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A machine vice whose length of the handle is 150mm and the coefficient of friction for thread and collar are 0.15 and 0.17 respectively has a force applied at handle of 125N. Also the outer and inner diameters of collar are 55mm and 45mm respectively. Find the clamping force W if nominal diameter=22mm and pitch=5mm.
  • a)
    4539.45N
  • b)
    2868.73N
  • c)
    3657.56N
  • d)
    2134.34N
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A machine vice whose length of the handle is 150mm and the coefficient...
Explanation: Net torque=M₁+M₂ or 125 x 150=2.286W + 4.25W or W=2868.73N.
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Most Upvoted Answer
A machine vice whose length of the handle is 150mm and the coefficient...
Given data:
Length of the handle (L) = 150mm
Coefficient of friction for thread (μt) = 0.15
Coefficient of friction for collar (μc) = 0.17
Force applied at handle (F) = 125N
Outer diameter of collar (D0) = 55mm
Inner diameter of collar (D1) = 45mm
Nominal diameter (d) = 22mm
Pitch (p) = 5mm

Formula used:
Clamping force (W) = [(2πμtL)/(p)]F + [(π/4)(D0² - D1²)μc]F

Calculation:
[(2πμtL)/(p)]F = [(2π×0.15×150)/(5)]×125 = 1413.72N
[(π/4)(D0² - D1²)μc]F = [(π/4)(55² - 45²)×0.17]×125 = 1455.01N
Clamping force (W) = 1413.72 + 1455.01 = 2868.73N

Therefore, the clamping force W is 2868.73N, which is option B.
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A machine vice whose length of the handle is 150mm and the coefficient of friction for thread and collar are 0.15 and 0.17 respectively has a force applied at handle of 125N. Also the outer and inner diameters of collar are 55mm and 45mm respectively. Find the clamping force W if nominal diameter=22mm and pitch=5mm.a)4539.45Nb)2868.73Nc)3657.56Nd)2134.34NCorrect answer is option 'B'. Can you explain this answer?
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