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# Friction [compatibility mode]

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### Friction [compatibility mode]

1. 1. Friction Cause of dry friction Contact between two surfaces. Hence first task in a friction problem is correct identification of contact surfaces Identify the surface, the normal and the tangential vectors. Also important is to get an idea of probable direction of relative motion The contact force acts along the normal. Gravity is the most common cause of normal force. Friction acts along the tangent plane opposite to the direction of relative motion Normal Relative velocity Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)
2. 2. Friction Normal Relative velocity Friction problems are essentially equilibrium problems with one f the forces being functions of another N Fr=f(N,V) Fr=f(N,V) N Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)
3. 3. The correct way of writing the dry friction force V ˆ Fr    N   N V V N=Normal force vector V=Relative velocity vector of the body m= coefficient of dry friction or Coulomb friction Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)
4. 4. Problem 1 Knowing that the coefficient of friction between the 13.5 kg block and the incline is ms = 0.25, determine a) the smallest value of P required to maintain the block in equilibrium, b) the corresponding value of b. Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)
5. 5. Problem 1 y N cos60  mg  f sin60  P sin   0  N sin60  f cos 60  P cos   0 f  N x mg sin60  f P sin  sin60  cos  cos 60 N cos60  mg   N sin60  P sin   0 mg  P sin  N cos 60   sin60  N sin60   N cos60  P cos   0 mg  P sin    sin60   cos 60   P cos   0 cos 60   sin60    mg  P sin   sin60   cos60   P cos   cos 60   sin60   0   mg  sin60   cos60   P sin   sin60   cos 60   P cos   cos60   sin60   0  P  sin60 sin    cos60 sin   cos60 cos    sin60 cos    mg  sin60   cos60   sin60   cos 60   P  mg cos  60      sin  60    1 cos  60      sin  60      mg P  sin60   cos 60  d 1 d  P   0  d   cos  60      sin  60      0 d     d   sin60 sin    cos 60 sin   cos60 cos    sin60 cos    0 d  sin60 cos    cos60 cos   cos 60 sin    sin60 sin   0  sin60   cos 60  o  tan      2.614    69  cos60   sin60  P  mg  sin 60   cos60   mg  0.866  0.125   0.72mg cos  60      sin  60    cos  9    sin  9  Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)
6. 6. Problem 2 Knowing that P = 110 N, determine the range of values value of q for which equilibrium of the 8 kg block is maintained. Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)
7. 7. Problem 2 y y x x mg mg  P cos   N  0  P cos   N  0 P sin  mg  f  0 P sin  mg  f  0 f  N f  N  N  P cos   N  P cos P sin  mg   N  0 P sin  mg   N  0  P sin  mg   P cos   0  P sin  mg   P cos   0  P  sin   cos    mg  P  sin   cos    mg mg mg P P sin   cos  sin   cos  Hence Hence upward movement will not start before downward movement will not start before mg mg P P sin   s cos  sin  k cos  Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)
8. 8. Problem 3 The coefficients of friction are ms = 0.40 and mk = 0.30 between all the surfaces of contact. Determine the force P for which motion of the 27 kg block is impending if cable a) is attached as shown, b) is removed Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)
9. 9. Problem 3 v T N1 f1 m1g f1 N1 T m2g f2 N2 T   N1  0 N 1  m1 g  0  T   N 1 , N 1  m1 g  T   m1 g T   N1   N 2  P  0 N 2  N 1  m2 g  0 N 2  N 1  m 2 g  0  N 2   m1  m 2  g T   N1   N 2  P  0   m 1 g   m1 g    m1  m 2  g  P  0  P  3  m1 g   m 2 g Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)
10. 10. Problem 3 v 0 N1 f1 m1g f1 N1 0 m2g f2 N2 T   N 1  m1 a Now T  0    N 1  m1 a N 1  m1 g  0  N 1  m1 g  N1   N2  P  0 N 2  N 1  m2 g  0 N 2  N 1  m 2 g  0  N 2   m1  m2  g  N1   N2  P  0   m1 g    m 1  m 2  g  P  0  P  2  m1 g   m2 g Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)
11. 11. Additional Problems The 8 kg block A and the 16 kg block B are at rest on an incline as shown. Knowing that the coefficient of static friction is 0.25 between all surfaces of contact, determine the value of q for which motion is impending. Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)
12. 12. Toppling The magnitude of the force P is slowly increased. Does the homogeneous box of mass m slip or tip first? State the value of P which would cause each occurrence. Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)
13. 13. Slip or topple? mg f1 f2 N1 N2 P cos 30   N 1   N 2  0 P sin 30  N 1  N 2  mg  0  P cos 30  d  mg  d   N 2  2d  0 P cos 30   N 1   N 2    N 1  N 2  P sin 30    N 1  N 2   mg P cos 30  mg  2  N 2  P sin 30     N 1  N 2    mg   P cos 30   mg  P sin 30  P cos 30   mg  P   sin 30  cos 30    mg  mg P  sin 30  cos 30 0.5mg P  0.448mg 0.25  0.866 Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)
14. 14. Slip or topple? mg f1 f2 N1 N2 P cos 30  f 2  0 P sin 30  N 2  mg  0  P cos 30  d  P sin 30  2d  mg  d  0 P cos 30  f 2 P sin 30   N 2  mg P cos 30  2P sin 30  mg  0  P  cos 30  2 sin 30   mg mg P cos 30  2 sin 30 mg P  0.536mg 0.866  1 Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)
15. 15. Additional Problems Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)
16. 16. Additional Problems Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)
17. 17. Additional Problems Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)
18. 18. Wedge f=mN P q N mg P  f cos   N sin  0 f sin  N cos   mg  0 f  N mg  N sin  N cos   mg  0  N   sin  cos  P   N cos   N sin  0  cos   sin  P    cos   sin  N  mg  sin  cos  P   tan   mg  tan  1 P tan  tan    tan     mg tan tan  1 Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)
19. 19. A screw thread is a wedge Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)
20. 20. A screw thread is a wedge M=Pa Q=Pa/r = equivalent force W Q N f=mN Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)
21. 21. A screw thread is a wedge W Q N q f=mN Q  f cos   N sin  0  f sin  N cos   W  0 f  N W   N sin  N cos   W  0  N    sin  cos  Q   N cos   N sin  0  cos   sin  Q    cos   sin  N  W   sin  cos  Q   tan   W 1   tan Pa   tan  Wr 1   tan 1 P if tan     W Screw locks Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)
22. 22. A screw thread is a wedge The W P N q f=mN 1 Pa tan    Screw locks  Wr 1 Wr   0 tan Pa There is a critical value of friction coefficient beyond which the thread does not move irrespective of the force applied. This happens when a screw is not maintained properly. Because of dirt and rust m becomes more than critical. Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)
23. 23. A screw thread is a wedge W N f=mN Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)
24. 24. A screw thread is a wedge W N q f For no movement f cos   N sin  0 f sin  N cos   W  0 Self locking f sin   N cos  sin   tan cos  Therefore after raising the load if we let go of the screw the load will not cause the screw to unscrew by itself. Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)
25. 25. Terminologies Lead (L) 2pr Pitch (p) Lead  L  np where n=no. of parallely running threads = starts L tan = 2 r Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)
26. 26. Turnbuckle T1 T2 Used to apply tension. The sleeve is rotated to pull the threads together. M   tan   T2  T1  r 1   tan Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)
27. 27. An improved screw jack W q q W T T T T 2T cos q Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)
28. 28. An improved screw jack W f f M=Pa W=2T cos f Pa  cos  sin  Wr   sin  cos  M  cos  sin   2Tr cos    sin  cos  Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)
29. 29. Worm gear MG M0/R R N M f MG M G  WR  W  R Pa   tan  Wr 1   tan M   tan   MG 1   tan r R MR   tan   M G r 1   tan Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)
30. 30. Belt drives Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)
31. 31. Belt drives   F x  0  T cos   F   T  T  cos 0 2 2    Fy  0  T sin 2   N  T  T  sin 2  0  F  s  N   T cos   F   T  T  cos 0 2 2     T cos   F  T cos  T cos 0 2 2 2      F  T cos  0  T cos   F  s  N 2 2   T sin   N   T  T  sin 0 2 2     T sin   N  T sin  T sin 0 2 2 2    2T sin   N  T sin 0 2 2    2T sin   N  0  2T sin  N 2 2 Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)
32. 32. Belt drives  T cos  s  N 2  2T sin  N 2    T cos  2T  s sin 2 2  sin 1 T  2  cos  s 2T  2   sin 1 T  2 Lim cos  Lim    0 ,T  0 2T  2  0 ,T 0 s  1 dT   1 s 2T d 2  dT   d T s Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)
33. 33. Belt drives Belt is just about to slide to the right dT   d T s T2 dT           s d T2 e s T e s T T T 10 2 1 T2  T1   lnT    T1    s 0   Torque required to drive the pulley  lnT  lnT2 1  s      T2  T1   e s  1 T  ln T    2    1  T 1 s T  2  es T 1 Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)
34. 34. Belt drives : Important points T  2  es T 1  Angle b must be expressed in radians  Smallest mb determines which pulley slips first  Larger tension occurs at that end of the belt where relative motion is about to begin or is already moving  T2 is used to denote the larger tension  A freely rotating pulley implies no friction  For a rotating pulley where slipping is about to start friction is ms since relative velocity between belt and pulley is zero.  Once slipping starts friction coefficient is dynamic or kinetic i.e. mk  If pulley does not rotate at all then rope has to slide and not slip, hence friction is mk Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)
35. 35. Belt drives A B q q A B q q Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)
36. 36. Belt drives T2 T2  exp      2   ???   T1 A Always check for mb value for each pulley in a system. The one with the T1 smallest mb q value will determine the Which expression is tensions. correct???? Is T2>T1 Or T1>T2. One pulley must slip. T2 Friction force is larger for the larger pulley since angle of wrap is larger. Hence smaller pulley B slips and determines the tension T1 q  exp      2   ???   T1 T2 Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)
37. 37. Band brakes Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)
38. 38. Band brakes TB T1 T3 B T1 T3  75  T2  T1 exp       T1 exp  0.25 T 75 4   T2  180   180  T2  1.39T1  135   135  T3  T4 exp      T4 exp  0.25   180   180  T4  0.55T3 M 0    T3  T4  T2  T1  R Consider the pin B F x  0  T1 cos 45  T3 cos 45  T1  T3 F y  0  T1 cos 45  T3 cos 45  TB  2T1 cos 45  TB  2T1  TB  T2  1.39T1 ,T3  T1 ,T4  0.55T1 Thus the largest tension is T2  5.6  T1  T3  4.03,T4  2.22,TB  5.7  M 0   5.6  2.22  R  3.38  0.16  0.54 KNm  540 Nm Taking moments about D 50TB  250 P  P  0.2TB  1.14KN Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)