What this chapter is about
This chapter introduces the fundamental principles that govern why objects move the way they do. While earlier chapters described motion using kinematics (how things move), this chapter answers the deeper question: what causes motion to change? The answer lies in forces and the three laws formulated by Isaac Newton in the 17th century.
A Class 11 student meets this chapter now because it forms the foundation for nearly all mechanics problems. Understanding forces, inertia, action-reaction pairs, and equilibrium allows you to analyse everything from a book resting on a table to a rocket launching into space. The concepts here connect directly to work-energy relations, rotational motion, and gravitation studied later.
After working through this chapter, you should be able to identify all forces acting on an object, draw free-body diagrams, apply Newton's laws to predict motion, understand friction and circular motion from a force perspective, and solve problems involving connected bodies, pulleys, and inclined planes.
Key ideas
- Newton's First Law (Law of Inertia): An object remains at rest or moves with constant velocity unless acted upon by a net external force. This defines inertia as the natural tendency to resist changes in motion.
- Newton's Second Law: The net force on an object equals the rate of change of its momentum. For constant mass, F = ma, where F is in newtons, m in kilograms, and a in metres per second squared.
- Newton's Third Law: When object A exerts a force on object B, object B exerts an equal and opposite force on object A. These forces act on different objects and never cancel each other.
- Momentum: Defined as p = mv, momentum is a vector quantity measured in kg m/s. The law of conservation of momentum states that total momentum of an isolated system remains constant.
- Friction: A contact force opposing relative motion between surfaces. Static friction (f_s ≤ μ_s N) prevents motion; kinetic friction (f_k = μ_k N) acts during sliding. Here μ is the coefficient of friction and N is the normal force.
- Free-body diagram: A sketch showing all forces acting on a single object, essential for correctly applying Newton's second law.
- Equilibrium: An object is in equilibrium when the net force on it is zero, meaning it either remains at rest or moves with constant velocity.
- Circular motion and centripetal force: An object moving in a circle requires a net inward force of magnitude mv²/r to maintain its curved path.
Formulas and facts to remember
- Newton's second law: F_net = ma (force in N, mass in kg, acceleration in m/s²)
- Momentum: p = mv; impulse = change in momentum = F × Δt
- Weight: W = mg, where g ≈ 9.8 m/s² near Earth's surface
- Static friction: f_s ≤ μ_s N (maximum value is μ_s N just before sliding begins)
- Kinetic friction: f_k = μ_k N (constant during sliding, typically μ_k < μ_s)
- Centripetal acceleration: a_c = v²/r directed toward the centre
- For a body on an incline at angle θ: component of weight along incline = mg sin θ; normal force N = mg cos θ
- Conservation of momentum: In the absence of external forces, m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Worked examples
Example 1: Applying Newton's second law
A 4 kg block rests on a frictionless horizontal surface. A horizontal force of 20 N is applied to it. Find the acceleration.
Solution: Since the surface is frictionless, the only horizontal force is the applied force. Using F = ma: 20 = 4 × a a = 20/4 = 5 m/s² The block accelerates at 5 m/s² in the direction of the applied force.
Example 2: Friction on a horizontal surface
A wooden crate of mass 10 kg is pushed along a floor with a horizontal force of 50 N. The coefficient of kinetic friction between crate and floor is 0.3. Find the acceleration.
Solution: Weight W = mg = 10 × 9.8 = 98 N On a horizontal surface, normal force N = W = 98 N Kinetic friction f_k = μ_k × N = 0.3 × 98 = 29.4 N Net horizontal force = Applied force − Friction = 50 − 29.4 = 20.6 N Acceleration a = F_net/m = 20.6/10 = 2.06 m/s²
Example 3: Two blocks connected by a string
Two blocks of masses 3 kg and 2 kg are connected by a light string passing over a frictionless pulley. The 3 kg block hangs vertically and the 2 kg block rests on a frictionless horizontal table. Find the acceleration of the system and tension in the string.
Solution: Let the acceleration be a and tension be T.
For the 3 kg block (moving downward): Weight pulls it down, tension pulls it up. 3g − T = 3a … (i)
For the 2 kg block (moving horizontally): Only tension accelerates it. T = 2a … (ii)
Substituting (ii) into (i): 3 × 9.8 − 2a = 3a 29.4 = 5a a = 5.88 m/s²
From (ii): T = 2 × 5.88 = 11.76 N
The system accelerates at 5.88 m/s² and the tension is 11.76 N.
Common mistakes
- Confusing mass and weight → Mass is the quantity of matter (kg); weight is the gravitational force (N) given by mg.
- Applying Newton's third law forces to the same object → Action and reaction act on different objects; they do not cancel when analysing one body.
- Forgetting that static friction adjusts itself → It only equals μ_s N at the point of impending motion; before that, it matches the applied force.
- Using weight instead of normal force to calculate friction on an incline → On an incline, N = mg cos θ, not mg.
- Ignoring direction when using F = ma → Force and acceleration are vectors; signs must reflect direction consistently.
Quick revision
- First law: no net force means no change in velocity.
- Second law: F_net = ma; use a free-body diagram first.
- Third law: equal and opposite forces on two different bodies.
- Friction opposes relative motion; f ≤ μN for static, f = μN for kinetic.
- Momentum is conserved when no external force acts.
- For circular motion, a centripetal force mv²/r must be provided by real forces like tension, gravity, or friction.