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. D FORMULAS: tand=u; M = Rrsin(0); M = rs=rW tan(±0), 0 = tan L -=tan 2m² np 2m²
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. D FORMULAS: tand=u; M = Rrsin(0); M = rs=rW tan(±0), 0 = tan L -=tan 2m² np 2m²
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![**Problem Statement:**
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 \( C = 1.35 \, \text{N} \cdot \text{m} \) is used to tighten the clamp, determine:
(a) the clamping force, and
(b) the torque required to loosen the clamp.
**Diagram Explanation:**
The diagram shows a C-clamp with a screw. The clamp is depicted with a threaded screw that tightens or loosens to exert force. The labeled part at the handle indicates where the torque is applied.
**Formulas Provided:**
\[
\begin{align*}
\tan \phi &= \mu \\
M &= Rr \sin(\phi) \\
M &= rs = rW \tan(\phi \pm \theta) \\
\theta &= \tan^{-1}\left(\frac{L}{2\pi m}\right) = \tan^{-1}\left(\frac{np}{2\pi}\right)
\end{align*}
\]
Where:
- \(\phi\) is the angle of friction.
- \(M\) represents moments related to force and geometry.
- \(R\) and \(r\) are linked to the geometry of the setup.
- \(W\) is the clamping force needed.
- \(\theta\) is related to the screw's thread angle, pitch \(p\), and lead \(L\).
- \(\mu\) is the coefficient of friction.
Use the formulas and given data to solve for the clamping force and the required torque to loosen the clamp.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1e4fcbe0-ed74-4447-b38a-a2af13f9fabc%2Fc00d1297-0f18-48b8-893b-490e68fc7563%2Ftu99gvl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
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 \( C = 1.35 \, \text{N} \cdot \text{m} \) is used to tighten the clamp, determine:
(a) the clamping force, and
(b) the torque required to loosen the clamp.
**Diagram Explanation:**
The diagram shows a C-clamp with a screw. The clamp is depicted with a threaded screw that tightens or loosens to exert force. The labeled part at the handle indicates where the torque is applied.
**Formulas Provided:**
\[
\begin{align*}
\tan \phi &= \mu \\
M &= Rr \sin(\phi) \\
M &= rs = rW \tan(\phi \pm \theta) \\
\theta &= \tan^{-1}\left(\frac{L}{2\pi m}\right) = \tan^{-1}\left(\frac{np}{2\pi}\right)
\end{align*}
\]
Where:
- \(\phi\) is the angle of friction.
- \(M\) represents moments related to force and geometry.
- \(R\) and \(r\) are linked to the geometry of the setup.
- \(W\) is the clamping force needed.
- \(\theta\) is related to the screw's thread angle, pitch \(p\), and lead \(L\).
- \(\mu\) is the coefficient of friction.
Use the formulas and given data to solve for the clamping force and the required torque to loosen the clamp.
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