1. Physics

2. Session Opener A recent space shuttle accident occurred because of failure of heat protecting devices. How was this heat generated ?

3. Session Objectives

4. 1.Friction & frictional force 2.Bodies connected in frictional surface 3.Kinetic friction 4.Static friction 5.Angle of friction 6.Coefficient of friction 7.Laws of friction Session Objective

5. Friction and Frictional Force Whenever an object moves or tends to move, contact surfaces of object resist the force tending to generate motion. This property of the surface is friction The resistive force is the friction force

6. An equal and opposite force f acts Consider the experiment : 1. F applied. ‘A’ does not move. 2. ‘A’ still at rest on increased F. f adjusts to match MBMB F MAMA A Bodies Connected in Frictional Surface f

7. 3. increased further F  A has acceleration. f has a maximum limiting value f during the motion is lower than at rest. 4. is reduced A moves uniformly Bodies Connected in Frictional Surface MBMB F f F f F f F f F f F f F f F f F f F f

8.  Friction retards motion  f adjusts to stay equal and opposite to  f has a limiting value Bodies Connected in Frictional Surface  moving friction force is less than friction force at rest.  Friction force (f) is zero if no external force ( ) exists

9. Friction force is a contact force, independent of area of contact. Bodies Connected in Frictional Surface Friction force is a non-conservative force as mechanical energy changes to heat during friction. That is why the space shuttle overheated and exploded F f

10. Kinetic Friction  appears when two objects in contact have relative motion.  Directed opposite the direction of motion  f k is a constant.  f k on B may start the motion of B. F fkfk A B F fkfk fkfk

11. Static Friction  F has component along contact surface, but there is no motion.  f = Fcos till limiting friction.  At limiting friction f(=f s ) is constant  f s > f k  f acts along the surface and opposite the force component F N mg fsfs Fcos

12. Class Exercise

13. Class Exercise - 1 Blocks A and B are pressed against a smooth wall and are in equilibrium by a horizontal force F as shown. Then friction force due to A on B (a) upward (d) system cannot be in equilibrium (c) dependent on relative mass of A and B (b) downward

14. Solution As wall is smooth, A and B will keep slipping down with acceleration g and cannot be in equilibrium. Hence answer is (d).

15. Coefficients of Friction Limiting static friction Kinetic friction  are coefficient of friction

16. Static region Graphically Kinetic region Limiting friction Coefficients of Friction

17. Laws of Friction 1. Bodies slip over each other 2. Direction of kinetic friction is opposite to the velocity 3. Bodies do not slip over each other N is called the limiting friction. do not depend upon the area of contact. 4.4.

18. Class Exercise

19. Class Exercise - 4 An object of mass m is pressed against a wall with a force F and is in equilibrium. The coefficient of friction between the wall and the object is . Then F must be equal to (c) mg (d) cannot be found (a) (b)

20. Solution As equilibrium exists, wall presses against the object by a force of F (equal and opposite to force exerted by object against the wall) At equilibrium

21. Class Exercise - 5 Three blocks of masses m a,m b and m c are arranged on a smooth horizontal surface as shown. Surface between m a and m b is smooth and the coefficient of friction between m b and m c is .Then the minimum force F required to keep m b from sliding down is: ab c d

22. Solution Let the force F give an acceleration a to the system To give an acceleration a to m c, m b presses against m c with force m c a, so m c gives a normal reaction m c a (to left) on m b. for m b not slipping down

23. Class Exercise - 6 A block of mass M is kept in a lift. Coefficient of friction between the block and the lift is . The force required to initiate horizontal motion of the block is maximum when (a) Lift is moving up with constant acceleration (b) Lift is moving down with constant acceleration (c) Lift is stationary (d) Lift is in free fall

24. Solution F is largest when N is maximum N is maximum when lift moves up with constant acceleration a [N = m(g + a)]. So F is maximum at that condition. Condition of motion of M:

25. Class Exercise - 8 A 70 Kg box is pulled along a horizontal surface by a 400 N force at an angle of 30° above horizontal. If the coefficient of friction is 0.50, what is the acceleration of the box? (g = 10 m/s 2 )

26. Solution Vertical: N = mg – F sin30 Horizontal: F cos 30 – f = ma Solving for a: a = 1.37 m/s 2

27. Angle of Friction f along surface (surface property) N normal to surface Angle of friction Static > kinetic

28. Class Exercise

29. Class Exercise - 2 The angle between the resultant cont- act force and the normal reaction force exerted by a body on the other when they are one on top of the other is . Then, what is the coefficient of friction, (a)  = tan  (b)  > tan  (c)  < tan  (d)  and  are not related Solution : Hence answer is (a)

30. Class Exercise - 3 A pushing force F is applied to a body of weight W, placed on a horizontal table, at an angle . The angle of friction is . The magnitude of F to initiate motion in the body is a. b. d.c.

31. Solution V ertical: Horizontal: (a = 0 at initiation of motion)

32. Angle of Repose Object at rest Object slips  increased.  is angle of repose N fsfs mg

33. Angle of Repose reduced At, object moves uniformly. : Angle of repose (kinetic friction) N fkfk mg

34. Class Exercise

35. Class Exercise - 9 A block ‘A’ slides from rest down an incline with a 30° inclination with horizontal. It covers 3 m in 5 seconds. What is the value of coefficient of kinetic friction? (g = 10 m/s 2 )

36. Solution Along y: N = mg cos Along x : mg sin – f = ma

37. Class Exercise - 10 A block of mass 2kg lies on a rough inclined plane of inclination 30° to the horizontal. Coefficient of friction between the block and the plane is 0.75. What minimum force will make the block move up the incline? (g = 10 m/s 2 )

38. Solution Along y: N = mg cos F is up the incline (along x) F – (mg sin  + f) = ma To start the motion a is put equal to zero

39. Thank you