Lecture on Static and Kinetic Friction

Introduction

• Focus on static friction and kinetic friction.
• Example: Pushing a box on a carpet floor.
  • Initial resistance to motion due to static friction.
  • Easier to maintain motion once the box starts sliding due to kinetic friction.

Definitions

• Static Friction: Force preventing movement when a force is applied.
  • Equation: ( f_s \leq \mu_s \cdot N )
  • ( f_s ) is variable, up to a maximum value.
• Kinetic Friction: Force opposing motion when sliding occurs.
  • Equation: ( f_k = \mu_k \cdot N )
  • ( f_k ) is a constant value.

Example Calculations

Example 1: 5 kg Box

• Normal Force Calculation:
  • Weight ( = 5 \times 9.8 = 49 \text{ N} )
• Coefficients: ( \mu_s = 0.4, \mu_k = 0.2 )
• Frictional Forces:
  • ( f_k = 0.2 \times 49 = 9.8 \text{ N} )
  • ( f_s \leq 0.4 \times 49 = 19.6 \text{ N} )

Force and Friction Table

• Applied Force:
  • 10 N < 19.6 N (static); box does not move.
  • Exceeds 19.6 N; kinetic friction applies, ( f_k = 9.8 \text{ N} ).

Example 2: 15 kg Box

• Minimum Force for Sliding: ( F = f_s = \mu_s \cdot mg )
  • ( F = 0.35 \times 15 \times 9.8 = 51.45 \text{ N} )
• Acceleration Calculation:
  • ( 90 \text{ N} ) applied: ( a = \frac{F - f_k}{m} )
  • ( f_k = 0.2 \times 15 \times 9.8 = 29.4 \text{ N} )
  • Acceleration: ( a = 4.04 \text{ m/s}^2 )

Example 3: 65 N Force for an 8 kg Box

• Static Friction Coefficient:
  • ( \mu_s = \frac{65}{8 \times 9.8} = 0.829 )
• Kinetic Friction Coefficient:
  • Acceleration: 1.4 m/s²
  • Solve for ( \mu_k ) using net force expression.
  • ( \mu_k = 0.686 )

Example 4: 150 N Pull on a 30 kg Box

• Normal Force Considerations:
  • Normal force changes due to vertical component of applied force.
  • ( F_N = mg - F_y )
• Acceleration Calculation:
  • Solve net force equation in x-direction.
  • ( a = 2.505 \text{ m/s}^2 )

Key Takeaways

• Static friction prevents initial motion; kinetic friction opposes motion when sliding.
• Static friction ( \mu_s ) is generally greater than kinetic friction ( \mu_k ).
• Normal force can change based on vertical components of applied force.