Friction PDF

228 Friction Chapter 5...

228 Friction Chapter 5 Friction Introduction (iii) Coefficient of static friction : (a)  s is called coefficient If we slide or try to slide a body over a surface, the motion is of static friction and is defined as the ratio of force of limiting resisted by a bonding between the body and the surface. This F friction and normal reaction  s  resistance is represented by a single force and is called friction force. R The force of friction is parallel to the surface and opposite to (b) Dimension : [M 0 L0T 0 ] the direction of intended motion. (c) Unit : It has no unit. Types of Friction (d) Value of  depends on material and nature of surfaces (1) Static friction : The opposing force that comes into in contact that means whether dry or wet ; rough or smooth play when one body tends to move over the surface of another, polished or non-polished. but the actual motion has yet not started is called static friction. (e) Value of  does not depend upon apparent area of (i) If applied force is P and the contact. R body remains at rest then static (3) Kinetic or dynamic friction : If the applied force is P friction F = P. increased further and sets the body in motion, the friction opposing F (ii) If a body is at rest and no the motion is called kinetic friction. pulling force is acting on it, force of mg (i) Kinetic friction depends upon the normal reaction. friction on it is zero. Fig. 5.1 Fk  R or Fk   k R where  k is called the coefficient of (iii) Static friction is a self- kinetic friction adjusting force because it changes itself in accordance with the (ii) Value of k depends upon the nature of surface in applied force and is always equal to net external force. contact. (2) Limiting friction : If the applied force is increased, the force of static friction also increases. If the applied force exceeds a (iii) Kinetic friction is always lesser than limiting friction certain (maximum) value, the body starts moving. This maximum Fk  Fl   k   s value of static friction upto which body does not move is called i.e. coefficient of kinetic friction is always less than coefficient limiting friction. of static friction. Thus we require more force to start a motion than (i) The magnitude of limiting friction between any two to maintain it against friction. This is because once the motion bodies in contact is directly proportional to the normal reaction starts actually ; inertia of rest has been overcome. Also when between them. motion has actually started, irregularities of one surface have little Fl  R or Fl   s R time to get locked again into the irregularities of the other surface. (iv) Kinetic friction does not depend upon the velocity of the (ii) Direction of the force of limiting friction is always opposite to the direction in which one body is at the verge of body. moving over the other (v) Types of kinetic friction  Friction 229 (a) Sliding friction : The opposing force that comes into (2) During cycling, the rear wheel moves by the force play when one body is actually sliding over the surface of the other communicated to it by pedalling while front wheel moves by body is called sliding friction. e.g. A flat block is moving over a itself. So, when pedalling a bicycle, the force exerted by rear horizontal table. wheel on ground makes force of friction act on it in the (b) Rolling friction : When objects such as a wheel (disc forward direction (like walking). Front wheel moving by itself or ring), sphere or a cylinder rolls over a surface, the force of experience force of friction in backward direction (like rolling friction that comes into play is called rolling friction. of a ball). [However, if pedalling is stopped both wheels move  Rolling friction is directly proportional to the normal by themselves and so experience force of friction in backward reaction (R) and inversely proportional to the radius (r) of the direction]. rolling cylinder or wheel. R Frolling   r r  r 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 While pedalling Pedalling is stoped sliding friction. That is why heavy loads are transported by placing Fig. 5.4 them on carts with wheels.  In rolling the surfaces at contact do not rub each other. (3) If a body is placed in a vehicle which is accelerating,  The velocity of point of contact with respect to the surface the force of friction is the cause of motion of the body along remains zero all the times although the centre of the wheel moves with the vehicle (i.e., the body will remain at rest in the forward. accelerating vehicle until ma   s mg). If there had been no Graph Between Applied Force and Force of Friction friction between body and vehicle, the body will not move along with the vehicle. (1) Part OA of the curve represents static friction (Fs ). Its value increases linearly with the applied force a smg ma (2) At point A the static friction is maximum. This represent limiting friction ( Fl ). A Force of friction (3) Beyond A, the B C force of friction is seen to Fs Fig. 5.5 decrease slightly. The Fl Fk portion BC of the curve From these examples it is clear that without friction motion represents the kinetic cannot be started, stopped or transferred from one body to the friction (Fk ). O Applied force other. (4) As the portion Fig. 5.2 BC of the curve is parallel to x-axis therefore kinetic friction does Advantages and Disadvantages of Friction not change with the applied force, it remains constant, whatever be (1) Advantages of friction the applied force. (i) Walking is possible due to friction. Friction is a Cause of Motion (ii) Two body sticks together due to friction. 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 and no motion. Fig. 5.6 Fig. 5.7 (iii) Brake works on the basis of friction. Friction (iv) Writing is not possible without friction.  Action (v) The transfer of motion from one part of a machine to Fig. 5.3 other part through belts is possible by friction.  230 Friction (2) Disadvantages of friction Angle of Repose (i) Friction always opposes the relative motion between any Angle of repose is defined as the angle of the inclined plane two bodies in contact. Therefore extra energy has to be spent in with horizontal such that a body placed on it is just begins to slide. over coming friction. This reduces the efficiency of machine. By definition,  is called the angle of repose. (ii) Friction causes wear and tear of the parts of In limiting condition F  mg sin and R  mg cos machinery in contact. Thus their lifetime reduces. R F (iii) Frictional force result in the production of heat, which causes damage to the machinery. mg sin  Methods of Changing Friction  mg cos  We can reduce friction  mg (1) By polishing. Fig. 5.9 (2) By lubrication. F So  tan  (3) By proper selection of material. R (4) By streamlining the shape of the body. F F    s  tan   tan  [As we know   s  tan  ] R R (5) By using ball bearing. Thus the coefficient of limiting friction is equal to the tangent Also we can increase friction by throwing some sand on of angle of repose. slippery ground. In the manufacturing of tyres, synthetic rubber    i.e. angle of repose = angle of friction. As well as is preferred because its coefficient of friction with the road is larger. Calculation of Required Force in Different Situation Angle of Friction If W = weight of the body,  = angle of friction,   tan   coefficient of friction Angle of friction may be defined as the angle which the resultant of limiting friction and normal reaction makes with the Then we can calculate required force for different situation normal reaction. R in the following manner : S By definition angle  is called  (1) Minimum pulling force P at an angle  from the F P P the angle of friction horizontal Fl  tan   R mg  tan  = s Fig. 5.8 F Fig. 5.10 [As we know l   s ] R By resolving P in horizontal and vertical direction (as shown in figure) R or   tan 1( L ) P sin Hence coefficient of static friction is equal to tangent of the F P cos angle of friction. Resultant Force Exerted by Surface on Block W 2 2 Fig. 5.11 In the above figure resultant force S  F  R For the condition of equilibrium S  (mg)2  (mg)2 F  P cos and R  W  P sin By substituting these value in F  R S  mg  2  1 P cos   (W  P sin ) when there is no friction (  0) S will be minimum sin  P cos  (W  P sin ) [As   tan  ] i.e. S = mg cos Hence the range of S can be given by, W sin  P mg  S  mg   1 2 cos(   )  Friction 231 (2) Minimum pushing force P at an angle  from the (4) Minimum force to move a body in downward direction horizontal along the surface of inclined plane P P   Fig. 5.12  By Resolving P in horizontal and vertical direction (as shown in the figure) R Fig. 5.16 By Resolving P in the direction of the plane and F P cos perpendicular to the plane (as shown in the figure) R + P sin F P sin P cos W + Fig. 5.13 W sin  W cos For the condition of equilibrium  W F  P cos and R  W  P sin Fig. 5.17 By substituting these value in F  R For the condition of equilibrium  P cos   (W  P sin ) R  P sin  W cos  sin  R  W cos   P sin and F  P cos  W sin   P cos  (W  P sin ) [As   tan  ] cos By substituting these values in F  R and solving we get W sin W sin(   )  P P cos(   ) cos (   ) (3) Minimum pulling force P to move the body up on (5) Minimum force to avoid sliding of a body down an inclined plane P on an inclined plane P     Fig. 5.14 Fig. 5.18 By Resolving P in the direction of the plane and By Resolving P in the direction of the plane and perpendicular to the plane (as shown in the figure) perpendicular to the plane (as shown in the figure) R + P sin R + P sin F + P cos P cos F + W sin   W cos W sin W cos  W  W Fig. 5.15 Fig. 5.19 For the condition of equilibrium For the condition of equilibrium R  P sin  W cos  R  P sin  W cos   R  W cos   P sin and F  W sin   P cos  R  W cos   P sin and P cos  F  W sin   F  P cos  W sin   F  W sin   P cos By substituting these values in F  R and solving we get By substituting these values in F  R and solving we get  sin (   )  W sin (   ) P W  P  cos (   )  cos (   )  232 Friction (6) Minimum force for motion along horizontal Acceleration of a Block Against Friction surface and its direction P (1) Acceleration of a block on horizontal surface  When body is moving under application of force P, then kinetic friction opposes its motion. R Let a is the net acceleration of the body ma Fig. 5.20 From the figure Fk P Let the force P be applied at an angle  with the horizontal. ma  P  Fk By resolving P in horizontal and vertical direction (as shown P  Fk in figure) R + P sin  a mg m Fig. 5.23 F (2) Acceleration of a block sliding down over a rough P cos inclined plane When angle of inclined plane is more than angle of repose, mg the body placed on the inclined plane slides down with an acceleration a. R Fig. 5.21 F ma From the figure ma  mg sin  F For vertical equilibrium  ma  mg sin  R R  P sin  mg  ma  mg sin   mg cos mg sin  mg cos  R  mg  P sin …(i)  mg and for horizontal motion  Acceleration a  g [sin    cos ] Fig. 5.24 P cos  F Note : For frictionless inclined plane   0 i.e. P cos  R …(ii)  a  g sin. Substituting value of R from (i) in (ii) (3) Retardation of a block sliding up over a rough P cos   (mg  P sin ) inclined plane  mg When angle of inclined plane is less than angle of repose, P …(iii) cos   sin then for the upward motion ma R For the force P to be minimum (cos    sin ) must be ma  mg sin  F maximum i.e. 2 ma  mg sin   mg cos d 1  [cos    sin ]  0 mg sin + F  d mg cos   mg Retardation a  g [sin    cos ]   sin   cos  0  1 Fig. 5.25  tan    Fig. 5.22 Note : For frictionless inclined plane   0 or   tan1()  angle of friction  a  g sin  i.e. For minimum value of P its angle from the horizontal Work done against friction should be equal to angle of friction (1) Work done over a rough inclined surface  As tan    so from the figure, sin  1 2 If a body of mass m is moved up slowly on a rough inclined plane through distance s, then 1 Work done = force  distance and cos  1 2 = ma  s = mg [sin +  cos ]s  mg s [sin    cos ] By substituting these value in equation (iii) ma R  mg  mg P  1 2 1 2  s 1 2 1 2 mg sin + F  mg cos mg  mg  Pmin  1 2 Fig. 5.26  Friction 233 (2) Work done over a horizontal surface (iii) If friction is present between A and B only and applied force is greater than limiting friction (F > Fl) In the above expression if we put  = 0 then In this condition the two bodies will move in the same direction Work done = force  distance = F  s =  mg s (i.e. of applied force) but with different acceleration. Here force of It is clear that work done depends upon kinetic friction  k mg will oppose the motion of A while cause the R motion of B. F P Free body diagram of A F  Fk  m a A maA F  Fk i.e. aA  s m mg A F Fig. 5.27 (F   k mg) (i) Weight of the body. aA  Fk m (ii) Material and nature of surface in contact. Free body diagram of B Fk  M a B (iii) Distance moved. MaB Fk i.e. aB  Motion of Two Bodies one Resting on the Other M FK When a body A of mass m is resting on a body B of mass B  k mg M then two conditions are possible  aB  M (1) A force F is applied to the upper body, (2) A force F is applied to the lower body m Note : As both the bodies are moving in the A F same direction. L M B Acceleration of body A relative to B will be MF   k mg (m  M ) a  a A  aB  Fig. 5.28 mM We will discuss above two cases one by one in the following So, A will fall from B after time manner : 2L 2 m ML t  (1) A force F is applied to the upper body, then a MF   k mg (m  M ) following four situations are possible (iv) If there is friction between B and floor (i) When there is no friction (where Fl    (M  m) g = limiting friction between B and (a) The body A will move on body B with acceleration (F/m). floor, Fk = kinetic friction between A and B) aA  F / m B will move only if Fk  Fl and then Fk  Fl  M a B (b) The body B will remain at rest MaB aB  0 FK B (c) If L is the length of B as shown in figure, A will fall from Fl B after time t Fig. 5.29 2L 2mL  1 2  However if B does not move then static friction will work t   As s  2 a t and a  F/m  a F   (not limiting friction) between body B and the floor i.e. friction (ii) If friction is present between A and B only and applied force = applied force (= Fk) not Fl. force is less than limiting friction (F < Fl) (2) A force F is applied to the lower body, then (F = Applied force on the upper body, Fl = limiting friction following four situations are possible between A and B, Fk = Kinetic friction between A and B) (i) When there is no friction (a) The body A will not slide on body B till F  Fl i.e. (a) B will move with acceleration (F/M) while A will remain F   s mg at rest (relative to ground) as there is no pulling force on A. (b) Combined system (m + M) will move together with F F a B    and a A  0 common acceleration a A  a B  M M m  234 Friction (b) As relative to B, A will move backwards with acceleration Negative sign implies that relative to B, A will move (F/M) and so will fall from it in time t. backwards and will fall it after time 2L 2 ML A m t  a F   k g(m  M ) L F M B (iv) If there is friction between B and floor and F > Fl : Fig. 5.30 (where Fl = s(m+M)g = limiting friction between body B 2L 2ML and surface)  t  a F The system will move only if F  Fl' ' then replacing F by (ii) If friction is present between A and B only and F < Fl F  Fl . The entire case (iii) will be valid. (where F = Pseudo force on body A and Fl = limiting However if F  F1  the system will not move and friction friction between body A and B) between B and floor will be F while between A and B is zero. (a) Both the body will move together with common Motion of an Insect in the Rough Bowl F acceleration a  The insect crawl up the bowl, up to a certain height h only till M m the component of its weight along the bowl is balanced by limiting (b) Pseudo force on the body A, frictional force. r O mF F   ma  and Fl   s mg  m M Fl R y mF A (c) F   Fl    s mg  F   s (m  M ) g m M mg sin h mg cos So both bodies will move together with acceleration mg Fig. 5.31 F a A  aB  if F   s [m  M ] g m M Let m = mass of the insect, r = radius of the bowl,  = coefficient of friction (iii) If friction is present between A and B only and F > Fl for limiting condition at point A (where Fl = s mg = limiting friction between body A and B) R  mg cos ......(i) and Fl  mg sin ......(ii) Both the body will move with different acceleration. Here Dividing (ii) by (i) F force of kinetic friction  k mg will oppose the motion of B while tan   l   As Fl  R R will cause the motion of A. r 2  y2 r   or y ma A   k mg Free body diagram of A y 1 2 maA     i.e. a A  k g A 1  ,  h  r 1  1 So h  r  y  r 1   Fk  1  2   1  2      F  Fk  Ma B Free body diagram of B Minimum Mass Hung from the String to Just MaB Start the Motion [F  kmg] (1) When a mass m1 placed on a rough horizontal plane i.e. aB  FK M F B Another mass m2 hung from the string connected by frictionless pulley, the tension (T) produced in string will try to start the motion of mass m1. R Note : As both the bodies are moving in the Fl m1 T same direction T Acceleration of body A relative to B will be m1g  F  k g(m  M )  a  a A  aB    m2  M  Fig. 5.32 m2g  Friction 235 At limiting condition T  Fl Coefficient of Friction Between a Body and Wedge  m2 g  R  m2 g   m1 g A body slides on a smooth wedge of angle  and its time of descent is t.  m2  m1 this is the minimum value of m 2 to start the motion. S S Note :  In the above condition Coefficient of Smooth wedge Rough wedge m2   friction   m1 Fig. 5.35 Fig. 5.36 (2) When a mass m1 placed on a rough inclined plane Another mass m 2 hung from the string connected by If the same wedge made rough then time taken by it to come down becomes n times more (i.e. nt) frictionless pulley, the tension (T) produced in string will try to The length of path in both the cases are same. start the motion of mass m1. 1 T For smooth wedge, S  u t  at 2 R 2 T 1 S (g sin  ) t 2 …(i) m1 2 m2 [ As u  0 and a  g sin  ] m1g sin + F  m1g cos m2g 1 2 For rough wedge, S  u t  at m1g 2 Fig. 5.33 1 At limiting condition S g (sin    cos  ) (nt)2 …(ii) 2 For m2 T  m2 g …(i) [As u  0 and a  g (sin    cos  )] For m1 T  m1 g sin   F From equation (i) and (ii)  T  m1 g sin   R 1 1 (g sin  ) t 2 = g (sin    cos  ) (nt)2  T  m1 g sin   m1 g cos  …(ii) 2 2 m2  m1[sin    cos  ]  sin  (sin   cos  ) n2 From equation (i) and (ii)  1 this is the minimum value of m2 to start the motion    tan  1  2   n  Note :  In the above condition Coefficient of Stopping of Block Due to Friction friction (1) On horizontal road  m2  (i) Distance travelled before coming to rest : A block of   tan   m  1 cos   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. Maximum Length of Hung Chain S A uniform chain of length l is placed on the table in such a v=0 u manner that its l ' part is hanging over the edge of table without sliding. Since the chain have uniform linear density therefore the Fig. 5.37 ratio of mass and ratio of length for any part of the chain will be Retarding force F  ma  R  ma   mg equal.  a  g m mass hanging from the table We know  2  From v 2  u 2  2aS  0  u2  2 g S m1 mass lying on the table [As v  0, a  g ]  For this case we can rewrite above expression in the 2 2 following manner u P  S or S length hanging from the table 2g 2m2 g  [As chain have uniform length lying on the table [As momentum P = mu] ( l – l ) linear density] (ii) Time taken to come to rest l From equation v  u  a t  0  u   g t   l [As v  0, a   g] l  l l u by solving l   t (  1) g Fig. 5.34  236 Friction (2) On inclined road : When block starts with velocity u its By the law of conservation of energy kinetic energy will be converted into potential energy and some 1 i.e. mv2  mgh  FL part of it goes against friction and after travelling distance S it 2 comes to rest i.e. v = 0. 2 We know that retardation a  g [sin    cos  ] v (mgh  FL) m By substituting the value of v and a in the following equation Sticking of a Block With Accelerated Cart v=0 When a cart moves with some acceleration toward right S then a pseudo force (ma) acts on block toward left. u This force (ma) is action force by a block on cart.  a F Fig. 5.38 v2  u2  2a S F ma m R M m 2  0  u  2g [sin   cos  ] S CART u2 mg  S Fig. 5.41 2g (sin    cos  ) Now block will remain static w.r.t. cart. If friction force Stopping of Two Blocks Due to Friction R  mg When two masses compressed towards each other and  ma  mg [As R  ma] suddenly released then energy acquired by each block will be g dissipated against friction and finally block comes to rest  a  i.e., F × S = E [Where F = Friction, S = Distance g covered by block, E = Initial kinetic energy of the block]  amin   A B This is the minimum acceleration of the cart so that block m1 m1 m2 m2 does not fall. and the minimum force to hold the block together S1 S2 Fmin  (M  m) amin Fig. 5.39 g P2 Fmin  (M  m)  FS  [Where P = momentum of block]  2m P2 Sticking of a Person with the Wall of Rotor  mg  S  [As F =  mg] 2m A person with a mass m stands in contact against the wall of P2 a cylindrical drum (rotor). The coefficient of friction between the  S 2m2 g wall and the clothing is . In the given condition P and  are same for both the blocks. If Rotor starts rotating about its axis, then person thrown 1 m  2 away from the centre due to centrifugal force at a particular speed S So, S  2 ;  1   2  , the person stuck to the wall even the floor is removed, because m S2  m1  friction force balances its weight in this condition. Velocity at the Bottom of Rough Wedge From the figure. 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. Friction force (F) = weight of person (mg) Loss of energy due to friction = FL (Work against  R = mg   Fc  mg F friction) u=0 [Here, Fc= centrifugal force] R FC m PE at point A = mgh A 2 mg L  mminr  mg 1 h KE at point B = mu 2 Fig. 5.42 2 m B g   min  v r Fig. 5.40  Friction 237 6. Pulling force making an angle  to the horizontal is applied on a block of weight W placed on a horizontal table. If the angle of friction is  , then the magnitude of force required to move the body is equal to [EAMCET 1987] W sin  W cos  (a) (b)  Force of friction is non-conservative force. g tan(   ) cos(   )  Force of friction always acts in a direction opposite to that W sin  W tan (c) (d) of the relative motion between the surfaces. cos(   ) sin(   )  Rolling friction is much less than the sliding friction. This 7. In the figure shown, a block of weight 10 N resting on a knowledge was used by man to invent the wheels. horizontal surface. The coefficient of static friction between the block and the surface  s  0.4. A force of 3.5 N will  The friction between two surfaces increases (rather than to keep the block in uniform motion, once it has been set in decrease), when the surfaces are made highly smooth. motion. A horizontal force of 3 N is applied to the block,  The atomic and molecular forces of attraction between the then the block will [MP PET 1993] two surfaces at the point of contact give rise to friction between the surfaces. T (a) Move over the surface with constant velocity (b) Move having accelerated motion over the surface (c) Not move Static and limiting friction (d) First it will move with a constant velocity for some time and then will have accelerated motion 1. The coefficient of friction  and the angle of friction  are related as 8. Two masses A and B of 10 kg and 5 kg respectively are (a) sin    (b) cos    connected with a string passing over a frictionless pulley fixed at the corner of a table as shown. The coefficient of (c) tan    (d) tan    static friction of A with table is 0.2. The minimum mass of C 2. A force of 98 N is required to just start moving a body of that may be placed on A to prevent it from moving is mass 100 kg over ice. The coefficient of static friction is [MP PET 1984] (a) 0.6 (b) 0.4 C (c) 0.2 (d) 0.1 (a) 15 kg A 3. A block weighs W is held against a vertical wall by applying (b) 10 kg a horizontal force F. The minimum value of F needed to B hold the block is [MP PMT 1993] (c) 5 kg (a) Less than W (b) Equal to W (d) 12 kg (c) Greater than W (d) Data is insufficient 9. The limiting friction is 4. The maximum static frictional force is (a) Always greater than the dynamic friction (a) Equal to twice the area of surface in contact (b) Always less than the dynamic friction (b) Independent of the area of surface in contact (c) Equal to the area of surface in contact (c) Equal to the dynamic friction (d) None of the above (d) Sometimes greater and sometimes less than the dynamic friction 5. Maximum value of static friction is called [BHU 1995; RPET 2000] 10. Which is a suitable method to decrease friction (a) Limiting friction (b) Rolling friction (a) Ball and bearings (b) Lubrication (c) Normal reaction (d) Coefficient of friction (c) Polishing (d) All the above  238 Friction 11. A uniform rope of length l lies on a table. If the coefficient of 17. A uniform chain of length L changes partly from a table friction is  , then the maximum length l1 of the part of this which is kept in equilibrium by friction. The maximum length that can withstand without slipping is l, then rope which can overhang from the edge of the table without coefficient of friction between the table and the chain is sliding down is [DPMT 2001] [EAMCET (Engg.) 1995] l l l l (a) (b) (a) (b)   l L Ll l L l l (c) (d) (c) (d) Ll Ll 1   1 18. When two surfaces are coated with a lubricant, then they 12. Which of the following statements is not true [AFMC 1998, 99; AIIMS 2001] [CMC Vellore 1989] (a) Stick to each other (b) Slide upon each other (c) Roll upon each other (d) None of these (a) The coefficient of friction between two surfaces increases as the surface in contact are made rough 19. A 20 kg block is initially at rest on a rough horizontal surface. A horizontal force of 75 N is required to set the (b) The force of friction acts in a direction opposite to the block in motion. After it is in motion, a horizontal force of applied force 60 N is required to keep the block moving with constant speed. The coefficient of static friction is [AMU 1999] (c) Rolling friction is greater than sliding friction (a) 0.38 (b) 0.44 (d) The coefficient of friction between wood and wood is less than 1 (c) 0.52 (d) 0.60 20. A block A with mass 100 kg is resting on another block B of 13. A block of 1 kg is stopped against a wall by applying a force mass 200 kg. As shown in figure a horizontal rope tied to a F perpendicular to the wall. If   0.2 then minimum value wall holds it. The coefficient of friction between A and B is of F will be [MP PMT 2003] 0.2 while coefficient of friction between B and the ground is 0.3. The minimum required force F to start moving B will be (a) 980 N (b) 49 N [RPET 1999] (c) 98 N (d) 490 N (a) 900 N 14. A heavy uniform chain lies on a horizontal table-top. If the (b) 100 N A coefficient of friction between the chain and table surface is (c) 1100 N B F 0.25, then the maximum fraction of length of the chain, that can hang over one edge of the table is [CBSE PMT 1990] (d) 1200 N 21. To avoid slipping while walking on ice, one should take (a) 20% (b) 25% smaller steps because of the [BHU 1999; BCECE 2004] (c) 35% (d) 15% (a) Friction of ice is large 15. The blocks A and B are arranged as shown in the figure. (b) Larger normal reaction The pulley is frictionless. The mass of A is 10 kg. The (c) Friction of ice is small coefficient of friction of A with the horizontal surface is 0.20. (d) Smaller normal reaction The minimum mass of B to start the motion will be 22. A box is lying on an inclined plane what is the coefficient of [MP PET 1994] static friction if the box starts sliding when an angle of inclination is 60o [KCET 2000] A (a) 1.173 (b) 1.732 (a) 2 kg (c) 2.732 (d) 1.677 (b) 0.2 kg B 23. A block of mass 2 kg is kept on the floor. The coefficient of static friction is 0.4. If a force F of 2.5 Newtons is applied on (c) 5 kg the block as shown in the figure, the frictional force between (d) 10 kg the block and the floor will be [MP PET 2000] 16. Work done by a frictional force is (a) 2.5 N (b) 5 N (a) Negative (b) Positive F (c) 7.84 N (c) Zero (d) All of the above (d) 10 N  Friction 239 24. Which one of the following is not used to reduce friction 30. A uniform metal chain is placed on a rough table such that [Kerala (Engg.) 2001] one end of chain hangs down over the edge of the table. When one-third of its length hangs over the edge, the chain (a) Oil (b) Ball bearings starts sliding. Then, the coefficient of static friction is (c) Sand (d) Graphite [Kerala PET 2005] 25. If a ladder weighing 250N is placed against a smooth 3 1 (a) (b) vertical wall having coefficient of friction between it and 4 4 floor is 0.3, then what is the maximum force of friction 2 1 (c) (d) available at the point of contact between the ladder and the 3 2 floor [AIIMS 2002] 31. A lift is moving downwards with an acceleration equal to (a) 75 N (b) 50 N acceleration due to gravity. A body of mass m kept on the floor of the lift is pulled horizontally. If the coefficient of (c) 35 N (d) 25 N friction is , then the frictional resistance offered by the 26. A body of mass 2 kg is kept by pressing to a vertical wall by body is [DPMT 2004] a force of 100 N. The coefficient of friction between wall (a) mg (b) mg and body is 0.3. Then the frictional force is equal to (c) 2mg (d) Zero [Orissa JEE 2003] 32. If a ladder weighing 250 N is placed against a smooth (a) 6 N (b) 20 N vertical wall having coefficient of friction between it and (c) 600 N (d) 700 N floor is 0.3, then what is the maximum force of friction available at the point of contact between the ladder and the 27. A horizontal force of 10 N is necessary to just hold a block floor [BHU 2004] stationary against a wall. The coefficient of friction between (a) 75 N (b) 50 N the block and the wall is 0.2. the weight of the block is (c) 35 N (d) 25 N [AIEEE 2003] (a) 2 N Kinetic Friction (b) 20 N 10 N 1. Which one of the following statements is correct (c) 50 N (a) Rolling friction is greater than sliding friction (d) 100 N (b) Rolling friction is less than sliding friction (c) Rolling friction is equal to sliding friction 28. The coefficient of static friction,  s , between block A of (d) Rolling friction and sliding friction are same mass 2 kg and the table as shown in the figure is 0.2. What 2. The maximum speed that can be achieved without skidding would be the maximum mass value of block B so that the by a car on a circular unbanked road of radius R and two blocks do not move? The string and the pulley are coefficient of static friction  , is [NCERT 1990] assumed to be smooth and massless. (g  10 m / s 2 ) (a) Rg (b) Rg  [CBSE PMT 2004] 2 kg (c)  Rg (d) Rg A 3. A car is moving along a straight horizontal road with a speed (a) 2.0 kg v0. If the coefficient of friction between the tyres and the (b) 4.0 kg road is  , the shortest distance in which the car can be B stopped is [MP PET 1985; BHU 2002] (c) 0.2 kg v02 v0 (d) 0.4 kg (a) (b) 2 g g 29. If mass of A  10 kg , coefficient of static friction = 0.2, 2  v0  v0 coefficient of kinetic friction = 0.2. Then mass of B to start (c)   (d)  g   motion is [Orissa PMT 2004] 10 kg 4. A block of mass 5 kg is on a rough horizontal surface and is A at rest. Now a force of 24 N is imparted to it with negligible (a) 2 kg impulse. If the coefficient of kinetic friction is 0.4 and (b) 2.2 kg g  9.8 m / s 2 , then the acceleration of the block is B (c) 4.8 kg (a) 0.26 m / s 2 (b) 0.39 m / s 2 (d) 200 gm (c) 0.69 m / s 2 (d) 0.88 m / s 2  240 Friction 5. A body of mass 2 kg is being dragged with uniform velocity 11. On a rough horizontal surface, a body of mass 2 kg is given of 2 m/s on a rough horizontal plane. The coefficient of a velocity of 10 m/s. If the coefficient of friction is 0.2 and friction between the body and the surface is 0.20. The g  10 m / s 2 , the body will stop after covering a distance of amount of heat generated in 5 sec is [MP PMT 1999] ( J  4.2 joule / cal and g  9.8 m / s 2 ) (a) 10 m (b) 25 m [MH CET (Med.) 2001] (c) 50 m (d) 250 m (a) 9.33 cal (b) 10.21 cal 12. A block of mass 50 kg can slide on a rough horizontal surface. The coefficient of friction between the block and the (c) 12.67 cal (d) 13.34 cal surface is 0.6. The least force of pull acting at an angle of 6. Two carts of masses 200 kg and 300 kg on horizontal rails 30° to the upward drawn vertical which causes the block to are pushed apart. Suppose the coefficient of friction just slide is [ISM Dhanbad 1994] between the carts and the rails are same. If the 200 kg cart (a) 29.43 N (b) 219.6 N travels a distance of 36 m and stops, then the distance travelled by the cart weighing 300 kg is (c) 21.96 N (d) 294.3 N [CPMT 1989; DPMT 2002] 13. A body of 10 kg is acted by a force of 129.4 N if g  9.8 m / sec 2. The acceleration of the block is 10 m / s 2. A B What is the coefficient of kinetic friction [EAMCET 1994] (a) 0.03 (b) 0.01 (c) 0.30 (d) 0.25 (a) 32 m (b) 24 m 14. Assuming the coefficient of friction between the road and (c) 16 m (d) 12 m tyres of a car to be 0.5, the maximum speed with which the 7. A body B lies on a smooth horizontal table and another car can move round a curve of 40.0 m radius without body A is placed on B. The coefficient of friction between A slipping, if the road is unbanked, should be [AMU 1995] and B is . What acceleration given to B will cause slipping (a) 25 m/s (b) 19 m/s to occur between A and B (c) 14 m/s (d) 11 m/s (a) g (b) g /  15. Consider a car moving along a straight horizontal road with a speed of 72 km/h. If the coefficient of kinetic friction (c) /g (d) g between the tyres and the road is 0.5, the shortest distance 8. A 60 kg body is pushed with just enough force to start it in which the car can be stopped is [g  10 ms2 ] moving across a floor and the same force continues to act [CBSE PMT 1992] afterwards. The coefficient of static friction and sliding friction are 0.5 and 0.4 respectively. The acceleration of the (a) 30 m (b) 40 m body is (c) 72 m (d) 20 m (a) 6 m / s 2 (b) 4.9 m / s 2 16. A 500 kg horse pulls a cart of mass 1500 kg along a level road with an acceleration of 1 ms 2. If the coefficient of 2 2 (c) 3.92 m / s (d) 1 m / s sliding friction is 0.2, then the force exerted by the horse in 9. A car turns a corner on a slippery road at a constant speed forward direction is [SCRA 1998] of 10 m / s. If the coefficient of friction is 0.5, the minimum (a) 3000 N (b) 4000 N radius of the arc in meter in which the car turns is (c) 5000 N (d) 6000 N (a) 20 (b) 10 17. The maximum speed of a car on a road turn of radius 30m; (c) 5 (d) 4 if the coefficient of friction between the tyres and the road is 0.4; will be [MH CET (Med.) 1999] 10. A motorcyclist of mass m is to negotiate a curve of radius r (a) 9.84 m/s (b) 10.84 m/s with a speed v. The minimum value of the coefficient of friction so that this negotiation may take place safely, is (c) 7.84 m/s (d) 5.84 m/s [Haryana CEE 1996] 18. A block of mass 50 kg slides over a horizontal distance of 1 m. If the coefficient of friction between their surfaces is 0.2, v2 then work done against friction is (a) v2rg (b) gr [BHU 2001; CBSE PMT 1999, 2000; AIIMS 2000] gr g (a) 98 J (b) 72J (c) (d) v2 v 2r (c) 56 J (d) 34 J  Friction 241 19. On the horizontal surface of a truck ( = 0.6), a block of 27. A 60 kg weight is dragged on a horizontal surface by a rope mass 1 kg is placed. If the truck is accelerating at the rate of upto 2 metres. If coefficient of friction is   0.5 , the angle 5m/sec2 then frictional force on the block will be of rope with the surface is 60° and g  9.8 m / sec 2 , then [CBSE PMT 2001] work done is [MP PET 1995] (a) 5 N (b) 6 N (a) 294 joules (b) 315 joules (c) 5.88 N (d) 8 N (c) 588 joules (d) 197 joules 20. A vehicle of mass m is moving on a rough horizontal road with momentum P. If the coefficient of friction between the 28. A car having a mass of 1000 kg is moving at a speed of 30 tyres and the road be , then the stopping distance is metres/sec. Brakes are applied to bring the car to rest. If the [CBSE PMT 2001] frictional force between the tyres and the road surface is 2 5000 newtons, the car will come to rest in [MP PMT 1995] P P (a) (b) (a) 5 seconds (b) 10 seconds 2 m g 2 m g (c) 12 seconds (d) 6 seconds P P2 (c) (d) 29. If  s ,  k and  r are coefficients of static friction, sliding 2 m 2 g 2 m 2 g friction and rolling friction, then [EAMCET (Engg.) 1995] 21. A body of weight 64 N is pushed with just enough force to start it moving across a horizontal floor and the same force (a)  s   k  r (b)  k   r   s continues to act afterwards. If the coefficients of static and (c) r  k   s (d) r  k   s dynamic friction are 0.6 and 0.4 respectively, the acceleration of the body will be (Acceleration due to gravity = g) 30. A body of mass 5kg rests on a rough horizontal surface of [EAMCET 2001] coefficient of friction 0.2. The body is pulled through a g distance of 10m by a horizontal force of 25 N. The kinetic (a) (b) 0.64 g energy acquired by it is (g = 10 ms2) 6. 4 [EAMCET (Med.) 2000] g (c) (d) 0.2 g 32 (a) 330 J (b) 150 J 22. When a body is moving on a surface, the force of friction is (c) 100 J (d) 50 J called [MP PET 2002] 31. A motorcycle is travelling on a curved track of radius 500m. If (a) Static friction (b) Dynamic friction the coefficient of friction between road and tyres is 0.5, the (c) Limiting friction (d) Rolling friction speed avoiding skidding will be [MH CET (Med.) 2001] 23. A block of mass 10 kg is placed on a rough horizontal (a) 50 m/s (b) 75 m/s surface having coefficient of friction  = 0.5. If a horizontal (c) 25 m/s (d) 35 m/s force of 100 N is acting on it, then acceleration of the block will be [AIIMS 2002] 32. A fireman of mass 60 kg slides down a pole. He is pressing (a) 0.5 m/s 2 (b) 5 m/s 2 the pole with a force of 600 N. The coefficient of friction between the hands and the pole is 0.5, with what (c) 10 m/s 2 (d) 15 m/s2 acceleration will the fireman slide down (g = 10 m/s2) 24. It is easier to roll a barrel than pull it along the road. This [Pb. PMT 2002] statement is [BVP 2003] (a) False (b) True (a) 1 m/s 2 (b) 2.5 m/s 2 (c) Uncertain (d) Not possible (c) 10 m/s2