Consider the wedge used to

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• Consider the wedge used to
• lift a block of weight W by
• applying a force P to the wedge

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• FBD of the block and the wedge
• Exclude the weight of the wedge since it is small
  compared to weight of the block

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• Frictional forces F1 and F2 must oppose the
  motion of the wedge
• Frictional force F3 of the wall on the block must
  act downward as to oppose the blocks upward
  motion
• Location of the resultant forces are not
  important since neither the block or the wedge
  will tip
• Moment equilibrium equations not considered
• 7 unknowns - 6 normal and frictional force and
  force P

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• 2 force equilibrium equations (?Fx 0, ?Fy 0)
  applied to the wedge and block (4 equations in
  total) and the frictional equation (F µN)
  applied at each surface of the contact (3
  equations in total)
• If the block is lowered, the frictional forces
  will act in a sense opposite to that shown
• Applied force P will act to the right if the
  coefficient of friction is small or the wedge
  angle ? is large

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• Otherwise, P may have the reverse sense of
  direction in order to pull the wedge to remove it
• If P is not applied or P 0, and friction forces
  hold the block in place, then the wedge is
  referred to as self-locking

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• Example 8.7
• The uniform stone has a mass of 500kg and is held
  
• in place in the horizontal position using a wedge
  at
• B. if the coefficient of static friction µs
  0.3, at the
• surfaces of contact, determine the minimum force
• P needed to remove the wedge. Is the wedge self-
• locking? Assume that the stone does not slip at A.

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• Solution
• Minimum force P requires F µs NA at the
  surfaces of contact with the wedge
• FBD of the stone and the wedge
• On the wedge, friction force opposes the motion
  and on the stone at A, FA µsNA, slipping does
  not occur

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• Solution
• 5 unknowns, 3 equilibrium equations for the stone
  and 2 for the wedge

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• Solution
• Since P is positive, the wedge must be pulled out
• If P is zero, the wedge would remain in place
  (self-locking) and the frictional forces
  developed at B and C would satisfy
• FB lt µsNB
• FC lt µsNC