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# lec4

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LEC 4
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 So: L 2 a  = tan α 2The resultant of the normal force and frictional force act directly ”below” thecenter of mass.EXAMPLEQ: For what range of   W  0  is the block in equilibrium?FBD   Case 1: Impending motion is down the plane.   F  x  = 0 T  1  + F  1 − W   sin α  = 0 F  y  = 0 N  1 − W   cos α  = 04  Case 2: Impending motion is up the plane.  F  x  = 0 T  2  + F  2 − W   sin α  = 0  F  y  = 0 N  2 − cos α  = 0What about  T  ?FBD of CableLook at diﬀerential element5         F  x  = 0 dθ dθT  ( θ )cos  − T  ( θ  + dθ )cos = 02 2 T  ( θ ) =  T  ( θ  + dθ ) =  T F  y  = 0 dθ dθdN   − T  ( θ )sin  − T  ( θ  + dθ )sin = 02 2 dθ dθdN   − T  ( θ )  − T  ( θ  + dθ ) = 02 2 Tdθ  =  dN  So: T   =  W  0 Back to block: T  1  =  T  2  =  W  0 N  1  =  N  2  =  N   =  W   cos α For case 1: F  1  =  µ s N   =  µ s W   cos αµ s W   cos α + W  0  + W   sin α  = 0 W  0  =  W   sin α − µ s W   cos α The block will be stable against downward motion when: W  0  =  W   sin α − µ s W   cos α For case 2: F  2  =  µ s N   =  µ s W   cos αµ s W   cos α + W  0  + W   sin α  = 0 W  0  =  W   sin α − µ s W   cos α 6        The block will be stable against downward motion when:   W  0  ≤ W   sin α + µ s W   cos α   So it is stable when:   W  (sin α − µ s  cos α ) ≤ W  0  ≤  W  (sin α + µ s  cos α )   What about pulley with friction?Look at a diﬀerential element.Recall a rope around a rod. F  x  = 0 T  ( θ )cos  dθ 2  − T  ( θ  + dθ )cos  dθ 2  − dF   = 0  F  y  = 0     dθ dθdN   − T  ( θ )sin  − T  ( θ  + dθ )sin  − dF   = 02 2 dθ dθ sin  ≈ 2 2 dθ cos  ≈ 12 dT   =  T  ( θ  + dθ ) − T  ( θ ) ⇒ T  ( θ  + dθ ) =  T  ( θ ) + dT   =  T   + dT  So: T  ( θ ) − T  ( θ  + dθ ) − dF   = 0 dT   = − dF Tdθ dθdN   −  + ( T   + dT  ) = 02 2 T   + dT   → 0 dN   − Tdθ  = 0With impending motion: dF   =  µ s dN dT   = − µ s dN
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