Newton’s Laws of Motion form the foundation of classical mechanics. These three laws describe the relationship between a body and the forces acting upon it, and its motion in response to those forces.
Overview
graph TD
A[Newton's Laws] --> B[First Law
Law of Inertia]
A --> C[Second Law
F = ma]
A --> D[Third Law
Action-Reaction]
B --> B1[Object at rest stays at rest]
B --> B2[Object in motion stays in motion]
C --> C1[Force causes acceleration]
C --> C2[a ∝ F, a ∝ 1/m]
D --> D1[Equal and opposite forces]
D --> D2[Act on different bodies]First Law: Law of Inertia
An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.
Key Concepts
- Inertia: The tendency of an object to resist changes in its state of motion
- Mass: A measure of inertia (more mass = more inertia)
- Reference Frames: The law is valid only in inertial (non-accelerating) reference frames
Mathematical Representation
If the net force $\vec{F}_{net} = 0$, then:
$$\vec{v} = \text{constant}$$This means either the object is at rest ($\vec{v} = 0$) or moving with constant velocity.
Second Law: Law of Acceleration
The rate of change of momentum of a body is directly proportional to the force applied and takes place in the direction of the force.
The Equation
$$\vec{F} = m\vec{a} = \frac{d\vec{p}}{dt}$$Where:
- $\vec{F}$ = Net force (in Newtons, N)
- $m$ = Mass (in kilograms, kg)
- $\vec{a}$ = Acceleration (in m/s²)
- $\vec{p} = m\vec{v}$ = Momentum
Component Form
In three dimensions:
$$F_x = ma_x, \quad F_y = ma_y, \quad F_z = ma_z$$Example Problem
A block of mass 5 kg is pushed with a force of 20 N on a frictionless surface. Find the acceleration.
Solution:
Given: $m = 5$ kg, $F = 20$ N
Using $F = ma$:
$$a = \frac{F}{m} = \frac{20}{5} = 4 \text{ m/s}^2$$Interactive Demo: Visualize Free Body Diagrams
Explore how to draw and analyze free body diagrams for various force scenarios.
Third Law: Action and Reaction
For every action, there is an equal and opposite reaction.
Key Points
| Property | Action Force | Reaction Force |
|---|---|---|
| Magnitude | F | F |
| Direction | One way | Opposite way |
| Acts on | Body A | Body B |
| Type | Same | Same |
Visualization
graph LR
subgraph "Person pushing wall"
A[Person] -->|F| B[Wall]
B -->|F| A
endApplications in JEE Problems
Connected Bodies
For two blocks connected by a string:
graph LR
A[Block 1
m₁] --- |Tension T| B[Block 2
m₂]
F[Force F] --> ASystem approach:
$$a = \frac{F}{m_1 + m_2}$$Individual approach:
- For Block 1: $F - T = m_1 a$
- For Block 2: $T = m_2 a$
Pulley Systems
For an Atwood machine with masses $m_1$ and $m_2$ (where $m_1 > m_2$):
$$a = \frac{(m_1 - m_2)g}{m_1 + m_2}$$ $$T = \frac{2m_1 m_2 g}{m_1 + m_2}$$Interactive Force Diagram
Free Body Diagram Builder
Below is an example of how you might embed an interactive element. In your actual content, you can include custom HTML, JavaScript, or embed iframes for simulations.
Free Body Diagram of a block on a surface with applied force
Summary
| Law | Statement | Equation |
|---|---|---|
| First | Objects maintain their state of motion | $\vec{F}_{net} = 0 \Rightarrow \vec{v} = \text{const}$ |
| Second | Force causes acceleration | $\vec{F} = m\vec{a}$ |
| Third | Action equals reaction | $\vec{F}_{12} = -\vec{F}_{21}$ |
Practice Problems
A 10 kg block is pulled with a force of 50 N at an angle of 30° to the horizontal. If $\mu = 0.2$, find the acceleration.
Two blocks of masses 3 kg and 5 kg are connected by a string over a frictionless pulley. Find the acceleration and tension.
A person weighing 60 kg stands in a lift accelerating upward at 2 m/s². Find the normal force.