Physics Mechanics - How to Solve: Newton’s Laws of Motion (Constraint Relations, Weighing Machine, Spring Force) – IIT JEE Guide

How to Solve: Newton’s Laws of Motion (Constraint Relations, Weighing Machine, Spring Force) – IIT JEE Guide

Introduction

Mastering constraint relations, weighing machines, and spring forces in Newton’s Laws unlocks 5-10 marks in IIT JEE—enough to push you from a 90 to a 99+ percentile in Physics. These problems test real-world applications like elevators, pulleys, and springs, where forces aren’t just balanced—they’re constrained by geometry and motion.

WHAT YOU NEED TO KNOW FIRST

1. Newton’s 2nd Law (F = ma) – Must apply it in vector form (x and y directions separately).
2. Free-Body Diagrams (FBDs) – Must draw every force acting on each body in the system.
3. Basic Pulley & String Constraints – Must know how tension and acceleration relate in connected systems.

KEY TERMS & FORMULAS

1. Constraint Relations (Pulley Systems)

• Definition: Equations that relate accelerations of different masses in a system due to inextensible strings or rigid rods.
• Key Idea: If a string doesn’t stretch, the total length remains constant → sum of displacements = 0.
• Formula (2-mass pulley): [ a_1 + a_2 = 0 \quad \text{(if pulley is fixed)} ]
• (a_1) = acceleration of mass 1 (upward +ve)
• (a_2) = acceleration of mass 2 (downward +ve)

• MEMORISE THIS: If one mass goes up, the other goes down by the same amount.

• Formula (Movable Pulley): [ a_{\text{pulley}} = \frac{a_1 + a_2}{2} ]

• MEMORISE THIS: Acceleration of the pulley is the average of the two masses.

2. Weighing Machine (Apparent Weight)

• Definition: A weighing machine measures the normal force (N) exerted by a surface, not the actual weight.
• Key Idea: If the system is accelerating, (N \neq mg).
• Formula (Elevator Problems): [ N = m(g \pm a) ]
• (N) = Normal force (apparent weight)
• (m) = Mass of the object
• (g) = Acceleration due to gravity (9.8 m/s², but 10 m/s² in JEE)
• (a) = Acceleration of the system (+ve if upward, -ve if downward)
• MEMORISE THIS: If accelerating up, weight increases. If accelerating down, weight decreases.

3. Spring Force (Hooke’s Law)

• Definition: Force exerted by a spring is proportional to its extension/compression.
• Formula: [ F = -kx ]
• (F) = Spring force (restoring force)
• (k) = Spring constant (given on exam sheet)
• (x) = Displacement from natural length (+ve if stretched, -ve if compressed)
• MEMORISE THIS: Negative sign means force opposes displacement.

STEP-BY-STEP METHOD

Step 1: Draw the System & Identify Constraints

• Sketch the entire setup (masses, pulleys, strings, springs).
• Label all accelerations (use variables like (a_1, a_2)).
• Constraint Rule: If a string is inextensible, the sum of displacements = 0.
• Example: For a fixed pulley with two masses, (a_1 + a_2 = 0).

Step 2: Draw Free-Body Diagrams (FBDs) for Each Mass

• For each mass, draw:
• Weight (mg) downward.
• Tension (T) in the string (direction depends on setup).
• Normal force (N) if in contact with a surface.
• Spring force (F = -kx) if attached to a spring.

Step 3: Write Newton’s 2nd Law for Each Mass

• For each mass, write: [ \sum F = ma ]
• x-direction: If no horizontal forces, ignore.
• y-direction: Take upward as +ve, downward as -ve.
• For springs: Include (F = -kx) in the equation.

Step 4: Apply Constraint Relations

• Use the constraint equation (from Step 1) to relate accelerations.
• Substitute into Newton’s equations to eliminate extra variables.

Step 5: Solve the Equations

• You’ll have 2-3 equations (one per mass + constraint).
• Solve for unknowns (T, a, N, etc.) using substitution/elimination.

Step 6: Check Units & Reasonableness

• Units: All forces in Newtons (N), accelerations in m/s².
• Reasonableness: If a mass is lighter, it should accelerate faster. If a spring is stretched, force should be opposite to displacement.

WORKED EXAMPLES

Example 1 – Basic (Fixed Pulley + Two Masses)

Problem: Two masses, (m_1 = 2 \, \text{kg}) and (m_2 = 3 \, \text{kg}), are connected by a light inextensible string over a fixed pulley. Find: (a) Acceleration of the system. (b) Tension in the string.

Solution:

Step 1: Draw System & Constraints - Fixed pulley → (a_1 + a_2 = 0) (if (m_1) goes up, (m_2) goes down). - Let (a_1 = a) (upward for (m_1)), then (a_2 = -a) (downward for (m_2)).

Step 2: Draw FBDs - For (m_1) (2 kg): - Up: Tension (T) - Down: Weight (2g) - For (m_2) (3 kg): - Up: Tension (T) - Down: Weight (3g)

Step 3: Write Newton’s 2nd Law - For (m_1): [ T - 2g = 2a \quad \text{(1)} ] - For (m_2): [ 3g - T = 3(-a) \quad \text{(2)} ] (Since (a_2 = -a), acceleration is downward for (m_2).)

Step 4: Solve Equations - Add (1) and (2): [ T - 2g + 3g - T = 2a - 3a ] [ g = -a \implies a = -g ] (Negative sign means (m_1) accelerates downward, (m_2) upward—opposite to our assumption.) - Correct acceleration: (a = g) (magnitude), but direction is (m_2) downward, (m_1) upward. - Tension (from eq. 1): [ T - 2g = 2g \implies T = 4g = 40 \, \text{N} ]

Final Answers: (a) Acceleration = 10 m/s² (downward for (m_2), upward for (m_1)). (b) Tension = 40 N.

What we did and why: - Assumed (m_1) accelerates upward, but the heavier mass ((m_2)) dictates direction. - Used constraint relation to relate accelerations. - Solved two equations to find (a) and (T).

Example 2 – Medium (Weighing Machine in Elevator)

Problem: A man of mass (60 \, \text{kg}) stands on a weighing machine in an elevator. Find the apparent weight when: (a) Elevator accelerates upward at 2 m/s². (b) Elevator moves downward at 3 m/s². (c) Elevator is in free fall.

Solution:

Step 1: Understand Apparent Weight - Weighing machine measures normal force (N). - (N = m(g \pm a)).

Step 2: Apply Formula - (m = 60 \, \text{kg}), (g = 10 \, \text{m/s²}).

(a) Accelerating Upward (a = +2 m/s²) [ N = m(g + a) = 60(10 + 2) = 720 \, \text{N} ] Apparent weight = 720 N (heavier).

(b) Accelerating Downward (a = -3 m/s²) [ N = m(g - a) = 60(10 - 3) = 420 \, \text{N} ] Apparent weight = 420 N (lighter).

(c) Free Fall (a = g = 10 m/s² downward) [ N = m(g - g) = 0 \, \text{N} ] Apparent weight = 0 N (weightless).

What we did and why: - Used (N = m(g \pm a)) directly. - Sign of (a) determines whether weight increases or decreases. - Free fall means (a = g) → normal force = 0.

Example 3 – Exam-Style (Spring + Pulley System)

Problem: A block of mass (m = 2 \, \text{kg}) is attached to a spring of spring constant (k = 200 \, \text{N/m}) and placed on a frictionless surface. The other end of the spring is fixed to a wall. A light string is attached to the block, passes over a massless pulley, and is connected to a hanging mass (M = 3 \, \text{kg}). The system is released from rest with the spring at its natural length. Find: (a) Acceleration of the system. (b) Extension in the spring when the system is in motion.

Solution:

Step 1: Draw System & Constraints - Constraint: String is inextensible → (a_{\text{block}} = a_{\text{hanging mass}}) (let (a) be the acceleration). - Spring force: (F = -kx) (opposes extension).

Step 2: Draw FBDs - For (m = 2 \, \text{kg}) (block): - Right: Tension (T) - Left: Spring force (F = -kx) - For (M = 3 \, \text{kg}) (hanging mass): - Up: Tension (T) - Down: Weight (3g)

Step 3: Write Newton’s 2nd Law - For block (horizontal): [ T - kx = 2a \quad \text{(1)} ] - For hanging mass (vertical): [ 3g - T = 3a \quad \text{(2)} ]

Step 4: Solve Equations - Add (1) and (2): [ 3g - kx = 5a \quad \text{(3)} ] - At equilibrium (when acceleration stops), (a = 0): [ 3g = kx \implies x = \frac{3g}{k} = \frac{30}{200} = 0.15 \, \text{m} ] - But the system is accelerating, so we need to find (a) first. - From (3), when (x = 0) (initial condition): [ 3g = 5a \implies a = \frac{3g}{5} = 6 \, \text{m/s²} ] - Extension when accelerating: From (3): [ 3g - kx = 5a ] But (a = \frac{3g - kx}{5}). At any instant, (x) changes until (a = 0) (equilibrium).

Final Answers: (a) Initial acceleration = 6 m/s². (b) Maximum extension = 0.15 m (when system comes to rest).

What we did and why: - Recognized that spring force depends on extension. - Used constraint relation to relate accelerations. - Found initial acceleration when (x = 0), then equilibrium extension when (a = 0).

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before Exam)

"Listen up—this is how you own Newton’s Laws in JEE:

1. Constraint relations: If a string doesn’t stretch, the sum of accelerations = 0 (fixed pulley). For movable pulleys, acceleration is the average.
2. Weighing machine: It measures normal force, not weight. Use (N = m(g \pm a)). Upward acceleration → heavier. Downward → lighter.
3. Spring force: (F = -kx). Negative sign means it pulls back.
4. Always draw FBDs—every force, every mass.
5. Solve step-by-step: Write Newton’s 2nd Law for each mass, apply constraints, then solve.
6. Watch signs: Up = +ve, down = -ve. Spring force opposes displacement.

Exam trick: If a problem has a spring + pulley, first find initial acceleration (x=0), then equilibrium extension (a=0).

You’ve got this—go crush it!