5. The screw has a mean radius of 5 mm and a lead or pitch of 1.85 mm. The coefficient of static friction between the threads is 0.25. If the torque C = 1.35 Nm is used to tighten the clamp, determine (a) the clamping force, and (b) the torque required to loosen the clamp. FORMULAS: tand=u; M = Rrsin(); M = rs =rW tan(±0), 0= tan L 2.m = tan np 2m
5. The screw has a mean radius of 5 mm and a lead or pitch of 1.85 mm. The coefficient of static friction between the threads is 0.25. If the torque C = 1.35 Nm is used to tighten the clamp, determine (a) the clamping force, and (b) the torque required to loosen the clamp. FORMULAS: tand=u; M = Rrsin(); M = rs =rW tan(±0), 0= tan L 2.m = tan np 2m
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Transcribed Image Text:**Problem 5:**
A screw has a mean radius of 5 mm and a lead (or pitch) of 1.85 mm. The coefficient of static friction between the threads is 0.25. If the torque C = 1.35 N·m is used to tighten the clamp, determine:
(a) The clamping force.
(b) The torque required to loosen the clamp.
**Diagram Explanation:**
The image includes a diagram of a C-clamp. The screw mechanism is shown in detail, illustrating how the torque is applied to tighten or loosen the clamp.
**Formulas Provided:**
The following formulas are given for calculations:
1. \( \tan \phi = \mu \)
2. \( M = Rr \sin(\phi) \)
3. \( M = rs = rW \tan(\phi \pm \theta) \)
4. \( \theta = \tan^{-1} \left(\frac{L}{2\pi m}\right) = \tan^{-1} \left(\frac{np}{2\pi}\right) \)
These equations can be used to calculate the clamping force and the torque required to loosen the clamp using the specified parameters.
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