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Friction: Important Notes

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Published in: IIT JEE Mains | NEET | Physics
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Friction: Important Notes Useful for the students preparing for IIT JEE and NEET exam.

H R / Chennai

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  1. Friction: Important Notes Introduction If we slide or try to slide a body over a surface, the motion is resisted by a bonding between the body and the surface. This resistance is represented by a single force and is called friction force. The force of friction is parallel to the surface and opposite to the direction of intended motion. Types of Friction (1) Static friction : The opposing force that comes into play when one body tends to move over the surface of another, but the actual motion has yet not started is called static friction. (i) If applied force is P and the body remains at rest then static friction F = P. (ii) If a body is at rest and no pulling force is acting on it, force of friction on it is zero. mg Fig. 5.1 because it changes itself in accordance with the applied force and is always equal to net external force. (2) Limiting friction : If the applied force is increased, the force of static friction also increases. If the applied force exceeds a certain (maximum) value, the body starts moving. This maximum value of static friction upto which body does not move is called limiting friction. (i) The magnitude of limiting friction between any two bodies in contact is directly proportional to the normal reaction between them. FIT R or (ii) Direction of the force of limiting friction is always opposite to the direction in which one body is at the verge of moving over the other (iii) Coefficient of static friction : (a) is called coefficient of static friction and is defined as the ratio of force of limiting friction and normal reaction (b) Dimension : [MO LOT 0] (c) Unit : It has no unit. (d) Value of g depends on material and nature of surfaces in contact that means whether dry or wet rough or smooth polished or non-polished. (e) Value of g does not depend upon apparent area of contact. (3) Kinetic or dynamic friction : If the applied force is increased further and sets the body in motion, the friction opposing the motion is called kinetic friction. (i) Kinetic friction depends upon the normal reaction. Fk r R or Fk = '1k R where Bk is called the coefficient of kinetic friction (ii) Value of depends upon the nature of surface in contact. (iii) Kinetic friction is always lesser than limiting friction Fk < Fl :. < Vs i.e. coefficient of kinetic friction is always less than coefficient of static friction. Thus we require more force to start a motion than to maintain it against friction. This is because once the motion starts actually inertia of rest has been overcome. Also when motion has actually started, irregularities of one surface have little time to get locked again into the irregularities of the other surface. (iv) Kinetic friction does not depend upon the velocity of the body. (v) Types of kinetic friction (a) Sliding friction : The opposing force that comes into play when one body is actually sliding over the surface of the other body is called sliding friction. e.g. A flat block is moving over a horizontal table. (b) Rolling friction : When objects such as a wheel (disc or ring), sphere or a cylinder rolls over a surface, the force of friction that comes into play is called rolling friction. Rolling friction is directly proportional to the normal reaction (R) and inversely proportional to the radius (r) of the rolling cylinder or wheel. rolling gr is called coefficient of rolling friction. It would have the dimensions of length and would be measured in metre. Rolling friction is often quite small as compared to the sliding friction. That is why heavy loads are transported by placing them on carts with wheels. In rolling the surfaces at contact do not rub each other. The velocity of point of contact with respect to the surface remains zero all the times although the centre of the wheel moves forward. Graph Between Applied Force and Force of Friction
  2. limiting represent friction (Fl) • (3) Beyond A, the force of friction is decrease to seen slightly. The portion BC of the curve represents the kinetic friction (Fk) . (1) Part OA of the curve represents static friction (F ) . Its value increases linearly with the applied force (2) At point A the static friction is maximum. This Fig. 5 4 c (3) If a body is placed in a vehicle which is accelerating, the force of friction is the cause of motion of the body along with the vehicle (i.e., the body will remain at rest in the accelerating vehicle until ma < gung). If there had been no friction between body and vehicle, the body will not move along with the vehicle. a gang Fig. 5-5 ma O Applied force Fig. 5.2 (4) As the portion BC of the curve is parallel to x- axis therefore kinetic friction does not change with the applied force, it remains constant, whatever be the applied force. Friction is a Cause of Motion It is a general misconception that friction always opposes the motion. No doubt friction opposes the motion of a moving body but in many cases it is also the cause of motion. For example . (1) While moving, a person or vehicle pushes the ground backwards (action) and the rough surface of ground reacts and exerts a forward force due to friction which causes the motion. If there had been no friction there will be slipping an no motion. Friction 0 Action Fig. 5-3 (2) During cycling, the rear wheel moves by the force communicated to it by pedalling while front wheel moves by itself. So, when pedalling a bicycle, the force exerted by rear wheel on ground makes force of friction act on it in the forward direction (like walking). Front wheel moving by itself experience force of friction in backward direction (like rolling of a ball). [However, if pedalling is stopped both wheels move by themselves and so experience force of friction in backward direction]. From these examples it is clear that without friction motion cannot be started, stopped or transferred from one body to the other. Advantages and Disadvantages of Friction (1) Advantages of friction (i) Walking is possible due to friction. (ii) Two body sticks together due to friction. Fig. 5.6 Fig. 5-7 While Pedalling is (iii) Brake works on the basis of friction. (iv) Writing is not possible without friction. (v) The transfer of motion from one part of a machine to other part through belts is possible by friction. (2) Disadvantages of friction (i) Friction always opposes the relative motion between any two bodies in contact. Therefore extra energy has to be spent in over coming friction. This reduces the efficiency of machine. (ii) Friction causes wear and tear of the parts of machinery in contact. Thus their lifetime reduces. (iii) Frictional force result in the production of heat, which causes damage to the machinery. Methods of Changing Friction
  3. Friction: Important Notes We can reduce friction (1) By polishing. (2) By lubrication. (3) By proper selection of material. (4) By streamlining the shape of the body. (5) By using ball bearing. so = tana Also we can increase friction by throwing some sand on slippery ground. In the manufacturing of = tano = tanu tyres, synthetic rubber is preferred because its coefficient of friction with the road is larger. = As = tano ] Angle of Friction Angle of friction may be defined as the angle which the resultant of limiting friction and normal we know reaction makes with the normal reaction. By definition angle 0 is called the angle of friction tan 0 = — tan 0 = 0 mg Fig. 5.8 we know or 0 = tan I(ßL) Hence coefficient of static friction is equal tangent of the angle of friction. Resultant Force Exerted by Surface on Block In the above figure resultant force S = to be by, Thus the coefficient of limiting friction is equal to the tangent of angle of repose. As well as a = 0 i.e. angle of repose — angle of friction. Calculation of Required Force in Different Situation If W = weight of the body, 0 = angle of friction, g = tan 0 = coefficient of friction Then we can calculate required force for different situation in the following manner : (1) Minimum pulling force P at an angle from the horizontal Fig. 5.10 By resolving P in horizontal and vertical direction (as shown in figure) P sinu P cosa S = (gmg)2 + (mg)2 when there is no minimum i.e. S = mg Hence the range mg < Song g +1 Angle of Repose friction S 2 will Of S can be given Fig. 5.11 For the condition of equilibrium F=Pcosa and R=W—Psina By substituting these value in F = AR sino (W - P sinu) Pcosu = cos0 W sin0 cos(a — 0) = tan 0 ] Angle of repose is defined as the angle of the inclined plane with horizontal such that a body placed on it is just begins to slide. By definition, u is called the angle of repose. In limiting condition F = mg Sina and R cosa mg sin a mg cos a mg (2) Minimum pushing force P at an angle from the horizontal a Fig. 5.12
  4. By Resolving P in horizontal and vertical direction (as shown in the figure) R P cosa P sinu Fig. 5.13 For the condition of equilibrium F=Pcosa and R=W+Psina By substituting these value in F = AIR = sinu) [As g Pcosu cos0 W sin0 cos(a + 0) (3) Minimum pulling force up on an inclined plane p Fig. 5.14 P to = tan 0 ] move the body (4) Minimum force to move a body in downward direction along the surface of inclined plane a Fig. 5.16 By Resolving P in the direction of the plane and perpendicular to the plane (as shown in the figure) R + P sinu P cosa W sin W cosZ Fig. 5.17 For the condition of equilibrium R + P sinu = WcosZ . R=WcosZ—Psina and F=Pcosa+WsinZ By substituting these values in F = AR and solving we get W sine -Z) cos(a — 0) (5) Minimum force to avoid sliding of a body down on an inclined plane P Fig. 5.18 By Resolving P in the direction of the plane and perpendicular to the plane (as shown in the figure) By Resolving P in the direction of the plane and perpendicular to the plane (as shown in the figure) R + P sinu P cosa W cosZ Fig. 5.15 For the condition of equilibrium R + P sinu = WcosZ R=WcosZ-Psinu and F+WsinZ=Pcosa F = Pcosa — W sinZ By substituting these values in F = 'IR and solving we get Wsin(O +1) cos(a — 0) R + P sina W sinl F + P cosa W cosZ Fig. 5.19 For the condition of equilibrium R + P sinu = WcosZ R and Pcosu+F=WsinZ . F=WsinZ—Pcosa By substituting these values in F=gR and solving we get
  5. Friction: Important Notes sin(Z —0) cos(0 + u) (6) Minimum force for motion along horizontal surface and its direction Fig. 5.20 Let the force P be applied at an angle a with the horizontal. By resolving P in horizontal and vertical direction g mg 1 2 g mg 2 2 2 gmg min 2 Acceleration of a Block Against Friction (1) Acceleration of a block on horizontal surface When body is moving under application of force P, then kinetic friction opposes its motion. Let a is the net acceleration of the bod ma (as shown in figure) R + P sinu P cosa mg From the figure ma = P-Fk P-Fk m mg Fig. 5.23 Fig. 5.21 For vertical equilibrium R + P sinu = mg . R = mg —P sinu and for horizontal motion Pcosu > F i.e. Pcosu>gR Substituting value of R from (i) in (ii) (2) Acceleration of a block sliding down over a rough inclined plane When angle of inclined plane is more than angle of repose, the body placed on the inclined plane slides down with an acceleration a. ma ...(i) ...(ii) ... (iii) 0 ma = mg sinO—gmg cosO mg sino mg 0 mg a = g [sin0 — gcosU] Fig. 5.24 frictionless inclined plane g = 0 (3) Retardation of a block sliding up over a rough inclined plane When angle of inclined plane is less than angle of repose, then for the upward motion ma =mg sin0+F ma = mg sin04-gmg cosO 0 mg sino + mg 0 mg Retardation a = g[sinO + gcosO] Fig. 5.25 az.e : For frictionless inclined plane g = 0 .a=g sino Work done against friction (1) Work done over a rough inclined surface If a body of mass m is moved up slowly on a rough inclined plane through distance s, then Work done = force x distance From the figure ma sino —F ma = mg sin0—gR . Acceleration aze : a For . a = g sino . Pcosu > "(mg — P sinu) g mg cos u + g sin u For the force P to be minimum be maximum i.e. — [cosa + g sinu] = 0 du —sinu + g cos a = 0 tana = g or u = tan (g) = angle of friction (cos u +gsinu) must 2 1 Fig. 5.22 i.e. For minimum value of P its angle from horizontal should be equal to angle of friction the As tana = g so from the figure, sinu = 1 and cos u = 2 By substituting these value in equation (iii) 2
  6. ma = mg s [sin0 + gcos0] mg sino + mg [sin 0 s 0 mg 0 mg Fig. 5.26 g cos0 ]s (2) Work done over a horizontal surface In the above expression if we put 0 = o then Work done = force x distance = F x s = g mg s It is clear that work done depends upon mg Fig. 5.27 (i) Weight of the body. (ii) Material and nature of surface in contact. (iii) Distance moved. Motion of Two Bodies one Resting on the Other When a body A of mass m is resting on a body B of mass M then two conditions are possible (1) A force F is applied to the upper body, (2) A force F is applied to the lower body Fig. 5.28 We will discuss above two cases one by one in the following manner : (1) A force F is applied to the upper body, then following four situations are possible (i) When there is no friction (a) The body A will move on body B with acceleration (F/ m). 2mL a 1 Ass — —at 2 and a = F/m 2 (ii) If friction is present between A and B only and applied force is less than limiting friction (F < FD (F = Applied force on the upper body, Fl = limiting friction between A and B, FIK = Kinetic friction between A and B) (a) The body A will not slide on body B till F < Fl i.e. F < gsmg (b) Combined system (m + M) will move together with common acceleration a A = (iii) If friction is present between A and B only and applied force is greater than limiting friction (F > FD In this condition the two bodies will move in the same direction (i.e. of applied force) but with different acceleration. Here force of kinetic friction gung will oppose the motion of A while cause the motion of B. i.e. F-Fk F-Fk m (F — gung) m gvng Free body diagram of A maA Free body diagram of B MaB aZe : OAS both the bodies are moving in the same direction. Acceleration of body A relative to B will be 14 MF - gung (m + M) mM (b) The body B will remain at rest (c) If L is the length of B as shown will fall from B after time t in figure, So, A will fall from B after time 2m ML a (iv) If there is friction between B and floor (where F/' = g' (M +m)g = limiting friction between B and floor, Fic = kinetic friction between A and B) B will move only if Fk > F/ and then Fk — F/ =
  7. Friction: Important Notes Fig. 5.29 However if B does not move then static friction will work (not limiting friction) between body B and the floor i.e. friction force = applied force ( = Fk) not F/ . (2) A force F is applied to the lower body, then Both the body will move with different acceleration. Here force of kinetic friction Amg will oppose the motion of B while will cause the motion of ma A i.e. a A following four situations are possible (i) When there is no friction (a) B will move with acceleration will remain at rest (relative to ground) pulling force on A. = "k mg [F - gung] Free body diagram of A maA Free body diagram of B MaB (F/ M) while A as there is no a and a A = 0 (b) As relative to B, A will move backwards with acceleration (F/ M) and so will fall from it in time t. Fig. 5-30 2ML a (ii) If friction is present between A and B only and az.e : a As both the bodies are moving in the same direction Acceleration of body A relative to B will be Negative sign implies that relative to B, A will move backwards and will fall it after time 2ML a (iv) If there is friction between B and floor and F'
  8. Let m = mass of the insect, r = radius of the bowl, g = coefficient of friction for limiting condition at point A R = mg cos 0 (ii) Dividing (ii) by (i) For T = nt2g T = ml g sino + AR T = ml g sinO-epn1g cos0 ...(i) ...(ii) (i) [ASE/ tan 0 = — — 2 2 So I and 2 or Y — 1 2 Fl = mg sino 1 2 Minimum Mass Hung from the String to Just Start the Motion (1) When a mass ml placed on a rough horizontal plane Another mass hung from the string connected by frictionless pulley, the tension (T) produced in string wi ltry to start the motion of mass m From equation (i) and (ii) = ml [sin0 + g cosO] this is the minimum value of m 2 to start the motion Naze : In the above condition Coefficient of friction — tan 0 ml COS0 Maximum Length of Hung Chain A uniform chain of length I is placed on the table in such a manner that its l' part is hanging over the edge of table without sliding. Since the chain have uniform linear density therefore the ratio of mass and ratio of length for any part of the chain will be equal. mass hanging from the table We know g mass lying on the table . For this case we can rewrite above expression mig Fig. 5-32 At limiting condition T = Fl 1712g in the following manner length han$ng from the table length lying on the table uniform linear density] by solving l' — chain have Fig. 5-34 Coefficient of Friction Between a Body and Wedge A body slides on a smooth wedge of angle 0 and start the motion. 2 friction g (2) When inclined plane this is the minimum value of to In the above condition Coefficient of a mass ml placed on a rough Another mass hung from the its time of descent i t. s Smooth wedge 0 Fig. 5-35 s Rough wedge 0 Fig. 5-36 string connected by frictionless pulley, the tension (T) produced in string will try to st the motion of mass m If the same wedge made rough then time taken by it to come down becomes n times more (i.e. nt) The length of path in both the cases are same. 2 For smooth wedge, S = u t+—at nilg sino + At limiting condition mig mig Fig. 5 33 1 S = —(g sin 0) t2 2 [Asu = 0 and a = g sin 0] 2 For rough wedge, S = u t+—at ...(i)
  9. From equation (i) and (ii) 1 1 —(g sin0)t2 = —g (sine —g cos 2 2 2 sin 0 =(sin 0 —g cos O)n 1 g = tan 0 1— 2 n Stopping of Block Due to Friction (1) On horizontal road (i) Distance travelled before coming to rest : A block of mass m is moving initially with velocity u on a rough surface and due to friction, it comes to rest after covering a distance S. Fig. 5-37 Retarding force F = ma = PR ma — gmg 2 —2aS 0=u2 —2ggS From v [As v = 0, a = gg] Friction: Important Notes -2aS 1 S = — g (sine 2 ...(ii) [Asu = 0 and a = g (sino — g cos O)] 2 u 2g (sin 0 + g cos 0) Stopping of Two Blocks Due to Friction When two masses compressed towards each other and suddenly released then energy acquired by each block will be dissipated against friction and finally block comes to rest i.e., F x S = E [Where F Friction, S Distance covered by block, E = Initial kinetic energy of the block] momentum of block] 2m gmg x S = 2m 2gm2g Fig. 5-39 [Where P — g mg] 2 u or S = In the given condition P and g are same for both 2gm2g [As momentum P = mu] the blocks. 1 SO, S m 2 m 2 (ii) Time taken to come to rest From equation v = u —at [ASV = 0, a = u (2) On inclined road : When block starts with velocity u its kinetic energy will be converted into potential energy and some part of it goes against friction and after travelling distance S it comes to rest i.e. v = o. We know that retardation a = g[sin0 + gcosO] By substituting the value of v and a in the following equation Velocity at the Bottom of Rough Wedge A body of mass m which is placed at the top of the wedge (of height h) starts moving downward on a rough inclined plane. Loss of energy due to friction against friction) PE at point A = mgh 1 KE at point B = —mu2 2 FL (Work Fig. 5-40 By the law of conservation of energy 1 2 i.e. mv =mgh-FL 2 2 —(mgh FL) 2 s u 0 Fig. 5-38 0 = u —2g [sin0+gcos 0] S
  10. Sticking of a Block With Accelerated Cart When a cart moves with some acceleration toward right then a pseudo force (ma) acts on block toward left. This force (ma) is action force by a block on cart. Fig. 5 42 ma a CART mg Fig. 5-41 Now block will remain static w.r.t. cart. If friction force AR 2 mg gna mg a min [ASR = ma] Force of friction is non-conservative force. Force of friction always acts in a direction opposite to that of the relative motion between the surfaces. Rolling friction is much less than the sliding friction. This knowledge was used by man to invent the wheels. The friction between two surfaces increases (rather than to decrease), when the surfaces are made highly smooth. The atomic and molecular forces of attraction between the two surfaces at the point of contact give rise to friction between the surfaces. This is the minimum acceleration of the cart so that block does not fall. and the minimum force to hold the block together = (M +m)amn Sticking of a Person with the Wall of Rotor A person with a mass m stands in contact against the wall of a cylindrical drum (rotor). The coefficient of friction between the wall and the clothing is g. If Rotor starts rotating about its axis, then person thrown away from the centre due to centrifugal force at a particular speed CD , the person stuck to the wall even the floor is removed, because friction force balances its Ord inaMThinklng— bjectiveQ1estions Static and limiting friction The coefficient of friction g and the angle of friction Z are related as 1. 2. 3. 4. (a) sinZ=g (c) tan Z (b) cos = g (d) tan g A force of 98 N is required to just start moving a body of mass 100 kg over ice. The coefficient of weight in this condition. From the figure. Friction force (F) weight person (mg) > gR = mg g FC = mg static friction is (a) 0.6 (c) 0.2 (b) 0.4 (d) 0.1 of A block weighs W is held against a vertical wall by applying a horizontal force F. The minimum value of F needed to hold the block is mg (a) Less than W (c) Greater than W (b) Equal to W (d) Data is insufficient force] CD m min [Here, m r = mg centrifugal The maximum static frictional force is (a) Equal to twice the area of surface in contact (b) Independent of the area of surface in contact (c) Equal to the area of surface in contact (d) None of the above I psg&— rick
  11. Friction: Important Notes 5. Maximum value of static friction is called [BHU 1995; RPET 2000] (a) Limiting friction (c) Normal reaction friction (b) Rolling friction (d) Coefficient of 9. 10. The limiting friction is (a) Always greater than the dynamic friction (b) Always less than the dynamic friction (c) Equal to the dynamic friction (d) Sometimes greater and sometimes less than the dynamic friction Which is a suitable method to decrease friction 6. 7. Pulling force making an angle 0 to the horizontal is applied on a block of weight W placed on a horizontal table. If the angle of friction is u , then the magnitude of force required to move the body is equal to W sin u (a) g tan(0 — u) W sin u (c) cos(0 — u) [EAMCET 1987] W cos u (b) cos(0 — u) W tan u (d) sin(0 — u) (a) Ball and bearings (c) Polishing (b) Lubrication (d) All the above In the figure shown, a block of weight 10 N resting on a horizontal surface. The coefficient of static friction between the block and the surface =0.4 . A force of 3-5 N will keep the block in uniform motion, once it has been set in motion. A horizontal force of 3 N is applied to the block, then the block will [MP PET 1993] (a) Move over the surface with constant velocity (b) Move having accelerated motion over the surface 8. (c) Not move (d) First it will move with a constant velocity for some time and then will have accelerated motion Two masses A and B of 10 kg and 5 kg respectively are connected with a string passing over a frictionless pulley fixed at the corner of a table as shown. The coefficient of static friction of A with table is 0.2. The minimum mass of C that may be placed on A to prevent it from moving is [MP PET 1984] (a) 15 kg (b) 10 kg (c) 5 kg (d) 12 kg

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