6161103 8.3 wedges

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6161103 8.3 wedges

  1. 1. 8.3 WedgesA simple machine used to transform an applied forceinto much larger forces, directed at approximatelyright angles to the applied forceUsed to give small displacements or adjustments toheavy loadConsider the wedge used tolift a block of weight W byapplying a force P to the wedge
  2. 2. 8.3 WedgesFBD of the block and the wedgeExclude the weight of the wedge since it is smallcompared to weight of the block
  3. 3. 8.3 WedgesFrictional forces F1 and F2 must oppose themotion of the wedgeFrictional force F3 of the wall on the block mustact downward as to oppose the block’s upwardmotionLocation of the resultant forces are not importantsince neither the block or the wedge will tipMoment equilibrium equations not considered7 unknowns - 6 normal and frictional force andforce P
  4. 4. 8.3 Wedges2 force equilibrium equations (∑Fx = 0, ∑Fy = 0)applied to the wedge and block (4 equations intotal) and the frictional equation (F = µN)applied at each surface of the contact (3equations in total)If the block is lowered, the frictional forces willact in a sense opposite to that shownApplied force P will act to the right if thecoefficient of friction is small or the wedge angleθ is large
  5. 5. 8.3 WedgesOtherwise, P may have the reversesense of direction in order to pull thewedge to remove itIf P is not applied or P = 0, and frictionforces hold the block in place, then thewedge is referred to as self-locking
  6. 6. 8.3 WedgesExample 8.7The uniform stone has a mass of 500kg and is heldin place in the horizontal position using a wedge atB. if the coefficient of static friction µs = 0.3, at thesurfaces of contact, determine the minimum forceP needed to remove the wedge. Is the wedge self-locking? Assume that the stone does not slip at A.
  7. 7. 8.3 WedgesSolution 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
  8. 8. 8.3 WedgesSolution 5 unknowns, 3 equilibrium equations for the stone and 2 for the wedge ∑ M A = 0; − 4905 N (0.5m) + ( N B cos 7 o N )(1m) + (0.3N B sin 7 o N )(1m) = 0 + → ∑ Fx = 0; 2383.1sin 7 o − 0.3(2383.1 cos 7 o ) + P − 0.3 N C = 0 + ↑ ∑ Fy = 0; N C − 2383.1 cos 7 o N − 0.3(2383.1sin 7 o ) = 0 N C = 2452.5 N , P = 1154.9 N = 1.15kN
  9. 9. 8.3 WedgesSolution 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 < µsNB FC < µsNC

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