Note on Newton's Laws of Motion
Newton's Laws
Introduction to Newton's Laws
- Date: 11 February 2026
- Key Concept: Newton's laws of motion describe the relationship between a body and the forces acting upon it, and its motion in response to those forces.
Newton’s First Law of Motion (Law of Inertia)
- Definition: An object will remain at rest or continue moving at a constant velocity unless acted upon by a non-zero net force.
- This suggests that if an object is not in motion, it will not start moving on its own. Similarly, if it is in motion, it won't stop or change direction unless a net force is applied.
Newton’s Second Law of Motion
- Statement: The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.
- Mathematical Representation:
F = ma
where: - F = net force
- m = mass of the object
- a = acceleration
Problem Example for Second Law
- Given:
- Mass (m) = 20 kg
- Force applied at angle θ = 30°
- Velocity (V) = 2 m/s
Force Calculation
- The sum of all forces acting on the object can be calculated as follows:
- General form:
F{net} = F{applied} + (-F{friction}) + (-F{gravity}) \ - Components of forces:
F{x} = F\cos(θ)
F{y} = F\sin(θ) - Results from forces acting on the slope (with friction):
F_{net} + (-18) + (20)(9.8)\sin(30°) = 2060
Resultant Force Calculation
- For calculations along the slope:
- Setting the forces equal to mass times acceleration gives:
F - 116 = 0 \ \ F = 116N \, \text{(up the slope)}
Newton’s Third Law of Motion (Action-Reaction Law)
- Definition: For every action, there is an equal and opposite reaction.
- This means that forces always occur in pairs; when one body exerts a force on another, the second body exerts a force of equal magnitude and in the opposite direction on the first body.
Classwork Corrections and Real-Life Application
- Include examples and calculations from classwork to demonstrate the application of Newton's laws in real-world scenarios.
Example of Force Application Problem
- Scenario: An object with a mass of 50 kg and an applied force of 180 N to the left and a resultant force of 300 N at an angle of 20° to the right.
- To find acceleration:
- F{net} = F{1} + (F_{2})
- Solve for acceleration using laws:
ma = F{app} - F{friction} \ \ 180 = 50a
- Rearranging gives:
- a = \frac{180}{50} = 3.6 \, m/s^2
Force Normal Calculation in Inclined Plane Problems
- The sum of vertical forces:
- F{N} + F{y} - F_{g} = 0
- Therefore:
- F{N} + F{applied} \sin(θ) - mg = 0
- Example Calculation: If F{applied} = 300 N at an angle of 20° and mass (m) = 50 kg:
- Calculate:
- F_{N} = 387.31 N \text{ (upward force)}