Physics Laws of Motion and Friction

Study Notes

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.

Description

This quiz explores the application of the laws of motion, focusing specifically on the forces of friction. It covers the concepts of static and kinetic friction, as well as how frictional forces oppose motion on surfaces. Test your understanding of these fundamental physics principles.