Physics Laws of Motion and Friction

Forces of Friction

• Friction opposes relative motion between two contact systems.
• Two main types of friction:
  • Static friction (fs): Occurs when systems are stationary relative to each other.
  • Kinetic friction (fk): Occurs when systems are in motion relative to each other.
• Direction of friction is opposite to intended motion, whether moving or stationary.

Magnitude of Friction

• Magnitude of friction is proportional to the normal force (N).
• Static friction formula: Ff = fs ≤ μsN, where μs is the coefficient of static friction.
• Kinetic friction formula: Ff = fk = μkN, where μk is the coefficient of kinetic friction.
• Coefficients of friction are generally independent of the area of contact.

Static Friction

• Static friction increases with the applied force (F) until it reaches its maximum value, when surfaces are about to slip.
• Maximum static friction is represented by the equation: fs ≤ μsN.

Kinetic Friction

• Kinetic friction acts when an object is already in motion.
• The force of kinetic friction neglects speed variations: fk = μkN.

Example 1: Crate on the Floor

• Given a 20.0 kg crate, static friction coefficient μs = 0.700, and kinetic friction coefficient μk = 0.600.
• Maximum static friction force: fs = 0.700 × (20.0 kg × 9.81 m/s²) = 137 N.
• If the applied force (P) is less than 137 N, static friction keeps the crate stationary.
• For P values:
  • P = 20.0 N: fs = 20.0 N.
  • P = 30.0 N: fs = 30.0 N.
  • P = 120.0 N: fs = 120.0 N.
  • P = 180.0 N: fk = 0.600 × (20.0 kg × 9.81 m/s²) = 118 N, resulting in acceleration: a = (180.0 N - 118 N)/20.0 kg = 3.10 m/s².

Example 2: Pulling a Crate

• Moving a crate with a rope at a 30° angle.
• To maintain constant velocity, T cos(30) = fk.
• Normal force: n = W - T sin(30).
• Static relationship yields: fk = μk(W - T sin(30)).
• Final tension calculation: T = μkW / (cos(30) + μk sin(30)).
• Resulting forces: Normal force n = 406 N and kinetic friction fk = 162.4 N.

Example 3: Block on a Ramp

• Block with mass 2.50 kg and static friction coefficient μs = 0.350 resting on a ramp.
• Maximum angle (θ) before slipping occurs determined using: tan(θ) = μs.
• Results in θ = tan⁻¹(0.350) = 19.3°.

Example 4: Snowboarder on a Slope

• Snowboarder glides down an incline at θ = 13° with μk = 0.20.
• Applying Newton’s second law results in: ax = g(sin θ - μk cos θ).
• Acceleration of the snowboarder calculated as: ax = 0.29 m/s².

Summary of Key Concepts

• The balance of forces along both x and y axes is essential to analyze motion and friction.
• Coefficients of friction play a critical role in determining the forces required to initiate and maintain motion.
• Different scenarios highlight the interplay between static and kinetic friction, showcasing practical applications of the laws of motion.