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ลิ่ม

ลิ่ม. เป็นเครื่องกลอย่างง่ายที่แปลงแรงให้เป็นแรงขนาดใหญ่ขึ้นในแนวตั้งฉาก ใช้ขยับวัตถุหนัก Consider the wedge used to lift a block of weight W by applying a force P to the wedge. FBD of the block and the wedge

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ลิ่ม

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  1. ลิ่ม • เป็นเครื่องกลอย่างง่ายที่แปลงแรงให้เป็นแรงขนาดใหญ่ขึ้นในแนวตั้งฉาก • ใช้ขยับวัตถุหนัก • Consider the wedge used to lift a block of weight W by applying a force P to the wedge

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

  3. 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 block’s 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

  4. 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 θ is large

  5. 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

  6. 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.

  7. 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

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

  9. 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 < μsNB FC < μsNC

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