Force and the language of dynamics

Kinematics describes how motion changes; dynamics explains those changes through interactions. A force is a push or pull arising from interaction between bodies or fields. It is a vector, so forces acting in different directions must be added vectorially. The net external force determines the acceleration of a body, not any single force considered alone. If forces cancel, the acceleration is zero even when the object is moving.

A free-body diagram (FBD) isolates a body and represents all the external forces acting on it: weight, normal reaction, tension, friction, spring force and applied loads. Do not put forces exerted by the selected body on that same FBD. For a stationary book on a table, the downward gravitational force and upward normal contact force balance. This is equilibrium, not the absence of gravity.

• Identify the object, select axes, mark external forces, resolve components, then apply equations.
• Force is measured in newtons (N), with 1 N = 1 kg·m/s².

Newton’s first law and inertia

Newton’s first law states that in an inertial reference frame a body remains at rest or continues in uniform straight-line motion unless acted upon by a nonzero net external force. It rejects the older misconception that a continuous force is necessary to sustain constant velocity. In ideal frictionless motion, zero net force means velocity remains unchanged; it does not require the body to stop.

Inertia is the resistance of matter to a change in velocity; for ordinary mechanics, inertial mass measures that resistance. When a moving bus brakes, passengers tend to continue moving forward relative to the bus. This is not a mysterious forward force; it is the tendency to preserve the previous state of motion. Seat belts provide the external force that safely changes the passenger’s velocity.

• An inertial frame is one in which Newton’s first law holds without fictitious forces.
• Greater mass generally means greater inertia at the same speed and conditions.

Newton’s second law and acceleration

The general expression for the net external force is the time derivative of total momentum: F_net = dp/dt. For constant mass, this reduces to the familiar F_net = ma. Acceleration points in the direction of net force, not necessarily in the direction of velocity. For example, friction may act opposite velocity and create an acceleration that reduces the speed.

When multiple forces act along one axis, assign signs consistently. A 6 kg object experiencing forces +18 N and −6 N has net force +12 N, so acceleration is +2 m/s². Dividing one component force by mass without including other forces is a common exam error. If net force doubles at fixed mass, acceleration doubles; if mass doubles at fixed net force, acceleration halves.

• Use vector addition; do not add magnitudes unless forces are collinear and appropriately signed.
• For changing mass systems, avoid blindly applying F = ma without careful momentum accounting.

Newton’s third law and action–reaction pairs

Newton’s third law says that if body A exerts a force on body B, body B exerts an equal-magnitude, opposite-direction force on A. The two forces act on different bodies and therefore never cancel each other on one body’s free-body diagram. This distinction resolves common confusions about walking, swimming, recoil, and rockets.

When walking, the foot pushes the ground backward and the ground exerts forward static friction on the walker. A rocket expels exhaust backward, receiving forward thrust through momentum exchange. Earth’s gravitational pull on a person and the person’s gravitational pull on Earth form an action–reaction pair; the person’s weight and the ground’s normal reaction, by contrast, both act on the person and are not the third-law pair.

• An action–reaction pair is equal and opposite regardless of the masses involved.
• The accelerations of two interacting bodies can be different because their masses differ.

Momentum, impulse and collisions

Linear momentum is p = mv, a vector in the velocity direction. An impulse J = ∫F dt changes momentum: J = Δp. This explains why airbags, safety mats and crumple zones reduce average contact force by extending the stopping time for a given momentum change. A force–time graph’s signed area gives impulse; a short high force and a longer lower force can produce the same momentum change.

For a system with zero net external impulse, total momentum is conserved. Internal forces redistribute momentum among constituents but cannot change the system’s total momentum. In a collision, mechanical kinetic energy may decrease even when momentum is conserved: deformation, sound and heat can absorb it. An isolated elastic collision conserves both momentum and kinetic energy; an isolated inelastic collision conserves momentum but not necessarily kinetic energy.

• Momentum conservation must be applied to the complete chosen system.
• In a perfectly inelastic collision, bodies stick together after impact.

Normal reaction, apparent weight and lifts

The normal force is the contact force perpendicular to a surface. It need not always equal mg. On a horizontal floor with no vertical acceleration and no other vertical load, N = mg. When a lift accelerates upward at a, N = m(g+a); when it accelerates downward at a, N = m(g−a). A weighing scale commonly reports normal force or an equivalent calibrated mass.

A passenger feels heavier in an upward-accelerating lift and lighter in a downward-accelerating lift. During ideal free fall, the normal contact force becomes zero and the passenger appears weightless even though gravity still acts. On an inclined plane, the normal force for a simple block without additional perpendicular forces is mg cosθ. Never assume N = mg on an incline.

• The normal reaction responds to geometry, acceleration and any additional applied forces.
• Apparent weightlessness does not mean gravitational field strength becomes zero.

Static, limiting, kinetic and rolling friction

Friction opposes relative motion or impending relative sliding of contact surfaces. Static friction is self-adjusting: it takes the value needed to prevent slipping, from zero up to the limiting value f_s,max = μ_sN in the simple dry-friction model. Kinetic sliding friction is often modelled as f_k = μ_kN, generally with μ_k less than μ_s for comparable surfaces.

Rolling resistance has a different mechanism from sliding friction; tyres roll with static friction at the contact patch when not skidding, but they also experience rolling losses due to deformation. The common classroom ordering ‘rolling less than sliding’ is a useful comparison in typical circumstances, not a universal equality of force coefficients. Friction can aid motion: without static friction, ordinary walking would be difficult.

• Static friction is not always equal to μsN: equality occurs at limiting equilibrium.
• Friction has no universal direction ‘against motion’; it opposes actual or impending relative sliding.

Inclined-plane mechanics

For a block on an incline of angle θ measured from horizontal, resolve weight mg into a down-slope component mg sinθ and a perpendicular component mg cosθ. For a smooth incline with no other applied force, the block accelerates downhill at g sinθ. If the block is at rest under static friction, the frictional force acts uphill and matches mg sinθ so long as this does not exceed the limiting static friction.

The onset of sliding occurs when mg sinθ = μ_smg cosθ, giving tanθ = μ_s at the critical angle of repose. When sliding downhill under the simple kinetic-friction model, a = g(sinθ − μ_kcosθ). Choose axes along and perpendicular to the slope; this dramatically simplifies the force components.

• The angle of repose relation assumes the basic dry-friction model and no extra forces.
• If tanθ exceeds μs, static equilibrium is not possible in this simple model.

Connected bodies, tension and pulleys

When blocks are connected by a light inextensible string over an ideal massless frictionless pulley, the tension is the same along that string and the constrained blocks share acceleration magnitude. The simplest Atwood machine has masses m₁ and m₂; if m₂ is heavier, its downward acceleration is a = (m₂−m₁)g/(m₁+m₂) and the string tension is 2m₁m₂g/(m₁+m₂).

These formulas rely on the ideal model. A pulley with rotational inertia or an accelerating support changes the tension relationships and dynamics. Always draw one free-body diagram for each separate mass. A frequent mistake is treating tension as equal to the heavier block’s weight even while the system accelerates. For an accelerating block, tension and weight differ.

• For ideal strings, tensions are equal along the same segment.
• For real pulleys with inertia, tensions on two sides can differ.

Centripetal force and turning motion

Circular motion requires an inward acceleration a_c = v²/r, supplied by the radial component of net force. On a level road, static friction between tyres and road commonly supplies the lateral inward force that allows a vehicle to turn. Thus the ideal no-skid limit for a flat road is v ≤ √(μ_srg), neglecting aerodynamic and other complications.

On a frictionless banked road of angle θ, tanθ = v²/(rg) for the design speed. In a vertical circle, tension or normal reaction can vary strongly with position. Centripetal force is not a new fundamental force and should never be drawn as an additional arrow on top of the physical forces that already provide it.

• If speed doubles at fixed radius, necessary inward net force becomes four times.
• At constant speed in a circle, work done by the inward net force is zero because it is perpendicular to displacement.

Pseudo-forces and non-inertial frames

Newton’s laws in simple F = ma form apply directly in inertial frames. A frame attached to a vehicle accelerating forward is non-inertial. To use Newtonian equations within that accelerating frame, an observer introduces a fictitious inertial force −ma_frame on each mass, opposite the frame’s acceleration. This accounts for the rearward apparent motion of passengers when the bus starts suddenly.

Centrifugal force in a rotating frame is similarly an inertial or pseudo-force, not an equal-and-opposite physical force on the same body. An observer on the ground instead describes a turning passenger through real forces, such as friction from the seat. Distinguishing descriptions by reference frame prevents double-counting forces in exam problems.

• Identify whether the observer’s reference frame accelerates.
• Do not add both a pseudo-force and an inconsistent real-force construction.

Applying Newton’s laws: reliable problem-solving protocol

Begin by defining the system and the coordinate axes. Identify interaction forces, draw an individual FBD for each body, then resolve forces along axes and write ΣF = ma. Apply constraints (same acceleration for ideal string systems, fixed contact where no separation occurs) and solve algebraically before substituting numbers. Last, check dimensions, limiting cases and the sign of any acceleration or tension.

Common limiting tests expose errors: an Atwood machine with equal masses must have zero acceleration, a frictionless incline becomes horizontal with θ = 0 giving zero acceleration, and the normal reaction in free fall should become zero. For multiple-choice questions, physical reasoning often excludes implausible answers before calculation.

• Check friction’s direction after identifying which relative sliding would occur.
• Use g ≈ 9.8 m/s² unless the problem specifies g = 10 m/s².

Formula sheet: definitions, conditions and units

Memorise these equations alongside their physical assumptions. The symbols used here follow the explanations in the theory sections.

Solved numericals and real applications

Work through the logic first and then inspect the arithmetic. These examples use idealised standard assumptions unless otherwise stated.

UPSC previous-paper concepts: indexed themes and explanations

The entries below identify independently indexed UPSC NDA General Ability Test question topics, year/session where available; question numbering can differ across booklet series and public indexes. The prompts are paraphrased topic summaries, not reproduced original question text or options. Their explanations are newly written for this chapter; exact question numbering and some session labels were not independently confirmed. Use the linked official UPSC question-paper archive to review the original question and options. NDA PYQs are relevant UPSC examination practice but are not presented as EPFO/APFC Recruitment Test PYQs.

UPSC-style MCQ practice: 20 questions

All questions below are original CrackTarget practice items, not past UPSC questions. Attempt them before opening the answer and explanation.

Exam traps and 30-minute revision

• Action and reaction act on different objects; never cancel them on one free-body diagram.
• Static friction equals the needed balancing force up to its limit, not automatically μsN.
• Net force zero is compatible with constant nonzero velocity.
• Apparent weightlessness is zero contact reaction, not necessarily zero gravity.
• For numerical work, preserve signs of force and acceleration.
• Centripetal force names the inward resultant from actual forces; do not double-count it.