Lecture Notes: Static and Dynamic Friction Problem

Introduction

• Discussed a problem involving a box, originally dynamic, now static due to an increase in horizontal force from 20N to 40N.
• Goal: Find various forces and angles.

Key Concepts

• Normal Force: Reaction force perpendicular to the surface.
• Incline Forces: Forces acting along the incline, both up and down.
• Friction Force: Maximum friction and actual friction required to maintain static equilibrium.

Calculating Forces

Normal Force

• Components:
  • Perpendicular component of weight: (100 \text{N} \times \cos(30^\circ))
  • Perpendicular component of applied force: (40 \text{N} \times \sin(30^\circ))
• Calculation:
  • (86.6 \text{N} + 20 \text{N} = 106.6 \text{N})

Maximum Friction Force

• Formula: (\text{Normal Force} \times \text{Coefficient of Friction})
• Calculation:
  • (106.6 \text{N} \times 0.2 = 21.32 \text{N})

Force Up the Incline

• Component: (F \times \cos(30^\circ))
• Calculation:
  • (40 \text{N} \times 0.866 = 34.64 \text{N})

Force Down the Incline

• Component: Weight (\times \sin(30^\circ))
• Calculation:
  • (100 \text{N} \times 0.5 = 50 \text{N})

Friction Force Required

• (\text{Force down incline} - \text{Force up incline} = 15.36 \text{N})
• Condition: Static situation as maximum friction ((21.32 \text{N})) is greater than required friction.

Angle Calculations

Resultant Force Angles

• Angle (\phi): Between normal and friction force
  • Calculation using arctangent of coefficient of friction:
    • (\tan^{-1}(0.2) = 11.31^\circ)

Angle (\alpha)

• Between resultant force and vertical:
  • (30^\circ - \phi = 18.7^\circ)

Additional Triangle Analysis

• Triangle Components:
  • Weight: 100N
  • Applied Force: 40N
  • Reaction Force
• Angle (\beta): Using actual and maximum friction force
  • (\tan^{-1}(0.4) = 21.8^\circ)

Conclusion

• Discussed difference between angles (\alpha) and (\beta) due to maximum vs actual friction forces.
• Verified static situation with all forces and angles calculated correctly.