plane inclined to the horizontal at an angle of 30. If the coefficient of friction is 0.2, what force must be applied (i) to just prevent the block from sliding down the inclined plane, (ii) to just move the block up the inclined plane, and (iii) to move it up the inclined plane with an acceleration of 20 cms2. Ans. (i) 1.6 N, (ii) 3.299 N, (iii) 3.399 N 21. A block A of mass 14 kg moves along an inclined plane that makes an angle of 30 with the horizontal, as shown in figure. Block A is connected to another block B of mass 14 kg by a taut massless string that runs around a frictionless, massless pulley. The block B moves downward with constant velocity. What is (i) the magnitude of the frictional force, and (ii) the coefficient of kinetic friction. Ans. (i) 68.6 N, (ii) 0.58 22. A wooden block of mass 100 kg rests on a flat wooden floor, the coefficient of friction between the two being 0.4. The block is pulled by a rope making an angle of 30 with the horizontal. What is the minimum tension along the rope that just makes the block sliding? Ans. 367.7 N 18
