Force, Motion, Pressure and Friction for CTET Paper II (Class 6–8)

intermediate 22 min read

Concept

Force is a push or pull that can change an object's state — its speed, direction, or shape. You cannot see force directly; you see its effects. When you kick a football, the ball starts moving. When you squeeze a rubber ball, its shape changes. Both are forces at work.

Motion is simply the change in position of an object over time. But here's where students trip up: motion is always relative. A person sitting in a moving train is at rest relative to the train but in motion relative to the platform. This relativity shows up in CTET questions dressed as conceptual traps.

Pressure is not the same as force. Look — a stiletto heel and a flat sandal might be worn by people of the same weight. The stiletto sinks into soft ground; the sandal does not. The difference is the area over which force is distributed. Pressure = Force ÷ Area. Smaller area, same force — higher pressure. This is why a sharp knife cuts better than a blunt one, why camels have broad padded feet for desert sand, and why building foundations are wide.

Friction is the force that opposes relative motion between surfaces in contact. It is not always the enemy. Without friction, you could not walk, a car could not brake, and a nail would not hold in a wall. But friction also wastes energy as heat in machines, which is why engineers lubricate moving parts.

Think of friction like a velcro handshake between surfaces. The rougher the surfaces, the more micro-hooks interlock, and the greater the friction. This is the analogy to hold onto when you encounter surface-comparison questions.

These four ideas — force, motion, pressure, friction — are deeply interconnected. A force causes or changes motion. That force, divided by area, gives pressure. When surfaces resist relative motion, friction is the force doing the resisting. CTET Paper II tests these as conceptual MCQs, surface-comparison questions, and numerical calculations. All of them are solvable if you have the right framework.


Deep Dive

Force: Types and Effects

Forces can be contact forces (friction, normal force, muscular force, spring force) or non-contact forces (gravity, magnetic force, electrostatic force). At the Class 6–8 level, the key classification is this one. CTET often tests whether a given scenario involves contact or non-contact force.

Effects of force:

A single force can produce multiple effects simultaneously. A cricket bat hitting a ball changes both its speed and direction.

Newton's Three Laws

First Law (Law of Inertia): An object at rest stays at rest, and an object in motion stays in motion at constant velocity, unless acted on by a net external force. Inertia is not a force — it is the tendency to resist change. Heavier objects have greater inertia, which is why you need a larger force to accelerate a truck compared to a bicycle.

Second Law: F = ma. Net force equals mass times acceleration. If mass doubles and force stays constant, acceleration halves. This law also tells you that when you apply the same force to different masses, the lighter object accelerates more — a core CTET concept.

Third Law: Every action has an equal and opposite reaction. Critically: these forces act on different bodies. When you push a wall with 50 N, the wall pushes you back with 50 N — but these two forces act on different objects (you and the wall), so they do not cancel each other. This is the most misunderstood aspect of Newton's Third Law and appears directly in CTET PYQs.

Pressure

Pressure = Force ÷ Area

The SI unit is Pascal (Pa), which equals N/m².

Unit conversion trap: CTET numerical questions often give area in cm² and expect the answer in Pa (which requires m²). You must convert: 1 cm² = 10⁻⁴ m², so 150 cm² = 0.015 m². Skipping this step is the single most common error in pressure numericals.

Pascal's Law: Pressure applied to a confined fluid is transmitted equally and undiminished in all directions. This is the operating principle of hydraulic systems.

For a hydraulic press:

P_small = P_large

F_small / A_small = F_large / A_large

Therefore: F_large = F_small × (A_large / A_small)

The force multiplication factor is simply the ratio of the larger area to the smaller area.

Applications of Pascal's Law: Hydraulic press, hydraulic brakes in vehicles, hydraulic lifts in garages.

Devices that measure pressure — barometer (atmospheric pressure), manometer (gas pressure) — do not themselves work on Pascal's Law. This distinction appears in CTET MCQs.

Friction: Types, Factors, Applications

Types of friction (in increasing order of magnitude):

Rolling friction < Fluid friction (drag) < Sliding friction < Static friction

Static friction is the friction that prevents an object from starting to move. It is the highest because it must be overcome to initiate motion. Once motion starts, the friction drops to kinetic (sliding) friction. This is why it takes more effort to get a heavy box moving than to keep it moving once it starts.

Rolling friction is much smaller than sliding friction — this is the entire engineering motivation behind wheels and ball bearings. Ball bearings convert sliding friction to rolling friction, dramatically reducing energy loss. Ball bearings reduce friction; they do not increase it.

Factors affecting friction:

  1. Nature of surfaces (roughness) — rougher surfaces, more friction
  2. Normal force pressing the surfaces together — heavier object, more friction
  3. Does NOT depend on the area of contact (a classic exam trap — a wider box does not have more friction than a narrow one, if the weight is the same)

Increasing friction (where useful): Treads on tyres, grooves on shoe soles, sand on icy roads, chalk on fingers for gymnastics.

Reducing friction (where harmful): Lubricants (oil, grease), ball bearings, polishing surfaces, streamlining (reducing fluid friction).

Streamlining is specifically about reducing fluid (air or water) friction by shaping objects so fluid flows smoothly around them — aeroplanes, cars, fish, and birds are naturally or deliberately streamlined. Fluid friction depends on shape, which is why this matters.


Memory Tricks & Shortcuts

patternFriction Ladder: Smooth to Rough

When comparing friction across surfaces, rank them by texture roughness from smoothest to roughest. Think of it as a ladder: Glass/Cellophane (bottom rung, least rough) → Newspaper → Sand paper → Carpet (top rung, most rough). Apply this pattern to any surface-comparison question: pick the sequence that goes from least rough to most rough for increasing friction order. Standard approach — reading and reasoning each surface: ~40s. Pattern recall: ~8s.

patternHydraulic Press Ratio Method

Skip the pressure equation entirely. For a hydraulic press, output force = input force × (big area ÷ small area). Example: Force = 50 N, areas 10 cm² and 200 cm². Ratio = 200 ÷ 10 = 20. Output = 50 × 20 = 1000 N. That is three steps instead of writing out pressure separately, computing it, then multiplying again — saves 2 steps and eliminates one rounding error opportunity.

patterncm² to m² Conversion: The 10⁻⁴ Rule

Every time you see area in cm² in a pressure question, immediately multiply by 10⁻⁴ to get m². Write it as a reflex before you do anything else. 150 cm² → write 150 × 10⁻⁴ = 0.015 m² in the margin first. Then proceed. This one-second habit prevents the most common error in CTET pressure numericals — which is computing pressure in N/cm² and reporting it as Pa. Standard method with ad-hoc conversion: error rate high. Fixed reflex conversion: error rate near zero.

eliminationNewton's Third Law: Two-Body Test

When a Third Law question has an option suggesting forces cancel, eliminate it immediately. Third Law force pairs always act on different bodies. Forces on the same body can cancel (that is equilibrium). Forces on different bodies never cancel each other — they are not even in the same equation. Spotting this eliminates one or two options in under 5 seconds, turning a 30-second reasoning question into a 10-second one.

eliminationBall Bearings Always Reduce

Any option stating ball bearings increase friction is incorrect. Ball bearings convert sliding friction to rolling friction. Rolling < Sliding — always. If a question has "ball bearings increase friction" as an option and asks for the incorrect statement, that is your answer. Zero calculation needed. Identification time: under 5 seconds.


Fast-Solving Framework

Use this decision tree in the exam hall:

Step 1 — Identify the concept cluster: Is the question about (a) friction comparison, (b) pressure calculation, (c) Newton's Laws, or (d) Pascal's Law devices?

Step 2 — Friction comparison questions: Rank surfaces by texture roughness. Smoother = less friction. Apply the friction ladder. Done.

Step 3 — Pressure calculation: Convert area to m² first (multiply cm² by 10⁻⁴). Then apply P = F/A. Watch for weight calculation: F = mg.

Step 4 — Hydraulic press: Use the ratio method. Output force = Input force × (A_large ÷ A_small).

Step 5 — Newton's Third Law questions: Check if forces are on the same body or different bodies. Options claiming Third Law pairs cancel → eliminate. Options claiming reaction force is unequal → eliminate.

Step 6 — Device identification questions: Hydraulic press = Pascal's Law. Barometer = atmospheric pressure measurement. Manometer = gas pressure measurement. Ball bearings = reduce friction (convert sliding to rolling).

If two options seem correct, ask: which one is more precisely stated? CTET rewards precision over generality.


Solved PYQs

Why this question: Tests ability to rank friction by surface texture — a direct application of the core concept that friction depends on surface roughness.

Previous Year Questionपिछले वर्ष का प्रश्न2019
Sameer rolls his marble on three different surfaces spread out on floor – taut cellophane sheet, carpet and newspaper. The force of friction acting on the marble in the increasing order is
  1. Newspaper, Cellophane sheet, Carpet
  2. Newspaper, Carpet, Cellophane sheet
  3. Cellophane sheet, Newspaper, Carpet
  4. Cellophane sheet, Carpet, Newspaper
Solutionसमाधान
Friction depends on surface roughness. Cellophane sheet is smoothest (least friction), newspaper is moderately rough, and carpet has highest friction due to its fibrous texture. So in increasing order of friction: Cellophane sheet < Newspaper < Carpet.

Solving path: Rank the three surfaces by texture roughness. Cellophane sheet is a smooth, taut plastic — minimal surface irregularity, least friction. Newspaper is slightly fibrous but relatively flat — moderate friction. Carpet has dense fibrous texture with maximum micro-interlocking — highest friction. Increasing order: Cellophane sheet → Newspaper → Carpet. That matches option C.


Why this question: A classic "identify the incorrect statement" format. Ball bearings appear in every friction chapter and this misconception is tested repeatedly.

Previous Year Questionपिछले वर्ष का प्रश्न2019
Identify the incorrect statement.
  1. Ball bearings are used to increase friction between parts of a machine.
  2. Friction between two surfaces in contact can never be eliminated on earth.
  3. Rolling friction is less than sliding friction.
  4. The friction force on an object moving through a fluid depends upon its shape.
Solutionसमाधान
Ball bearings are used to REDUCE friction, not increase it, by converting sliding friction into rolling friction which is much smaller. The other statements are all correct: friction cannot be eliminated entirely, rolling friction is less than sliding friction, and fluid friction depends on shape (streamlining).

Solving path: Apply the Two-Body Test and Ball Bearings Always Reduce tricks together. Statement A says ball bearings increase friction — that is the opposite of their purpose. Ball bearings reduce friction by converting sliding to rolling. Eliminate A as incorrect. Verify the others: friction cannot be fully eliminated (true), rolling < sliding (true), fluid friction depends on shape — streamlining (true). Answer is A.


Why this question: Tests the specific conceptual distinction between static friction and other forces — a Third Law and friction type question combined.

Previous Year Questionपिछले वर्ष का प्रश्न2021
In order to slide a huge box lying on the ground in her room, Reshma should apply a force which is greater than which of the following forces?
  1. Static friction
  2. Gravitational force
  3. Normal force
  4. Muscular force
Solutionसमाधान
To start sliding a stationary object, the applied force must overcome the static friction between the object and the surface. Static friction acts to oppose the initiation of motion. Gravitational and normal forces act vertically and do not oppose horizontal sliding. Muscular force is the force Reshma applies, not the opposing force. Hence, the applied force must exceed the static friction to start the box moving.

Solving path: The box is stationary (at rest). To start horizontal sliding, the applied force must exceed the force opposing the start of motion. That force is static friction — the friction resisting initiation of motion. Gravitational force and normal force are vertical, not horizontal. Muscular force is Reshma's own applied force, not the opposing force. The answer is A: Static friction.


Why this question: Newton's Third Law with a stationary object — a direct test of whether students understand that reaction exists even when there is no motion.

Previous Year Questionपिछले वर्ष का प्रश्न
A student pushes a wall with a force of 50 N but the wall does not move. According to Newton's Third Law, which statement is correct?
एक छात्र दीवार को 50 N बल से धकेलता है लेकिन दीवार नहीं हिलती। न्यूटन के तीसरे नियम के अनुसार कौन सा कथन सही है?
  1. The wall does not exert any force because it is stationary
  2. The forces cancel each other, so no force exists in the system
  3. The wall exerts a force of 50 N on the student in the opposite direction
  4. The wall exerts a force greater than 50 N to keep itself from moving
  1. दीवार कोई बल नहीं लगाती क्योंकि वह स्थिर है
  2. बल एक-दूसरे को निरस्त कर देते हैं, इसलिए सिस्टम में कोई बल नहीं होता
  3. दीवार छात्र पर 50 N बल विपरीत दिशा में लगाती है
  4. दीवार खुद को हिलने से रोकने के लिए 50 N से अधिक बल लगाती है
Solutionसमाधान
Newton's Third Law states that for every action, there is an equal and opposite reaction. When the student pushes the wall with 50 N, the wall simultaneously pushes back on the student with exactly 50 N in the opposite direction. This reaction force acts on the student (different body), so the forces do not cancel. The wall being stationary is due to external support (foundation), not absence of force.
न्यूटन का तीसरा नियम कहता है कि प्रत्येक क्रिया के लिए एक समान और विपरीत प्रतिक्रिया होती है। जब छात्र दीवार को 50 N से धकेलता है, तो दीवार उसी समय छात्र पर ठीक 50 N विपरीत दिशा में धकेलती है। यह प्रतिक्रिया बल छात्र पर (अलग वस्तु पर) कार्य करता है, इसलिए बल निरस्त नहीं होते। दीवार का स्थिर रहना बाहरी आधार (नींव) के कारण है, न कि बल की अनुपस्थिति के कारण।

Solving path: Apply the Two-Body Test. The student pushes with 50 N (action). The wall pushes back with exactly 50 N in the opposite direction (reaction). These forces act on different bodies — student and wall — so they do not cancel. The wall stays stationary not because there is no force, but because its foundation provides support. Option C is correct.


Why this question: Pascal's Law numerical — the most common pressure calculation type. Tests unit handling and the ratio method.

Previous Year Questionपिछले वर्ष का प्रश्न
A hydraulic press has a small piston of area 10 cm² and a large piston of area 200 cm². If a force of 50 N is applied on the small piston, what force is exerted by the large piston?
एक हाइड्रोलिक प्रेस में एक छोटे पिस्टन का क्षेत्रफल 10 cm² और एक बड़े पिस्टन का क्षेत्रफल 200 cm² है। यदि छोटे पिस्टन पर 50 N बल लगाया जाए, तो बड़े पिस्टन द्वारा कितना बल लगाया जाएगा?
  1. 500 N
  2. 1000 N
  3. 2000 N
  4. 100 N
  1. 500 N
  2. 1000 N
  3. 2000 N
  4. 100 N
Solutionसमाधान
By Pascal's Law, pressure is transmitted equally throughout the liquid. Pressure on small piston = 50 N ÷ 10 cm² = 5 N/cm². This same pressure acts on large piston: Force = Pressure × Area = 5 N/cm² × 200 cm² = 1000 N. The hydraulic press multiplies force in the ratio of areas (200:10 = 20), so output force = 50 × 20 = 1000 N.
पास्कल के नियम के अनुसार, दाब पूरे तरल में समान रूप से संचारित होता है। छोटे पिस्टन पर दाब = 50 N ÷ 10 cm² = 5 N/cm²। यही दाब बड़े पिस्टन पर कार्य करता है: बल = दाब × क्षेत्रफल = 5 N/cm² × 200 cm² = 1000 N। हाइड्रोलिक प्रेस क्षेत्रफलों के अनुपात (200:10 = 20) में बल को गुणित करती है, इसलिए आउटपुट बल = 50 × 20 = 1000 N।

Solving path: Use the ratio method. Area ratio = 200 ÷ 10 = 20. Output force = 50 × 20 = 1000 N. Cross-check: Pressure on small piston = 50 ÷ 10 = 5 N/cm². Force on large piston = 5 × 200 = 1000 N. Both methods confirm: 1000 N, option B.


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