Laws of Motion

The ancient Greek philosopher Aristotle proposed that a body needs a continuous external force acting on it to keep moving with constant velocity, and that a body's natural state is rest. This idea seems to match everyday experience — a cart stops if you stop pushing it, a ball rolling on the ground eventually stops. But this reasoning is flawed, and it is known as Aristotle's fallacy.

The flaw lies in ignoring friction. The cart and the ball stop not because motion itself needs a force to sustain it, but because an opposing force — friction between the moving object and the surface (and air resistance) — is constantly acting on them and decelerating them. If a body could move on a perfectly smooth, frictionless, infinite surface, it would continue moving forever at the same velocity, with no force required at all. Galileo was the first to correctly identify this through his thought experiments on inclined planes: he observed that a ball rolling down one incline onto a second incline rises to nearly the same height it started from, and reasoned that if the second incline were made flatter and flatter (eventually horizontal and frictionless), the ball would keep moving forever at constant speed, never coming to rest by itself.

This insight was formalised by Newton as his First Law of Motion, also called the Law of Inertia: every body continues in its state of rest or of uniform motion in a straight line, unless it is compelled to change that state by an external unbalanced (net) force acting on it.

Inertia is the natural tendency of a body to resist any change in its state of rest or of uniform motion. Mass is the quantitative measure of inertia — the greater the mass of a body, the greater its inertia, i.e., the harder it is to start it moving, stop it, or change its direction. Inertia is commonly classified into three types: inertia of rest (tendency to remain at rest, e.g. a coin on a card flicked away stays behind and falls into a glass), inertia of motion (tendency to keep moving, e.g. a passenger lurches forward when a bus suddenly brakes), and inertia of direction (tendency to keep moving along the same straight line, e.g. mud flies off a spinning wheel tangentially).

The first law also gives an operational (qualitative) definition of force: force is that external agency which is required to change a body's state of rest or uniform motion, i.e., to produce acceleration in it. If no net force acts, velocity stays constant — this includes the special case of zero velocity (rest) as well as any case of several forces balancing out to zero net force.

Worked Example:

Given: A metal ball of mass 2 kg lies on a horizontal, frictionless table. Two horizontal forces act on it simultaneously: 15 N towards the East and 15 N towards the West. The ball is initially at rest.

Formula: By Newton's first law, the state of motion changes only if the net (resultant) external force is non-zero: ΣF = F₁ + F₂

Substitution: Taking East as positive, ΣF = (+15 N) + (−15 N) = 0 N

Result: Since the net external force on the ball is zero, by Newton's first law the ball continues in its original state of rest — it does not accelerate in any direction. This illustrates that it is the net unbalanced force, not the mere presence of forces, that changes a body's state of motion.

Linear momentum (p) of a body is defined as the product of its mass and velocity: it is a vector quantity, pointing in the same direction as the velocity, with SI unit kg·m/s (equivalently N·s). Momentum is a more fundamental quantity than velocity alone because it combines both 'how much mass' and 'how fast', and it is momentum — not velocity — that Newton's second law directly relates to force.

Newton's Second Law of Motion states: the rate of change of momentum of a body is directly proportional to the applied net external force, and takes place in the direction in which the force acts. Mathematically, F = dp/dt. For a body of constant mass m, since p = mv, this becomes F = m(dv/dt) = ma, where a is the acceleration produced. This is the single most-used equation in mechanics.

The SI unit of force, the newton (N), is defined directly from this law: one newton is the force that produces an acceleration of 1 m/s² in a body of mass 1 kg (1 N = 1 kg·m/s²). The second law is a vector law — it holds independently along each of the mutually perpendicular directions (x, y, z), which is why forces and accelerations are resolved into components while solving problems.

An important related idea is impulse. When a large force acts for a very short time (like a bat hitting a ball, or a collision), we often cannot measure the force and the exact time separately, but their product — the impulse J = F·Δt — equals the total change in momentum produced, Δp = m(v − u). This is called the impulse-momentum theorem and is extremely useful for collision and impact problems.

Worked Example:

Given: A constant force acts on a body of mass 4 kg, initially at rest (u = 0), and gives it a velocity of 12 m/s after 3 s.

Formula: (a) a = (v − u)/t (b) F = ma (c) p = mv

Substitution: (a) a = (12 − 0)/3 = 4 m/s². (b) F = 4 kg × 4 m/s² = 16 N. (c) p = 4 kg × 12 m/s = 48 kg·m/s.

Result: The acceleration is 4 m/s², the applied force is 16 N, and the final momentum is 48 kg·m/s. As a check, using the impulse-momentum theorem: J = F·t = 16 N × 3 s = 48 kg·m/s = Δp, which matches, confirming the answer is consistent.

Newton's Third Law of Motion states: to every action, there is always an equal and opposite reaction; the action and reaction forces act on two different bodies (never on the same body), and they act simultaneously — neither force exists without the other, so neither can be called the 'cause' of the other. For example, when a book presses down on a table (action), the table pushes up on the book with an equal and opposite force (reaction, the normal force); when a swimmer pushes water backward, the water pushes the swimmer forward with equal force.

Because action and reaction act on different bodies, they never cancel each other out for the system as a whole, even though they are equal in magnitude and opposite in direction. This is a common source of confusion — action-reaction pairs never produce equilibrium of a single body by themselves; equilibrium of a body requires all the forces acting on that one body to balance.

A major consequence of the third law, combined with the second law, is the Law of Conservation of Linear Momentum: if no external force acts on a system of bodies (an isolated system), the total momentum of the system remains constant, however the bodies of the system might interact among themselves (collisions, explosions, etc.). This follows because the mutual (internal) forces between any two bodies of the system, being action-reaction pairs, are equal and opposite, so their effects on the total momentum cancel out exactly, leaving the total momentum unchanged over time.

For a two-body collision or interaction with no external force, this is written as: total momentum before = total momentum after, i.e., m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂. A classic application is the recoil of a gun: before firing, the gun-bullet system is at rest (total momentum zero); after firing, the forward momentum of the bullet must be exactly balanced by an equal and opposite (backward) momentum of the gun, so that the total remains zero.

Worked Example:

Given: A gun of mass M = 5 kg, initially at rest, fires a bullet of mass m = 0.02 kg with a muzzle velocity of 400 m/s.

Formula: Conservation of momentum (system initially at rest): total momentum before = total momentum after ⇒ 0 = mv_bullet + MV_gun, so MV_gun = −mv_bullet (opposite direction), or in magnitude: mv_bullet = MV_gun

Substitution: 0.02 kg × 400 m/s = 5 kg × V_gun ⇒ 8 = 5 × V_gun

Result: V_gun = 8/5 = 1.6 m/s, directed opposite to the bullet's motion (recoil). The much larger mass of the gun means it recoils with a much smaller speed than the bullet's, even though both have equal and opposite momenta (0.02 × 400 = 5 × 1.6 = 8 kg·m/s on each side).

A particle (or a body treated as a point mass) is said to be in equilibrium when the vector sum of all the external forces acting on it is zero: ΣF = 0. By Newton's first/second law, this means the particle has zero acceleration — it is either at rest or moving with constant velocity. Equilibrium under two forces requires them to be equal, opposite, and collinear. Equilibrium under three concurrent, coplanar forces is common in problems (e.g., a lamp hanging from two strings) and is conveniently solved using Lami's theorem: if three concurrent forces F₁, F₂, F₃ keep a particle in equilibrium, then each force is proportional to the sine of the angle between the other two: F₁/sin α = F₂/sin β = F₃/sin γ, where α is the angle between F₂ and F₃, and so on.

Some forces that appear repeatedly in mechanics problems:

• Weight (W = mg): the gravitational force on a body, always directed vertically downward, towards the centre of the Earth.
• Normal reaction (N): the contact force a surface exerts on a body perpendicular to the surface, preventing the body from penetrating it; its magnitude adjusts automatically to whatever is needed for equilibrium in the perpendicular direction (up to physical limits).
• Tension (T): the pulling force transmitted through a string, rope, or cable, directed along the string, away from the body, and (for an ideal massless, inextensible string) equal in magnitude throughout the string.
• Spring force: given by Hooke's Law, F = −kx, where k is the spring constant (stiffness, unit N/m) and x is the displacement from the natural (unstretched) length; the negative sign shows the force always opposes the displacement (restoring force).

To solve equilibrium problems, forces are resolved along two perpendicular axes (usually horizontal and vertical), and the sum of components along each axis is separately set to zero: ΣF_x = 0 and ΣF_y = 0.

Worked Example:

Given: A signboard of weight W = 100 N hangs in equilibrium from a point O, supported by two strings OA and OB fixed to a ceiling. String OA makes an angle of 30° with the horizontal ceiling, and string OB makes an angle of 60° with the horizontal ceiling, on the other side (so OA is perpendicular to OB). Find the tensions T₁ (in OA) and T₂ (in OB).

Formula: Resolve forces at O. Horizontal equilibrium: T₁ cos30° = T₂ cos60°. Vertical equilibrium: T₁ sin30° + T₂ sin60° = W.

Substitution: From the horizontal equation: T₁(√3/2) = T₂(1/2) ⇒ T₂ = √3 T₁. Substituting into the vertical equation: T₁(1/2) + √3T₁(√3/2) = 100 ⇒ T₁/2 + 3T₁/2 = 100 ⇒ 2T₁ = 100.

Result: T₁ = 50 N, and T₂ = √3 × 50 ≈ 86.6 N. The string making the smaller angle with the horizontal (OA) carries less tension than the steeper string (OB), which supports more of the vertical load.

Friction is the opposing (tangential) force that arises at the surface of contact between two bodies when one body moves or tends to move relative to the other. Friction always acts along the surface of contact, opposite to the direction of relative motion (or tendency of motion).

Static friction (f_s) acts when there is no relative sliding — it is a self-adjusting force that exactly balances any applied force up to a certain maximum, keeping the body at rest. If the applied force exceeds this maximum, the body begins to slide. The maximum value is called limiting friction, f_s(max) = μ_sN, where μ_s is the coefficient of static friction and N is the normal reaction. In general, f_s ≤ μ_sN.

Kinetic (sliding) friction (f_k) acts once the body is actually sliding over the surface, and is given by f_k = μ_kN, where μ_k is the coefficient of kinetic friction. Experimentally, μ_k is found to be slightly less than μ_s for the same pair of surfaces — this is why it takes a bit more force to start an object moving than to keep it moving at constant velocity.

The laws of friction (empirical, approximately valid) state that: (i) the magnitude of limiting/kinetic friction is directly proportional to the normal reaction N, and the ratio f/N (= μ) is called the coefficient of friction, a dimensionless number depending on the nature and roughness of the two surfaces in contact; (ii) friction is approximately independent of the apparent area of contact (for given N); and (iii) kinetic friction is nearly independent of the relative speed of sliding, over ordinary speed ranges. Rolling friction (for a body rolling without slipping) is much smaller than sliding/kinetic friction, which is why wheels and ball-bearings are used to reduce friction in machinery.

The angle of friction (φ) is the angle the resultant of the normal reaction and the limiting friction force makes with the normal; it satisfies tanφ = μ_s. Closely related is the angle of repose — the maximum angle of inclination of a rough surface at which a body placed on it just remains on the verge of sliding down under gravity — which is numerically equal to the angle of friction: tanθ = μ_s.

Worked Example:

Given: A block of mass 4 kg rests on a rough horizontal surface. The coefficient of static friction μ_s = 0.5 and the coefficient of kinetic friction μ_k = 0.4. Take g = 9.8 m/s².

Formula: Normal reaction on horizontal surface: N = mg. Maximum static friction (force needed to just start motion): f_s(max) = μ_sN. Kinetic friction once sliding (force needed to keep it moving at constant velocity): f_k = μ_kN.

Substitution: N = 4 × 9.8 = 39.2 N. f_s(max) = 0.5 × 39.2 = 19.6 N. f_k = 0.4 × 39.2 = 15.68 N.

Result: A minimum horizontal force of 19.6 N is required to just start the block sliding, but once it is sliding, a smaller force of only 15.68 N is enough to keep it moving at constant velocity (any extra force beyond this would accelerate it, by the second law).

In uniform circular motion, a body moves along a circular path at constant speed, but its velocity direction is continuously changing, so it is always accelerating. This acceleration, called centripetal acceleration, is directed radially inward, toward the centre of the circle, with magnitude a_c = v²/r = ω²r, where v is the speed, r is the radius of the circular path, and ω is the angular speed (v = ωr).

By Newton's second law, a net inward force must act on the body to produce this acceleration — this is the centripetal force, F_c = mv²/r = mω²r. It is important to understand that 'centripetal force' is not a new, separate type of force; it is simply the name given to whatever net real force (tension, gravity, normal reaction, friction, or a combination) happens to be directed toward the centre and is responsible for the circular motion. For a stone whirled on a string, it is the string's tension; for a satellite, it is gravity; for a car turning on a flat road, it is friction between the tyres and road.

On a flat (unbanked) road, only friction between the tyres and the road can supply the centripetal force, which limits the maximum safe speed for turning (since friction has a maximum value μ_smg). To allow higher, safer speeds — and to reduce dependence on friction, which can be unreliable in wet or icy conditions — roads and railway tracks are banked: the outer edge of the curved road is raised above the inner edge, tilting the road surface at an angle θ to the horizontal. This tilts the normal reaction N, giving it a horizontal component that can itself provide (all or part of) the required centripetal force.

For an ideal banked road, designed so that a vehicle moving at the design speed v needs no friction at all to negotiate the curve safely, balancing vertical and horizontal components of N gives: N cosθ = mg (vertical) and N sinθ = mv²/r (horizontal, centripetal). Dividing these two equations eliminates both N and m: tanθ = v²/(rg). This is the standard relation used to design the banking angle for a given radius and design speed. (When friction is also present, the maximum safe speed becomes v_max = √[rg(μ_s + tanθ)/(1 − μ_stanθ)], showing friction extends the safe speed range beyond the frictionless design speed.)

Worked Example 1 (Centripetal force):

Given: A stone of mass 0.5 kg, tied to a string of length 1 m, is whirled in a horizontal circle at a constant speed of 4 m/s.

Formula: F_c = mv²/r

Substitution: F_c = 0.5 × (4)² / 1 = 0.5 × 16 / 1

Result: F_c = 8 N. This is the tension the string must supply, directed towards the centre of the circle, to keep the stone moving in its circular path.

Worked Example 2 (Banking of roads):

Given: A road is banked at an angle θ = 15° for a curve of radius r = 100 m. Find the safe speed at which a car can travel around the curve without depending on friction. Take g = 9.8 m/s² (tan15° ≈ 0.268).

Formula: tanθ = v²/(rg) ⇒ v = √(rg tanθ)

Substitution: v = √(100 × 9.8 × 0.268) = √262.6

Result: v ≈ 16.2 m/s (about 58.3 km/h). This is the ideal, frictionless design speed for this banked curve; a car moving exactly at this speed needs no friction at all to safely go around the turn.