Chapter 5 — Laws Of Motion

Fig 1 — Educational Diagram: Introduction and basic concepts

Overview

"Laws of Motion" begins with basic concepts needed to understand how forces change motion. Core ideas: force, mass, inertia, momentum, and system. A force is a push or pull that tends to change the state of motion of a body. Motion is described by position, velocity and acceleration, and forces are vectors (they have magnitude and direction).

Inertia and Mass

Inertia is the property of a body that resists any change in its state of motion. Mass is a quantitative measure of inertia: larger mass → greater resistance to acceleration. Mass is a scalar and is measured in kilograms (kg).

Types of Forces

Common forces encountered in problems: gravitational (weight), normal, friction (static and kinetic), tension, applied/contact forces, and non-contact forces (like electrostatic and gravitational). Distinguish contact forces (require physical contact) from non-contact (act at a distance).

Net Force and Equilibrium

The net force on a body is the vector sum of all forces acting on it. If the net force is zero the body is in translational equilibrium: either at rest or moving with constant velocity (no acceleration).

Newton's Laws — qualitative summary

• First law (Inertia): If net force = 0, velocity is constant.
• Second law: Net force causes acceleration; acceleration is proportional to net force and inversely proportional to mass.
• Third law: Forces occur in action–reaction pairs: if A exerts a force on B, B exerts an equal and opposite force on A.

Momentum and Impulse

Linear momentum p = m v. Impulse is the change in momentum and equals the time integral (or product for constant force) of force: J = Δp = ∫F dt ≈ FΔt. This links force and duration to changes in motion.

Free-Body Diagrams and Problem Strategy

To solve motion problems: (1) identify the system, (2) draw a free-body diagram showing all forces, (3) write vector sum of forces = ma (Newton's second law), (4) solve for unknowns and check units and limiting behavior.

Units and Vector Nature

SI units: force in newton (N), where 1 N = 1 kg·m/s²; mass in kg; acceleration in m/s²; momentum in kg·m/s. Treat forces as vectors — resolve into components (usually horizontal and vertical) and apply Newton's laws separately in each direction.

Fig 2 — Educational Diagram: Inertia and mass

Inertia is the property of a body that makes it resist any change in its state of rest or uniform motion in a straight line. This concept is expressed by Newton's First Law (the law of inertia): a body remains at rest or continues to move with constant velocity unless acted upon by a net external force.

Inertia has three common aspects:

• Inertia of rest – tendency to remain at rest.
• Inertia of motion – tendency to continue moving.
• Inertia of direction – tendency to maintain the current direction of motion.

Mass is the quantitative measure of inertia. It tells how strongly an object resists acceleration when a force is applied. Mass (m) is a scalar quantity; its SI unit is kilogram (kg). The larger the mass, the greater the inertia.

Relationship with Newton's Second Law: when a net force F acts on a body of mass m, it produces acceleration a according to F = ma. Rearranged, a = F/m shows that for the same applied force, larger mass gives smaller acceleration — this is the quantitative statement that mass measures inertia.

Mass vs Weight: mass is an intrinsic property (measured in kg); weight is the gravitational force on the mass, W = mg, and depends on the acceleration due to gravity g (varies with location).

Inertial mass (measured by F and a) and gravitational mass (appearing in gravitational force) are experimentally equal to high precision — a key empirical fact used in physics.

Common demonstrations: tablecloth trick (objects on table remain nearly at rest while cloth is pulled), passengers lurch forward when a bus stops (inertia of motion), and heavier objects are harder to start or stop than lighter ones (inertia ∝ mass).

Fig 3.1 — Educational Diagram: Newton's Laws of Motion & Free Body Diagrams

Statement: Newton's First Law (Law of Inertia) — An object continues in its state of rest, or of uniform motion in a straight line, unless acted upon by a net external force.

Meaning and detailed explanation:

• Inertia is the property of a body that resists any change in its state of motion (rest or uniform straight-line motion). The greater the mass of a body, the greater its inertia.
• The law identifies a special class of reference frames called inertial frames: frames in which a free particle (no net external force) moves with constant velocity. If you observe acceleration without any applied force, you are in a non-inertial frame.
• Mathematically, if the net external force ΣF = 0 on a particle of mass m, then acceleration a = 0 and the velocity v is constant. This is a special case of Newton's Second Law (F = ma).
• Consequences: equilibrium (static or dynamic) occurs when ΣF = 0. Conservation of linear momentum for an isolated system follows because if ΣF_ext = 0, the total momentum p = Σ m v is constant in time.

Key points:

• Mass is a quantitative measure of inertia: larger mass → larger resistance to change in motion.
• The law defines the behavior of objects when no unbalanced force acts; it does not explain the cause of motion.
• Valid only in inertial frames — accelerating elevators or rotating frames require fictitious forces to apply similar reasoning.

Simple demonstrations and thought experiments:

• Book on a table stays at rest until a net horizontal force is applied.
• A puck on a frictionless ice surface keeps moving in a straight line at constant speed once pushed.
• Pulling a tablecloth quickly from under dishes: dishes tend to remain at rest (inertia) so they don’t move much if the cloth is removed rapidly.

Practical importance: It lets us identify when forces are present (e.g., a change in velocity signals a net force), and it provides the conceptual foundation for dynamics and for introducing Newton's Second Law and conservation laws.

Fig 4.1 — Educational Diagram: Newton's Laws of Motion & Free Body Diagrams

Statement: In an inertial frame, the net external force acting on a body is equal to the time rate of change of its linear momentum: F_net = dp/dt. For constant mass this reduces to the familiar form F_net = m a, where F_net and a are vectors.

Meaning and implications:

• Force causes change in motion: a nonzero net force produces acceleration (change of velocity).
• Proportionality: for a given mass, acceleration is directly proportional to the net force: a ∝ F_net.
• Mass dependence: for a given net force, acceleration is inversely proportional to mass: a = F_net / m.
• Vector nature: directions matter — the direction of acceleration is the direction of the net force.
• Valid only in inertial frames (frames not accelerating). In non-inertial frames fictitious forces must be introduced.

Derivation (brief): Momentum p = m v. In full generality, F_net = dp/dt = m dv/dt + v dm/dt. For systems with constant mass (typical rigid-body problems in Class 11) dm/dt = 0, so dp/dt = m (dv/dt) = m a, giving F_net = m a.

Interpretation in problem solving: Draw a free-body diagram, sum all external forces (vector sum) to get F_net, then use F_net = m a to find acceleration, or use dp/dt for problems with changing mass or when impulse is considered.

Related concepts: Impulse J = ∫F dt = Δp, conservation of momentum (when net external force = 0), and work-energy connections when force acts over a distance.

Fig 5 — Educational Diagram: Momentum and Impulse

Momentum is the product of the mass and velocity of a body: p = mv. It is a vector (same direction as velocity). SI unit: kg·m/s. Momentum measures how hard it is to stop a moving object.

Impulse is the effect of a force acting over a short time. For a force F(t) acting on a body during time interval t1 to t2, the impulse J is

J = \int_{t1}^{t2} F(t)\,dt.

The impulse–momentum theorem states that impulse equals change in momentum:

J = \Delta p = p_{final} - p_{initial}.

Using Newton’s second law in its general form: F_{net} = dp/dt. For constant mass this reduces to F_{net} = m a.

Conservation of Momentum: For an isolated system (no external net force), total linear momentum is conserved: \Sigma p_{initial} = \Sigma p_{final}. For a system of particles the total momentum equals the total mass times the velocity of the center of mass: P_{total} = M V_{cm}.

Collisions: In collisions between bodies, momentum is always conserved (if external forces negligible). Collisions are classified as:

• Elastic: both momentum and kinetic energy conserved.
• Inelastic: momentum conserved, kinetic energy not conserved (some lost as heat, deformation).
• Perfectly inelastic (completely inelastic): colliding bodies stick together after impact.

For one-dimensional two-body collision, conservation of momentum gives: m1 u1 + m2 u2 = m1 v1 + m2 v2. The coefficient of restitution e is defined as the ratio of relative speed of separation to relative speed of approach: e = (v2 - v1)/(u1 - u2) (signs chosen consistently). e = 1 for perfectly elastic, 0 for perfectly inelastic.

Practical idea: Increasing the time of impact (e.g., airbags, padded surfaces) reduces the average force because for the same impulse J = \Delta p the average force F_{avg} = J/\Delta t is smaller when \Delta t is larger.

Fig 6.1 — Educational Diagram: Newton's Laws of Motion & Free Body Diagrams

Statement: For every action there is an equal and opposite reaction. More precisely: if body A exerts a force F on body B, then body B simultaneously exerts a force −F on body A.

Vector form: F_AB = −F_BA. These forces have the same magnitude, opposite direction and act on different bodies along the same line of action.

Key points and consequences:

• Action and reaction act on different bodies, so they do not cancel each other out on the same body.
• Action–reaction pairs are of the same nature (both contact forces or both field forces).
• In an isolated system of particles the internal action–reaction pairs cancel pairwise, which leads to conservation of momentum (total momentum remains constant).
• If only two bodies interact (and no external forces), m1 a1 = −m2 a2, so accelerations are inversely proportional to masses.

Relation to momentum: Using impulse-momentum theorem, ∫F_AB dt = −∫F_BA dt, so the impulses (change of momentum) on the two bodies are equal and opposite; hence total momentum of the isolated system is conserved.

Common misconceptions: The action–reaction forces do not cancel because they act on different bodies. Also, the phrase "equal and opposite" refers to force vectors, not to displacements or energies.

Experimental demonstrations: Examples include recoil of a gun, launching of a rocket (expelled gases push rocket forward), pushing off a boat to reach the shore, and collisions where two carts exert equal and opposite forces during contact.

Fig 7 — Educational Diagram: Conservation of Linear Momentum

Definition. The total linear momentum of an isolated system (one on which net external force is zero) remains constant in time. In symbols: if F_ext = 0, then dP_total/dt = 0 ⇒ P_total = constant.

Why it holds (derivation for two particles). For two interacting particles, Newton's second law gives dp1/dt = F_12 + F1_ext and dp2/dt = F_21 + F2_ext. By Newton's third law F_12 = −F_21, so internal forces cancel: d(p1 + p2)/dt = F1_ext + F2_ext. If external forces sum to zero, d(p1 + p2)/dt = 0, hence p1 + p2 = constant. This generalizes to any number of particles.

Impulse–momentum theorem. The change in momentum of a body equals the impulse delivered: J = ∫ F dt = Δp. During short collisions external impulses may be negligible, so momentum of the colliding bodies is conserved.

Applications to collisions (1D). For two bodies of masses m1, m2 with initial velocities u1, u2 and final velocities v1, v2, momentum conservation gives m1 u1 + m2 u2 = m1 v1 + m2 v2 (vector form or component-wise). Special cases:

• Perfectly inelastic (stick together): v = (m1 u1 + m2 u2)/(m1 + m2).
• Perfectly elastic: both momentum and kinetic energy conserved; for head‑on collisions the final velocities are v1 = ((m1 - m2)/(m1 + m2)) u1 + (2 m2/(m1 + m2)) u2, and v2 = (2 m1/(m1 + m2)) u1 + ((m2 - m1)/(m1 + m2)) u2.

Relation to centre of mass. Total momentum P_total = M V_cm, where M = Σ m_i and V_cm is the velocity of the centre of mass. If no external force acts, V_cm is constant and the centre of mass moves with uniform velocity.

Important conditions. Conservation holds for an isolated system or when the net external impulse during the interaction is negligible (typical approximation for short collisions). Momentum is a vector — conservation must hold component-wise.

Physical intuition. Momentum conservation is a mathematical statement of Newton's third law: when two bodies push on each other, equal and opposite changes in their momenta leave the total unchanged. It governs everything from billiard-ball collisions to rocket motion (consider rocket + expelled fuel as the system).

Fig 8 — Educational Diagram: Collisions and Impact

Collision is an event in which two bodies exert forces on each other for a short time resulting in change of their velocities. In Class 11 (Laws of Motion) we study one-dimensional (head-on) collisions and key quantities: momentum, impulse, kinetic energy, and the coefficient of restitution.

Conservation of linear momentum: For two interacting bodies in an isolated system (no external net impulse), total linear momentum before collision equals total linear momentum after collision.

Impulse: During collision a large internal force acts for a short time. The impulse J is the integral of force over contact time and equals the change in momentum of a body: J = m(v - u). The average force can be written F_avg = J / Δt.

Coefficient of restitution (e): A measure of elasticity of collision defined (for one-dimensional case) as the ratio of relative speed of separation to relative speed of approach: e = (relative speed after collision)/(relative speed before collision) = (v2 - v1)/(u1 - u2). 0 ≤ e ≤ 1: e = 1 perfectly elastic (no kinetic energy lost), 0 < e < 1 partially inelastic, e = 0 perfectly inelastic (bodies stick together).

Solving two-body head-on collisions (m1, m2; initial velocities u1, u2; final velocities v1, v2): Use 1) conservation of momentum: m1 u1 + m2 u2 = m1 v1 + m2 v2, and 2) restitution: v2 - v1 = e (u1 - u2). From these you get

v1 = [m1 u1 + m2 u2 - m2 e (u1 - u2)] / (m1 + m2), v2 = [m1 u1 + m2 u2 + m1 e (u1 - u2)] / (m1 + m2).

Special cases: - Elastic (e = 1): v1 = [(m1 - m2) u1 + 2 m2 u2] / (m1 + m2), v2 = [2 m1 u1 + (m2 - m1) u2] / (m1 + m2). - Perfectly inelastic (e = 0): bodies stick and move with common velocity v = (m1 u1 + m2 u2) / (m1 + m2).

Energy loss: Kinetic energy is not conserved for e < 1. The kinetic energy loss during collision is ΔK = (1/2) (m1 m2/(m1 + m2)) (1 - e^2) (u1 - u2)^2, where ΔK = K_before - K_after (positive when some energy is lost).

Center of mass (COM) frame: Velocity of COM Vcm = (m1 u1 + m2 u2)/(m1 + m2). In the COM frame initial velocities are u1' = u1 - Vcm, u2' = u2 - Vcm, and after collision v1' = - e u1', v2' = - e u2' (for one-dimensional collisions). This shows speeds are reduced by factor e and directions reverse in COM frame.

Important remarks: - Momentum is always conserved in isolated collisions; kinetic energy is conserved only for elastic collisions. - Sign conventions matter: the restitution formula uses relative velocities with consistent direction signs. - Real collisions also involve deformation and sound/heat; e quantifies how “bouncy” the collision is.

Fig 9.1 — Educational Diagram: Newton's Laws of Motion & Free Body Diagrams

Definition: Friction is the resistive force that acts tangentially between two surfaces in contact and opposes relative motion (or the tendency to move) of the surfaces.

Types:

• Static friction (f_s): Acts when surfaces tend to move but remain at rest relative to each other. It adjusts up to a maximum value to prevent motion.
• Kinetic (sliding) friction (f_k): Acts when surfaces slide relative to each other; usually smaller than limiting static friction.
• Rolling friction (rolling resistance): Small resistive torque when a body rolls because of deformation of surfaces.
• Fluid friction (viscous): Resistance when an object moves through a fluid (liquid or gas).

Empirical laws (Amontons' laws):

• Friction is approximately proportional to the normal reaction: f ∝ N.
• Friction is (approximately) independent of contact area for a given pair of surfaces and normal force.
• Limiting static friction is greater than kinetic friction: μ_s > μ_k.

Microscopic origin: Real surfaces have asperities (microscopic bumps). Contact occurs at asperity tips producing adhesion and interlocking; deformation and breaking of these contacts consume mechanical energy and produce heat.

Direction and nature: Friction acts tangent to the surface and opposite to (or opposing) the relative motion or impending motion. It is a non-conservative force: work done by friction depends on the path and converts mechanical energy into thermal energy.

Common applications and consequences: Friction enables walking, driving (tyres on road), and braking; it causes wear, heating and energy loss in machines. Lubrication reduces friction by introducing a fluid layer between surfaces.

Fig 10.1 — Educational Diagram: Newton's Laws of Motion & Free Body Diagrams

Overview
Free-body diagrams (FBDs) are vector drawings that show all forces acting on a single object. They are the primary tool for applying Newton's laws to solve mechanics problems. Use FBDs to set up the vector equations ΣF = ma (Newton's second law) and Στ = Iα when rotations are involved.

Steps to draw a correct Free-body Diagram

• Isolate the body: imagine the object alone, separated from surroundings.
• Represent all contact and non-contact forces by arrows from the object's center or point of contact. Label each force (W or mg, N, T, f, F_applied, R, etc.).
• Choose convenient axes (often one axis parallel to an inclined plane and the other perpendicular). Resolve forces into components along these axes.
• Indicate known directions of motion or acceleration. If direction unknown, assume and solve; sign will tell actual direction.
• Write Newton's second law for each axis: ΣF_x = m a_x, ΣF_y = m a_y. Solve for unknowns.

Common forces to include

• Weight (W or mg): force due to gravity acting at the center of mass, vertically downward.
• Normal force (N): contact force perpendicular to the surface.
• Tension (T): pull in a string, assumed along the string direction.
• Friction (f): tangential contact force opposing relative motion. Static friction f_s ≤ μ_s N; kinetic friction f_k = μ_k N opposite the direction of motion.
• Applied force (F): any external push or pull.
• Air resistance or drag: often velocity dependent; neglected in ideal problems unless specified.

Using Newton's laws with FBDs

• First law (inertia): If ΣF = 0 then a = 0 (object at rest or moving with constant velocity). Useful for equilibrium problems.
• Second law: ΣF = m a. Apply separately along chosen axes after resolving forces. This gives equations to solve for a, T, N, f, etc.
• Third law: Forces come in action–reaction pairs (e.g., block pushes table downward with mg, table pushes block upward with N). Action and reaction act on different bodies and are equal in magnitude and opposite in direction.

Problem-solving strategies

• Always draw FBD before writing equations.
• Pick axes to simplify (parallel/perpendicular to plane or to acceleration).
• Use ΣF = m a separately in each axis; if no acceleration in an axis, set ΣF = 0 to find reaction forces.
• For systems (connected masses, pulleys), draw FBD for each body and relate accelerations and tensions via constraints.

Example setups often covered: block on a horizontal surface with friction, block on an inclined plane, Atwood (two masses connected over a pulley), elevator problems (apparent weight), and blocks connected on horizontal/ inclined surfaces.

Short derivation examples shown in words

• Block on incline (angle θ), mass m, no friction: FBD shows mg downward split into components mg sinθ (down the plane) and mg cosθ (into plane). Equation along plane: m a = mg sinθ, so a = g sinθ.
• Block on incline with kinetic friction μ_k: friction = μ_k N = μ_k mg cosθ acting up the plane if block slides down. Then m a = mg sinθ - μ_k mg cosθ → a = g(sinθ - μ_k cosθ).
• Atwood machine (m1 and m2, m1 > m2): Draw FBD for each mass: m1g - T = m1 a; T - m2 g = m2 a. Solve to get a = (m1 - m2)g/(m1 + m2) and T = 2 m1 m2 g/(m1 + m2) (or solve for T from one equation).

Common mistakes to avoid: forgetting to include friction or normal force, mixing forces on different bodies, applying action–reaction forces to the same body, choosing poor axes that complicate algebra.

Conclusion
Mastery of free-body diagrams plus Newton's laws allows systematic solution of many mechanics problems. Practice drawing FBDs, resolving components, and writing ΣF = m a equations for each axis and each body in a system.

Fig 11.1 — Educational Diagram: Newton's Laws of Motion & Free Body Diagrams

What is a system of particles?
A system of particles is any collection of material points (particles) considered together as one object. Each particle experiences forces from other particles in the system (internal forces) and from agents outside the system (external forces).

Centre of mass (CM)
The position vector of the centre of mass R_cm of N particles (m_i at r_i) is

R_cm = (Σ m_i r_i) / M, where M = Σ m_i.

The velocity and acceleration of the CM are

V_cm = (Σ m_i v_i) / M, A_cm = (Σ m_i a_i) / M.

Newton's second law for a system (resultant external force)
For each particle: m_i a_i = Σ_j F_ij + F_i^ext, where F_ij are internal forces on i from j and F_i^ext are external forces on i. Summing over all particles gives

Σ m_i a_i = Σ_i Σ_j F_ij + Σ F_i^ext.

Internal forces come in action–reaction pairs (Newton's third law). For central forces F_ij = −F_ji so Σ_i Σ_j F_ij = 0. Therefore

M A_cm = Σ F_i^ext = F_ext (net external force).

Thus the centre of mass moves as if the whole mass M were concentrated at CM and acted on by the net external force.

Momentum and its conservation
Total linear momentum P of the system is

P = Σ m_i v_i = M V_cm.

Differentiate: dP/dt = M A_cm = F_ext. If net external force is zero, dP/dt = 0 and total momentum is conserved. This underlies collisions, explosions, and recoil phenomena.

Impulse
Change of total momentum equals impulse of net external force:

ΔP = ∫ F_ext dt.

Kinetic energy: CM and internal motion
Total kinetic energy K_total can be split into motion of CM plus motion about CM:

K_total = (1/2) M V_cm^2 + Σ (1/2) m_i v_i,rel^2,

where v_i,rel = v_i − V_cm. The first term is kinetic energy of CM motion; the second is internal (relative) kinetic energy.

Physical meaning and consequences

• Internal forces cannot change the total momentum of an isolated system; they only redistribute momentum among particles.
• Net external force controls the motion of the CM (not necessarily each part of the system).
• In collisions or explosions, analyze using CM frame and conservation of momentum when external impulse is negligible.

Limitations/notes
Newton's third-law cancellation of internal forces assumes the internal forces are equal and opposite (true for central forces) and no external fields break pair symmetry; variable-mass systems (rockets) require careful bookkeeping of mass flow and application of momentum conservation.

Fig 12.1 — Educational Diagram: Newton's Laws of Motion & Free Body Diagrams

Definition: A non-inertial frame is a reference frame that is accelerating (translationally and/or rotationally) with respect to an inertial frame. In such frames Newton's 2nd law does not hold in its simple form unless extra, apparent forces — called pseudo or fictitious forces — are introduced.

Why pseudo forces? If you analyze motion from a frame that itself accelerates, objects that are free (no real force in that direction) appear to accelerate. To restore Newton's law in the accelerating frame, we add a pseudo force on each mass equal to minus mass times the acceleration of the frame (or other kinematic acceleration terms for rotating frames).

Translationally accelerating frame (simple case):

• Let A_frame be the acceleration of the non-inertial frame relative to an inertial frame. For a particle of mass m, the pseudo force is
  F_pseudo = - m A_frame
• Then in the non-inertial frame Newton's 2nd law becomes
  m a' = ΣF_real + F_pseudo where a' is acceleration measured in the accelerating frame and ΣF_real are real forces (gravity, normal, friction, etc.).

Example derivation (elevator): If the elevator accelerates upward with acceleration a, then a_frame = +a (up). For a mass at rest in the elevator (a' = 0):

• Real forces: Normal N (up), weight mg (down).
• Pseudo force: F_pseudo = -m a (down).
• 0 = N - mg - m a ⇒ N = m(g + a). This is the apparent weight.

Rotating frames — additional pseudo forces: For a frame rotating with angular velocity ω (and possibly changing ω), the acceleration relation between inertial and rotating-frame quantities introduces three extra kinematic acceleration terms. For a particle at position r' with velocity v' measured in the rotating frame, the inertial acceleration a_inertial is

a_inertial = a' + 2 (ω × v') + ω × (ω × r') + (dω/dt × r') + A_frame

Rewriting Newton's law in the rotating (non-inertial) frame gives the pseudo forces (negative of the kinematic accelerations times m):

• Centrifugal force: F_cf = -m [ω × (ω × r')] (in practice this equals +m ω² r_radial outward).
• Coriolis force: F_cor = -2 m (ω × v') (causes deflection of moving objects in rotating frames).
• Euler force: F_E = -m (dω/dt × r') (appears if rotation rate changes).
• Plus the translational pseudo: -m A_frame, if the rotating frame also translates.

How to apply in problems: 1) Choose the non-inertial frame and find its acceleration(s). 2) Draw real forces on the body. 3) Add pseudo forces (equal to −m times the appropriate kinematic accelerations) acting on the body. 4) Apply m a' = ΣF_real + ΣF_pseudo and solve (for bodies at rest in the non-inertial frame, a'=0).

Physical meaning / real-life intuition: Pseudo forces are not caused by physical interactions; they are artifacts of observing from an accelerating frame. They allow you to use Newton's laws within that frame. For rotating frames, centrifugal force behaves like a radially outward push, while Coriolis force causes sideways deflection of moving objects.

Fig 13 — Educational Diagram: Equilibrium and Conditions of Equilibrium

Introduction
Equilibrium describes a state in which a particle or a rigid body remains at rest or moves with constant velocity because no net effect (translation or rotation) acts to change its motion. In mechanics we distinguish equilibrium of a particle and a rigid body:

• Particle: All forces acting on it add vectorially to zero, so acceleration is zero.
• Rigid body: Both net force and net torque (moment) on the body must be zero so it has no linear acceleration and no angular acceleration.

Conditions of equilibrium (general)
For a rigid body in equilibrium the following must hold simultaneously:

• 1. Translational equilibrium: The vector sum of all external forces is zero: ΣF = 0. In components (2D): ΣFx = 0 and ΣFy = 0.
• 2. Rotational equilibrium: The sum of all moments (torques) about any chosen point is zero: Στ = 0. (If ΣF = 0, it is sufficient to require Στ = 0 about any one point; if ΣF ≠ 0, Στ depends on the point chosen.)

Torque (moment)
The torque produced by a force F about a point is given by the vector cross product τ = r × F. Its magnitude is τ = r F sinθ, where r is the perpendicular distance from the point to the line of action of the force and θ is the angle between r and F. For a simple pair of equal and opposite forces (a couple) separated by perpendicular distance d, the moment is τ = F d and is independent of the choice of origin.

Types of equilibrium for a rigid body

• Stable equilibrium: After a small displacement the body experiences restoring forces/torques that bring it back to equilibrium. Graphically, potential energy U has a local minimum: dU/dx = 0 and d2U/dx2 > 0.
• Unstable equilibrium: A small displacement leads to forces/torques that move the body further away. U has a local maximum: d2U/dx2 < 0.
• Neutral equilibrium: A small displacement neither restores nor moves it further; U is flat (d2U/dx2 = 0).

Stability criterion using center of gravity and base of support
A rigid body supported on a surface is stable if the vertical line through its center of gravity (center of mass) falls inside its base of support. If the line passes outside the base, the body tips (unstable).

Practical notes and sign conventions
Choose a consistent sign convention for moments (clockwise negative/positive). When writing Στ = 0 take care to use perpendicular lever arm distances. For 2D problems you normally take moments about one convenient point to eliminate unknown forces whose lines of action pass through that point.

Fig 14 — Educational Diagram: Experimental verification and problem-solving techniques

Overview

"Experimental verification and problem-solving techniques" in the Laws of Motion means (1) how to design and perform simple experiments to verify Newton's laws and frictional laws, and (2) the systematic methods used to solve mechanics problems reliably.

Experimental verification — key experiments and what they show

• Verification of Newton's Second Law (a ∝ F, a ∝ 1/m): Use a dynamics trolley on a low-friction track attached to hanging masses (Atwood-like setup) or apply known forces with calibrated weights/springs. Measure acceleration for different applied forces and masses; plots of a vs F (at fixed m) and a vs 1/m (at fixed F) should be linear through the origin.
• Atwood machine: Two masses m1 and m2 over a pulley verify a = (m1−m2)g/(m1+m2). Measure acceleration and compare with theoretical value.
• Verification of friction laws: Use a block on a horizontal surface: gradually increase horizontal pull until motion starts to get static friction μs from f_max = μsN. For kinetic friction, pull at constant speed and measure force to find μk via f_k = μk N. Alternatively use an inclined plane to find the angle θ where block just slides: μs = tanθ.
• Third law demonstration: Use two spring balances attached to each other—each shows equal and opposite forces. Colliding carts with force sensors also show equal-and-opposite interaction forces.
• Impulse and momentum: Use carts on a track with a force sensor or collision bumper to measure force vs time. Area under F–t curve gives impulse J = Δp; compare with measured change in momentum.

Typical measurements and error sources

• Measure time using photogates or motion sensors for accurate acceleration.
• Reduce friction and pulley friction or correct for them in calculations.
• Account for mass of string/pulley if significant.

Problem-solving techniques — stepwise method

1. Read and sketch: Draw the physical situation and label all given quantities and unknowns.
2. Choose system and free-body diagrams (FBD): Draw FBDs for each body (or the combined system). Show all forces: gravity, normal, tension, applied forces, friction, and pseudo-forces if using a non-inertial frame.
3. Pick axes: Choose convenient axes (parallel and perpendicular to surfaces or along motion) and sign convention.
4. Write equations: Apply ΣF = ma along each axis. For rotational problems (if any) include torques. Use kinematic relations only after writing dynamics equations when necessary.
5. Include constraints: For connected bodies, relate accelerations (e.g., same magnitude, or a = rα for pulley constraints). For limiting friction use f ≤ μsN and check if static friction requirement exceeds μsN.
6. Solve algebraically: Solve the set of equations for unknowns, keeping units consistent.
7. Check special cases and dimensions: Verify limiting behaviour (e.g., m→0, F→0) and check units/dimensions and signs. Compare with intuition (e.g., heavier mass → smaller acceleration for same force).

Tips and common pitfalls

• Always draw FBDs even for simple problems.
• Distinguish carefully between static and kinetic friction.
• When applying Newton's laws to systems, internal forces cancel; use external forces for the system equation.
• In collisions, use conservation of momentum (if no external impulse) and energy only if elastic.
• For non-inertial frames include pseudo (inertial) forces on each mass: F_pseudo = −ma_frame.