Transcript ลิ่ม - Suranaree University of Technology

ลิม่
• เป็ นเครื่ องกลอย่างง่ายที่แปลงแรงให้เป็ นแรงขนาด
ใหญ่ข้ ึนในแนวตั้งฉาก
• 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
• Exclude the weight of the wedge since it is
small compared to weight of the block
• 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
• 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
• 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
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 selflocking? Assume that the stone does not slip at A.
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
Solution
• 5 unknowns, 3 equilibrium equations for the
stone and 2 for the wedge
 M A  0;
 4905N (0.5m)  ( N B cos7  N )(1m)  (0.3N B sin 7  N )(1m)  0
   Fx  0;
2383.1sin 7   0.3(2383.1cos7  )  P  0.3N C  0
   Fy  0;
N C  2383.1cos7 N  0.3(2383.1sin 7 )  0

N C  2452.5 N , P  1154.9 N  1.15kN

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