![Problem 1 y
N cos60 − mg + f sin60 + P sin β = 0
− N sin60 + f cos 60 + P cos β = 0
f = µN x
mg sin60 − f
⇒P=
sin β sin60 + cos β cos 60
N cos60 − mg + µ N sin60 + P sin β = 0
mg − P sin β
⇒N=
cos 60 − µ sin60
− N sin60 + µ N cos 60 + P cos β = 0
mg − P sin β
⇒− ( sin60 − µ cos 60 ) + P cos β =0
cos60 − µ sin60
⇒ − ( mg − P sin β )( sin60 − µ cos 60 ) + P cos β ( cos 60 − µ sin60 ) =
0
⇒ − mg ( sin60 − µ cos 60 ) + P sin β ( sin60 − µ cos 60 ) + P cos β ( cos 60 − µ sin60 ) =
0
⇒ P ( sin60 sin β − µ cos 60 sin β + cos 60 cos β − µ sin60 cos β ) = ( sin60 − µ cos 60 )
mg
( sin60 − µ cos 60 )
⇒P=
mg
cos ( 60 − β ) − µ sin ( 60 − β )
1 cos ( 60 − β ) − µ sin ( 60 − β )
⇒ =mg
P ( sin60 − µ cos 60 )
d 1 d
=⇒0 cos ( 60 − β ) − µ sin ( 60 − β ) =
dβ P dβ 0
d
⇒ [ sin60 sin β − µ cos 60 sin β + cos 60 cos β − µ sin60 cos β ] =
0
dβ
⇒ sin60 cos β − µ cos 60 cos β − cos 60 sin β + µ sin60 sin β = 0
sin60 − µ cos 60
β
⇒ tan = β
= 2.614 ⇒ = 69 o
cos 60 − µ sin60
( sin 60 − µ cos 60 ) ( 0.866 − 0.125 )
P = mg = mg = 0.72mg
cos ( 60 − β ) − µ sin ( 60 − β ) cos ( −9 ) − µ sin ( −9 )](https://image.slidesharecdn.com/friction-100415053918-phpapp01/85/Friction-5-320.jpg)
Friction
- 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
- 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
- 3. The correct way of writing the dry friction force V −µ N Fr = =ˆ −µ N V V N=Normal force vector V=Relative velocity vector of the body µ= coefficient of dry friction or Coulomb friction
- 4. Problem 1 Knowing that the coefficient of friction between the 13.5 kg block and the incline is µs = 0.25, determine a) the smallest value of P required to maintain the block in equilibrium, b) the corresponding value of β.
- 5. Problem 1 y N cos60 − mg + f sin60 + P sin β = 0 − N sin60 + f cos 60 + P cos β = 0 f = µN x mg sin60 − f ⇒P= sin β sin60 + cos β cos 60 N cos60 − mg + µ N sin60 + P sin β = 0 mg − P sin β ⇒N= cos 60 − µ sin60 − N sin60 + µ N cos 60 + P cos β = 0 mg − P sin β ⇒− ( sin60 − µ cos 60 ) + P cos β =0 cos60 − µ sin60 ⇒ − ( mg − P sin β )( sin60 − µ cos 60 ) + P cos β ( cos 60 − µ sin60 ) = 0 ⇒ − mg ( sin60 − µ cos 60 ) + P sin β ( sin60 − µ cos 60 ) + P cos β ( cos 60 − µ sin60 ) = 0 ⇒ P ( sin60 sin β − µ cos 60 sin β + cos 60 cos β − µ sin60 cos β ) = ( sin60 − µ cos 60 ) mg ( sin60 − µ cos 60 ) ⇒P= mg cos ( 60 − β ) − µ sin ( 60 − β ) 1 cos ( 60 − β ) − µ sin ( 60 − β ) ⇒ =mg P ( sin60 − µ cos 60 ) d 1 d =⇒0 cos ( 60 − β ) − µ sin ( 60 − β ) = dβ P dβ 0 d ⇒ [ sin60 sin β − µ cos 60 sin β + cos 60 cos β − µ sin60 cos β ] = 0 dβ ⇒ sin60 cos β − µ cos 60 cos β − cos 60 sin β + µ sin60 sin β = 0 sin60 − µ cos 60 β ⇒ tan = β = 2.614 ⇒ = 69 o cos 60 − µ sin60 ( sin 60 − µ cos 60 ) ( 0.866 − 0.125 ) P = mg = mg = 0.72mg cos ( 60 − β ) − µ sin ( 60 − β ) cos ( −9 ) − µ sin ( −9 )
- 6. Problem 2 Knowing that P = 110 N, determine the range of values value of θ for which equilibrium of the 8 kg block is maintained.
- 7. Problem 2 y y x x mg mg − P cos θ + N =0 − P cos θ + N =0 P sinθ − mg − f = 0 P sinθ − mg + f = 0 f = µN f = µN ⇒N= θ P cos ⇒N= θ P cos P sinθ − mg − µ N = 0 P sinθ − mg + µ N = 0 ⇒ P sinθ − mg − µ P cos θ = 0 ⇒ P sinθ − mg + µ P cos θ = 0 ⇒ P ( sinθ − µ cos θ ) = mg ⇒ P ( sinθ + µ cos θ ) = mg mg mg ⇒P= ⇒P= sinθ − µ cos θ sinθ + µ cos θ Hence Hence upward movement will not start before downward movement will not start before mg mg P= P= sinθ − µ s cos θ sinθ + µ k cos θ
- 8. Problem 3 The coefficients of friction are µs = 0.40 and µk = 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
- 9. Problem 3 v T N1 f1 m1g f1 N1 T m2g f2 N2 T − µ N1 = 0 N 1 − m1 g = 0 = µ N= m1 g ⇒T 1 , N1 µm ⇒ T = 1g T + µ N1 + µ N2 − P = 0 N 2 − N 1 − m2 g = 0 N 2 − N 1 − m2 g = ⇒ N 2 =m1 + m2 ) g 0 ( T + µ N1 + µ N2 − P = 0 ⇒ µ m1 g + µ m1 g + µ ( m1 + m 2 ) g − P = 0 ⇒ P 3 µ m1 g + µ m 2 g =
- 10. Problem 3 v 0 N1 f1 m1g f1 N1 0 m2g f2 N2 T − µ N1 = m1 a Now T = 0 ⇒ − µ N 1 = m1 a N 1 − m1 g = 0 ⇒ N1 = m1 g µ N1 + µ N2 − P = 0 N 2 − N 1 − m2 g = 0 N 2 − N 1 − m2 g = ⇒ N 2 =m1 + m2 ) g 0 ( µ N1 + µ N2 − P = 0 ⇒ µ m1 g + µ ( m1 + m 2 ) g − P = 0 ⇒ P 2 µ m1 g + µ m 2 g =
- 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 θ for which motion is impending.
- 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.
- 13. Slip or topple? mg f1 f2 N1 N2 P cos 30 − µ N 1 − µ N 2 = 0 P sin 30 + N 1 + N 2 − mg = 0 − P cos 30 × d − mg × d + µ N 2 × 2d = 0 P cos 30 = µ N 1 + µ N 2 = µ ( N 1 + N 2 ) P sin 30 = 1 + N 2 ) + mg −( N 2µ N P cos 30 + mg = 2 µ P sin 30 = N 1 + N 2 ) + µ mg = 30 + µ mg −µ ( − P cos µ P sin 30 + P cos 30 =mg ⇒ P ( µ sin 30 + cos 30 ) =mg µ µ µ mg ⇒P= µ sin 30 + cos 30 0.5mg ⇒P= = 0.448mg 0.25 + 0.866
- 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 = 2 + mg −N 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
- 18. Wedge f=µN P θ N mg P − f cos θ + N sinθ =0 f sinθ + N cos θ − mg = 0 f = µN mg µ N sinθ + N cos θ − mg = ⇒ N = 0 µ sinθ + cos θ P − µ N cos θ + N sinθ = 0 µ cos θ − sinθ ⇒ P ( µ cos θ − sinθ= mg = )N µ sinθ + cos θ P µ − tanθ ⇒ = mg µ tanθ + 1 P tanφ − tanθ ⇒ = = tan (φ − θ ) mg tanφ tanθ + 1
- 19. A screw thread is a wedge
- 20. A screw thread is a wedge M=Pa Q=Pa/r = equivalent force W Q N f=µN
- 21. A screw thread is a wedge W Q N θ f=µN Q − f cos θ − N sinθ = 0 − f sinθ + N cos θ − W = 0 f = µN W − µ N sinθ + N cos θ − W = ⇒ N = 0 − µ sinθ + cos θ Q − µ N cos θ − N sinθ = 0 µ cos θ + sinθ ⇒ Q ( µ cos θ + sinθ= W = )N − µ sinθ + cos θ Q µ + tanθ ⇒ = W 1 − µ tanθ Pa µ + tanθ = Wr 1 − µ tanθ 1 P if tanθ =⇒ = ∞ µ W Screw locks
- 22. A screw thread is a wedge The W P N θ f=µN 1 Pa tanθ = ⇒ = ∞ Screw locks µ Wr 1 Wr µ = = = 0 tanθ 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 µ becomes more than critical.
- 23. A screw thread is a wedge W N f=µN
- 24. A screw thread is a wedge W N θ f For no movement f cos θ − N sinθ = 0 f sinθ + N cos θ − W = 0 Self locking f sinθ ⇒ = N cos θ sinθ ⇒= µ = tanθ cos θ Therefore after raising the load if we let go of the screw the load will not cause the screw to unscrew by itself.
- 25. Terminologies Lead (L) 2πr Pitch (p) = = Lead L np where n=no. of parallely running threads = starts L tanθ = 2π r
- 26. Turnbuckle T1 T2 Used to apply tension. The sleeve is rotated to pull the threads together. M µ + tanθ = ( T2 − T1 ) r 1 − µ tanθ
- 27. An improved screw jack W θ θ W T T T T 2T cos θ
- 28. An improved screw jack W φ φ M=Pa W=2T cos φ Pa µ cos θ + sinθ = Wr − µ sinθ + cos θ M µ cos θ + sinθ ⇒ = 2Tr cos φ − µ sinθ + cos θ
- 29. Worm gear MG M0/R R N M f MG M G = WR ⇒ W = R Pa µ + tanθ = Wr 1 − µ tanθ M µ + tanθ ⇒ = MG 1 − µ tanθ r R MR µ + tanθ ⇒ = M G r 1 − µ tanθ
- 30. Belt drives
- 31. Belt drives ∆θ ∆θ ∑F x = 0 ⇒ −T cos − ∆ F + ( T + ∆T ) cos = 0 2 2 ∆θ ∆θ ∑F y = 0 ⇒ −T sin − ∆ N − ( T + ∆T ) sin = 0 2 2 ∆ F = µs ∆ N ∆θ ∆θ −T cos − ∆ F + ( T + ∆T ) cos = 0 2 2 ∆θ ∆θ ∆θ ⇒ −T cos − ∆ F + T cos + ∆T cos =0 2 2 2 ∆θ ∆θ ⇒ − ∆ F + ∆T cos = 0 ⇒ ∆T cos = ∆ F = µs ∆ N 2 2 ∆θ ∆θ −T sin − ∆ N − ( T + ∆T ) sin = 0 2 2 ∆θ ∆θ ∆θ ⇒ −T sin + ∆ N − T sin − ∆T sin =0 2 2 2 ∆θ ∆θ ⇒ −2T sin + ∆ N − ∆T sin =0 2 2 ∆θ ∆θ ⇒ −2T sin + ∆ N = 0 ⇒ 2T sin = ∆N 2 2
- 32. Belt drives ∆θ ∆T cos = µs ∆ N 2 ∆θ 2T sin = ∆N 2 ∆θ ∆θ ⇒ ∆T cos 2T µ s sin = 2 2 ∆θ sin 1 ∆T ∆θ 2 ⇒ cos µs = 2T ∆θ 2 ∆θ ∆θ sin 1 ∆T ∆θ 2 Lim cos = Lim µ ∆θ → 0 ,∆T → 0 2T ∆θ 2 ∆θ →0 ,∆T →0 s ∆θ 1 dT µs ⇒ 1= 2T dθ 2 ⇒ dT = µ dθ T s
- 33. Belt drives Belt is just about to slide to the right dT = µ dθ T s dT β T2 µ β µ β ⇒ ∫ = ∫ µ s dθ T2 = e s ⇒T e s T = T T 1 0 2 1 T1 ⇒ lnT β T2 = µ sθ 0 T1 Torque required to drive the pulley ⇒ lnT − lnT 2 1 = µs β µ β T2 − T1 e = s − 1 T ⇒ ln T = µ β 2 1 T 1 s 2 = e µsβ T ⇒ T 1
- 34. Belt drives : Important points 2 = e µsβ T ⇒ T 1 Angle β must be expressed in radians Smallest µβ 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 µs since relative velocity between belt and pulley is zero. Once slipping starts friction coefficient is dynamic or kinetic i.e. µk If pulley does not rotate at all then rope has to slide and not slip, hence friction is µk
- 35. Belt drives A B θ θ A B θ θ
- 36. Belt drives T2 T2 = exp µ ( π + 2θ ) ??? T1 A Always check for µβ value for each pulley in a system. The one with the T1 smallest µβ θ 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 θ exp µ ( π − 2θ ) ??? T1 T2
- 37. Band brakes
- 38. Band brakes TB T1 T3 B T1 T3 75 75 4 Tπ T2 T1 exp µ π = T1 exp 0.25 180 180 T2 T2 = 1.39T1 135 135 T3 T4 exp µ π = T4 exp 0.25 π 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 = cos 45 + T3 cos 45 = 1 cos 45 = ⇒ 2T1 = 0 ⇒ T1 TB ⇒ 2T TB 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 ∴ M0 = ( 5.6 − 2.22 ) R = 3.38 × 0.16 = 0.54 KNm = 540 Nm Taking moments about D 50TB 250 P ⇒ = 0.2TB 1.14 KN = P =