Lecture on Static and Kinetic Friction

Key Concepts

• Static Friction: Prevents an object from moving. Represented by an inequality ( f_s \leq \mu_s \times N ).
• Kinetic Friction: Acts when an object is in motion. Represented by ( f_k = \mu_k \times N ).
• Coefficients:
  • ( \mu_s ) (static friction coefficient) is usually greater than ( \mu_k ) (kinetic friction coefficient).
  • Typical values: ( \mu_s = 0.4, \mu_k = 0.2 ).

Calculation Examples

Example 1: 5 kg Box

• Weight Calculation: Normal force ( N = mg = 5 \times 9.8 = 49 ) N.
• Frictional Forces:
  • Static: ( f_s \leq 0.4 \times 49 = 19.6 ) N.
  • Kinetic: ( f_k = 0.2 \times 49 = 9.8 ) N.

Example 2: 15 kg Box

• Static Friction Calculation:
  • Minimum force to move box: ( F = \mu_s \times mg = 0.35 \times 15 \times 9.8 = 51.45 ) N.
• Acceleration Calculation:
  • Applied force: 90 N.
  • Net force: ( F - f_k = 90 - 0.2 \times 15 \times 9.8 ).
  • Acceleration: ( a = \frac{60.6}{15} = 4.04 \text{ m/s}^2 ).

Example 3: 8 kg Box

• Static Friction:
  • Force required to start moving: 65 N.
  • Coefficient: ( \mu_s = \frac{65}{78.4} \approx 0.829 ).
• Kinetic Friction and Acceleration:
  • Acceleration: 1.4 m/s².
  • Coefficient: ( \mu_k = \frac{53.8}{78.4} \approx 0.686 ).

Example 4: 30 kg Box (Pulled with 150 N)

• Components of Force:
  • Horizontal component ( F_x = 150 \cos(30) ).
  • Normal force calculation includes reduction by lifting component.
• Acceleration:
  • Horizontal acceleration ( a = \frac{75.15}{30} = 2.505 \text{ m/s}^2 ).

General Observations

• Static friction adjusts to match applied force until it reaches maximum value.
• Kinetic friction remains constant once the object begins to move.
• Normal force adjustments are crucial when forces act at angles, affecting friction calculations.

Practical Application

• Use these principles to determine minimum force needed to move objects, calculate frictional forces in various scenarios, and solve problems related to motion on surfaces.