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Physics

Physics. Session Opener. A recent space shuttle accident occurred because of failure of heat protecting devices. How was this heat generated ?. Session Objectives. Session Objective. Friction & frictional force Bodies connected in frictional surface Kinetic friction Static friction

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Physics

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  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. Session Objective • Friction & frictional force • Bodies connected in frictional surface • Kinetic friction • Static friction • Angle of friction • Coefficient of friction • Laws of friction

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

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

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

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

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

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

  12. Class Exercise

  13. 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 (b) downward (c) dependent on relative mass of A and B (d) system cannot be in equilibrium Class Exercise - 1

  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. Coefficients of Friction Limiting friction Static region Kinetic region Graphically

  17. Laws of Friction 1. Bodies slip over each other 3. Bodies do not slip over each other N is called the limiting friction. 4. do not depend upon the area of contact. 2. Direction of kinetic friction is opposite to the velocity

  18. Class Exercise

  19. 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 m. Then F must be equal to (a) (b) (c) mg (d) cannot be found Class Exercise - 4

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

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

  23. A block of mass M is kept in a lift. Coefficient of friction between the block and the lift is m. 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 Class Exercise - 6

  24. Condition of motion of M: 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.

  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/s2)

  26. Solution Vertical: N = mg – F sin30 Horizontal: F cos 30 – f = ma Solving for a: a = 1.37 m/s2

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

  28. Class Exercise

  29. 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 f. Then, what is the coefficient of friction, (a)  = tan  (b)  > tan  (d)  and  are not related (c)  < tan  Class Exercise - 2 Solution : Hence answer is (a)

  30. 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 b. a. c. d. Class Exercise - 3

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

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

  33. Angle of Repose reduced At , object moves uniformly. : Angle of repose (kinetic friction) N fk 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/s2)

  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/s2)

  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

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