Lecture Notes on Screw Mechanical Advantage

Introduction

• Discussion on practical applications of screws.
• Focus on square-threaded screws connected to clamps.
• Objective: Determine the applied force to the blocks using given parameters.

Parameters Given

• Force applied (F): 400 Newtons
• Handle distance from screw center: 10 cm (100 mm)
• Lead of screw threading: 4 mm
• Screw diameter: 10 mm (radius = 5 mm)
• Static coefficient of friction (μ): 0.3
• Angle of friction (φ): 16.7 degrees

Forces and Calculations

1. Applied Force on the Screw:

   • Relationship between force F and moment:
   • Moment arm (R) = 10 cm
   • Calculation:
     • ( F = \frac{R}{\text{Radius of screw}} \times F_P )
     • ( F = \frac{100 \text{ mm}}{5 \text{ mm}} \times 400 \text{ N} = 8000 \text{ N} )

2. Triangle of Forces:

   • Components: Weight (W), Applied force (F), and Reaction force.
   • Reaction force: Sum of friction and normal forces.
   • Angle between reaction force and vertical = ( θ + φ )

3. Calculation of Lead Angle (θ):

   • ( θ = \tan^{-1} \left( \frac{\text{Lead}}{2\pi \times \text{Radius}} \right) )
   • ( θ = \tan^{-1} \left( \frac{4}{2\pi \times 5} \right) = 7.26 \text{ degrees} )

4. Sum of Angles:

   • ( θ + φ = 7.26 + 16.7 = 23.96 \text{ degrees} )

5. Determining the Weight:

   • Use tangent of angle:
   • ( \tan(θ + φ) = \frac{F}{W} )
   • Solving for W:
     • ( W = \frac{F}{\tan(θ + φ)} = \frac{8000 \text{ N}}{\tan(23.96)} )
     • Resulting in ( W ≈ 18,000 \text{ N} )

Conclusion

• Mechanical Advantage: Small force (400 N) results in a large clamping force (18,000 N).
• Demonstrates the mechanical advantage provided by a screw.
• Part B will address the torque required to loosen the clamp in a subsequent video.