Class 9 – CBSE Science
Force & Laws of Motion
“Force is a push or a pull that changes the state of motion of an object.”
1. Simple Explanation
| Concept | What it means | Example |
|---|---|---|
| Force | Any interaction that changes an object’s speed, direction or shape. | A hand pushing a door. |
| Newton’s First Law | An object stays at rest or moves with constant velocity until a net force acts on it. | A book lying on a table—no net force, so it stays still. |
| Newton’s Second Law | (F = ma) – the net force on an object equals its mass times its acceleration. | A 5 kg box accelerated at 2 m/s² requires 10 N of net force. |
| Newton’s Third Law | For every action, there is an equal and opposite reaction. | When you jump, you push the Earth downward; Earth pushes you upward. |
2. Key Points
2.1 Force
- Definition – A vector quantity that can change motion.
- Units – Newton (N); 1 N = 1 kg·m/s².
- Types
- Contact forces – friction, tension, normal, applied.
- Non‑contact forces – gravity, magnetic, electric.
2.2 Newton’s Laws of Motion
| Law | Statement | Significance |
|---|---|---|
| 1st | Inertia – An object’s motion remains unchanged unless acted upon by an external force. | Predicts static & uniform motion. |
| 2nd | F = ma – Net force causes acceleration. | Quantifies force‑acceleration relationship. |
| 3rd | Action‑reaction – Forces come in pairs. | Explains why rockets launch, why you feel push when jumping. |
2.3 Important Concepts
- Net Force – Vector sum of all forces on an object.
- Free‑Body Diagram (FBD) – A diagram showing all forces acting on an object.
- Friction – Resistive force between surfaces; static (prevent motion) vs kinetic (act during motion).
- Weight – Gravitational force; (W = mg).
- Normal Force – Perpendicular reaction from a surface.
2.4 Common Formulas
| Quantity | Formula | Units |
|---|---|---|
| Weight | (W = mg) | N |
| Gravitational force between two masses | (F_g = G\frac{m_1m_2}{r^2}) | N |
| Net force (vector) | (\vec F_{\text{net}} = \sum \vec F_i) | N |
| Acceleration | (a = \frac{F_{\text{net}}}{m}) | m/s² |
| Friction (kinetic) | (f_k = \mu_k N) | N |
| Friction (static) | (f_s \le \mu_s N) | N |
3. Worked Example
Problem
A 10 kg crate is pulled across a horizontal floor by a rope at an angle of (30^\circ) above the horizontal with a tension of 60 N. The coefficient of kinetic friction between the crate and floor is (0.3).
Find:
- The net horizontal force on the crate.
- The acceleration of the crate.
Solution
-
Resolve the tension into horizontal & vertical components
[ T_x = T \cos 30^\circ = 60 \times 0.866 = 51.96 \text{ N} ] [ T_y = T \sin 30^\circ = 60 \times 0.5 = 30 \text{ N} ] -
Weight & normal force
Weight: (W = mg = 10 \times 9.8 = 98 \text{ N}).
Normal force is reduced by the upward component of tension:
[ N = W - T_y = 98 - 30 = 68 \text{ N} ] -
Friction force
[ f_k = \mu_k N = 0.3 \times 68 = 20.4 \text{ N} ] -
Net horizontal force
[ F_{\text{net}} = T_x - f_k = 51.96 - 20.4 = 31.56 \text{ N} ] -
Acceleration
[ a = \frac{F_{\text{net}}}{m} = \frac{31.56}{10} = 3.16 \text{ m/s}^2 ]
Answer
- Net horizontal force ≈ 32 N to the right.
- Acceleration ≈ 3.2 m/s² to the right.
4. Quick Revision Summary
| Topic | Main Idea | Key Formula / Concept |
|---|---|---|
| Force | Interaction changing motion | (F = ma) |
| Units | Newton (N) | 1 N = 1 kg·m/s² |
| Newton’s 1st Law | Inertia | No net force → no change |
| Newton’s 2nd Law | Quantitative force‑acceleration | (F_{\text{net}} = ma) |
| Newton’s 3rd Law | Action–reaction pairs | (F_{AB} = -F_{BA}) |
| Free‑Body Diagram | Visualising forces | Draw all vectors |
| Friction | Resistive force | (f_k = \mu_k N) (kinetic) |
| Weight | Gravitational force | (W = mg) |
| Normal Force | Reaction perpendicular to surface | (N = mg - T_y) (if |