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Newton's Laws & Forces

Why things move: forces, friction, and the laws that govern them. 50 topics, from beginner to advanced. Open the interactive version with labs and quizzes

Dynamics asks why objects move the way they do. Newton's three laws link forces to changes in motion, introduce inertia and mass, and explain how contact forces such as friction, tension and normal force act on bodies. Free-body diagrams are the main tool for turning a physical situation into equations you can solve.

How the topics connect

Dynamics asks why things move as they do. Newton's laws connect forces to changes in motion, and this area builds from single forces and free-body diagrams to friction, circular motion and rotating frames. Kinematics tells you what happens; dynamics explains why.

Real-world uses

Structural and vehicle design

Calculate the forces in cables, supports and brakes so that a structure or vehicle stays in equilibrium or decelerates safely.

Sport and biomechanics

Explain why a sprinter pushes backwards on the blocks to move forward, and how friction limits grip in a turn.

Lifts and pulley systems

Work out the tension in ropes and the acceleration of a lift or counterweight using Newton's second law.

Safety and crash analysis

Use inertia and force to understand why seatbelts and crumple zones reduce injury in a collision.

Vehicle design

Engineers use friction, banking and centripetal force to predict how fast a car can take a bend safely.

Structures and machines

Civil and mechanical engineers draw free-body diagrams to check that every support, cable and joint can carry its loads.

Beginner

What is a force?

A push or pull that can change an object's motion or shape.

Key relationF = ma

A force is an interaction that can change a body's velocity or deform it. It is a vector, measured in newtons, and it always comes from another object. The net force, the vector sum of all forces, determines the acceleration.

Worked example

A 2 N push and a 3 N push in the same direction give 5 N; in opposite directions, 1 N.

Common mistake

Forces are not carried by moving objects; "force of motion" is not a thing.

Beginner

Contact and non-contact forces

Contact forces need touching; gravity, electric and magnetic act at a distance.

Contact forces (normal, friction, tension, drag) act where bodies touch. Field forces (gravity, electric, magnetic) act across empty space through fields. Microscopically, contact forces are electromagnetic.

Worked example

Gravity pulls a falling apple with no contact, while a table supports a book by contact.

Common mistake

Contact force on a surface has two components: normal (perpendicular) and friction (parallel).

Beginner

Newton's first law

A body keeps its state of rest or uniform motion unless acted on by a net force.

A body at rest stays at rest and a body in motion keeps moving at constant velocity unless a net external force acts. Motion doesn't need a force; changing motion does.

Worked example

A hockey puck on smooth ice keeps gliding almost unchanged.

Common mistake

Everyday objects stop because friction is a force, not because "motion fades".

Beginner

Inertia

The resistance of a body to changes in its motion; measured by mass.

Inertia is the tendency to resist changes in motion. Mass is its measure: the larger the mass, the larger the force needed for a given acceleration.

Worked example

A loaded truck needs a longer distance to stop than an empty one at the same speed.

Common mistake

Inertia isn't a force; it's a property.

Beginner

Newton's second law

Net force equals mass times acceleration.

Key relationΣF = ma

The net force on a body equals its mass times its acceleration, ΣF = ma, and acceleration points along the net force. Apply it separately to each axis.

Worked example

A 5 kg block pushed by 20 N with 5 N friction: a = 15/5 = 3 m/s².

Common mistake

Use the net force, including friction, not just the applied force.

Beginner

Units of force

One newton accelerates one kilogram at one metre per second squared.

Key relation1 N = 1 kg·m/s²

One newton gives 1 kg an acceleration of 1 m/s². Weight of an apple (~0.1 kg) is about 1 N. A kilogram-force is 9.8 N, an older unit.

Worked example

A 70 kg person weighs about 686 N.

Common mistake

Mass in grams must be converted to kg before using F = ma.

Beginner

Newton's third law

Forces come in equal and opposite pairs acting on different bodies.

Key relationF_AB = −F_BA

If A pushes on B, B pushes back on A equally and oppositely. The two forces are of the same type and act on different bodies, so they never cancel in a single free-body diagram.

Worked example

A rocket pushes gas back; the gas pushes the rocket forward.

Common mistake

Weight and the normal force on a resting book are not a third-law pair.

Beginner

Weight and normal force

Weight pulls down; the surface pushes back perpendicular to it.

Key relationN = mg

Weight W = mg pulls down. The normal force is the surface's push perpendicular to the contact, adjusting to whatever stops penetration. On a level surface at rest N = mg, but it changes on slopes and in accelerating lifts.

Worked example

On a 30° slope a 10 kg block has N = 98 cos30° ≈ 85 N.

Common mistake

N isn't always equal to mg.

Beginner

Free-body diagrams

Sketch all forces on one body to apply Newton's laws.

Draw the body alone as a dot and draw every force acting on it with arrows, then choose axes and write ΣF = ma for each. Include only forces on that body, not forces it exerts.

Worked example

A block on a slope: weight, normal, friction; resolve along and perpendicular to the slope.

Common mistake

Do not draw the acceleration or "force of motion" as a force.

Beginner

Net force and equilibrium

Zero net force means zero acceleration.

Key relationΣF = 0

If the net force is zero the acceleration is zero: the body is at rest or moving at constant velocity (translational equilibrium). All forces form a closed polygon.

Worked example

A 10 kg lamp hung by two cables is in equilibrium when the vertical components of the tensions sum to 98 N.

Common mistake

Equilibrium doesn't mean no forces, only balanced forces.

Beginner

Tension

Pulling force transmitted along a rope or string.

Tension is the pulling force transmitted through a rope or string. For a light, inextensible rope over a smooth pulley the tension is the same everywhere along it. It pulls on whatever the rope is attached to.

Worked example

A 3 kg mass hung at rest has T = 29.4 N.

Common mistake

Tension changes if the rope has mass or rubs a rough pulley.

Intermediate

Atwood machine

Two masses over a pulley accelerate by the difference in weights.

Key relationa = (m₁−m₂)g/(m₁+m₂)

Two masses on either side of a light pulley accelerate at a = (m₁ − m₂)g/(m₁ + m₂), the heavier going down. The tension is T = 2m₁m₂g/(m₁ + m₂).

Worked example

3 kg and 1 kg: a = 2·9.8/4 = 4.9 m/s², T = 14.7 N.

Common mistake

T is not equal to either weight when accelerating.

Intermediate

Connected bodies

Bodies linked by strings share acceleration in the same direction of motion.

For bodies joined by a rope, they share the same acceleration magnitude. Write ΣF = ma for each body separately, or for the whole system when the internal tensions can be ignored.

Worked example

A 4 kg block pulls a 2 kg block with 12 N: a = 12/6 = 2 m/s², tension on the second block 4 N.

Common mistake

Don't forget that the tension acts on both bodies in opposite directions.

Intermediate

Elevator problems

Apparent weight changes with the elevator's acceleration.

Key relationN = m(g+a)

Apparent weight is the normal force on you: N = m(g + a) for upward acceleration a, m(g − a) for downward. In free fall N = 0 (weightlessness).

Worked example

70 kg in a lift accelerating up at 2 m/s²: N = 70·11.8 = 826 N.

Common mistake

Your mass doesn't change; your apparent weight does.

Intermediate

Pulley systems

Ideal pulleys change force direction; movable pulleys give mechanical advantage.

Ideal pulleys are frictionless and massless: they change the direction of the tension but not its size. A movable pulley supports a load with two rope segments, halving the force but doubling the rope pulled.

Worked example

A system with a movable pulley lifts 100 N with 50 N, pulling twice the length.

Common mistake

Energy is not gained: force advantage costs distance.

Intermediate

Motion on an inclined plane

Component of weight along the plane drives motion.

Key relationa = g sinθ

Resolve weight into mg sinθ along the slope and mg cosθ perpendicular. Without friction a = g sinθ; with friction a = g(sinθ − μ cosθ).

Worked example

At 30° frictionless: a = 4.9 m/s².

Common mistake

N = mg cosθ, not mg.

Beginner

Static friction

Friction that prevents sliding, up to a maximum value.

Key relationf_s ≤ μ_s N

Static friction opposes impending sliding and adjusts up to a maximum fs,max = μₛN. Below that value it exactly balances the applied force parallel to the surface.

Worked example

A 10 kg crate with μₛ = 0.4 needs more than 39 N to start moving.

Common mistake

Static friction is not always μN; it equals the applied force until it reaches the limit.

Beginner

Kinetic friction

Friction during sliding, usually less than the static maximum.

Key relationf_k = μ_k N

When sliding, friction is fk = μₖN, usually smaller than the static maximum, and independent of speed and contact area (to a good approximation). It always opposes the relative motion.

Worked example

A 10 kg crate with μₖ = 0.3 feels 29.4 N while sliding.

Common mistake

Friction does not depend on area in the simple model.

Intermediate

Angle of friction and repose

On an incline, sliding starts when tanθ exceeds μ.

Key relationtanθ = μ

The angle of repose is the steepest slope at which an object stays at rest: tanθ = μₛ. Beyond it the block slides. It tells you μ by tilting a plane.

Worked example

A block starts sliding at 25°: μₛ = tan25° ≈ 0.47.

Common mistake

Mass doesn't affect the angle.

Intermediate

Rolling friction

Smaller resistance of rolling wheels compared with sliding.

A rolling wheel deforms the surface slightly, giving a resisting force much smaller than sliding friction (Frr = μᵣN with μᵣ ≈ 0.01). This is why wheels are so useful.

Worked example

A car tyre on road: μᵣ ≈ 0.015.

Common mistake

Pure rolling without slipping involves static friction at the contact point that does no work.

Intermediate

Friction on inclines with applied force

Include friction direction opposing impending motion.

Pushing a block up a slope, friction points down the slope; pushing it down, friction acts up the slope. Write along-slope balance: F = mg sinθ ± μmg cosθ.

Worked example

To push a 10 kg block at constant speed up a 30° slope with μ = 0.2: F = 49 + 17 = 66 N.

Common mistake

The sign of friction depends on the direction of motion or tendency.

Advanced

Block on block problems

Friction between stacked blocks sets maximum common acceleration.

For stacked blocks, friction between them provides the top block's acceleration, limited by μₛ m g. The top block slips when the common acceleration needed exceeds μₛg.

Worked example

Top 2 kg, bottom 4 kg, μₛ = 0.5: the maximum shared acceleration is 4.9 m/s².

Common mistake

Analyse each block separately; friction pairs are equal and opposite.

Intermediate

Impulse

Force integrated over time, equal to change in momentum.

Key relationJ = ∫F dt = Δp

Impulse is the force integrated over time and equals the change in momentum: J = ∫F dt = Δp. For constant force J = FΔt. The area under an F–t graph gives it.

Worked example

A 0.15 kg ball hit from +20 to −30 m/s: Δp = 7.5 N·s.

Common mistake

Direction matters: reversing direction doubles the change.

Beginner

Linear momentum

Mass times velocity.

Key relationp = mv

Momentum p = mv is a vector measuring quantity of motion. It's what makes heavy, fast things hard to stop.

Worked example

A 1000 kg car at 20 m/s has p = 20,000 kg·m/s.

Common mistake

Momentum depends on velocity, not speed.

Intermediate

Second law as rate of momentum change

Force is the time rate of change of momentum.

Key relationF = dp/dt

Newton's original statement is F = dp/dt. It reduces to ma for constant mass but also covers changing mass (rockets).

Worked example

A force of 10 N acting for 3 s changes the momentum by 30 kg·m/s.

Common mistake

F = ma only when m is constant.

Advanced

Variable mass systems

Rockets and conveyor belts require the full dp/dt form.

Key relationF = m dv/dt + v dm/dt

When mass enters or leaves a system, apply F = dp/dt carefully to a defined system including the added mass. Examples: a rocket, a leaking cart, rain falling in a moving wagon.

Worked example

A coal-loaded belt needs extra force F = v dm/dt to accelerate new coal to belt speed.

Common mistake

Do not simply use m dv/dt without accounting for the mass change.

Advanced

Rocket propulsion

Exhaust gas gives thrust and velocity change.

Key relationΔv = v_e ln(m₀/m)

A rocket expels gas backward at speed vₑ relative to itself, gaining forward momentum. Thrust is F = vₑ dm/dt, and the final speed from the rocket equation is Δv = vₑ ln(m₀/m).

Worked example

With vₑ = 3 km/s and mass ratio 10: Δv ≈ 6.9 km/s.

Common mistake

Rockets work in vacuum; they don't push against air.

Intermediate

Conservation of momentum

Total momentum of an isolated system stays constant.

Key relationΣp_i = Σp_f

In an isolated system the total momentum is constant, because internal forces cancel in pairs. It applies to collisions, explosions and recoil, even when energy isn't conserved.

Worked example

A 60 kg skater pushes a 40 kg skater: velocities are in ratio 2:3 opposite.

Common mistake

External forces such as friction must be negligible during the short collision.

Intermediate

Circular motion dynamics

Net inward force provides centripetal acceleration.

Key relationF = mv²/r

For circular motion, the net force toward the centre is mv²/r. Identify which actual force(s) supply it (tension, friction, normal force or gravity) and set their inward sum equal to mv²/r.

Worked example

A 0.5 kg stone on a 1 m string at 4 m/s: T = 0.5·16/1 = 8 N.

Common mistake

Centripetal force isn't a new force; it's the name for the net inward one.

Intermediate

Banking of roads

Bank angle supplies centripetal force without friction at design speed.

Key relationtanθ = v²/rg

On a banked curve, the horizontal component of the normal force provides the centripetal force. With no friction, tanθ = v²/(rg); with friction, vmax = √(rg(μ + tanθ)/(1 − μ tanθ)).

Worked example

For v = 20 m/s and r = 100 m: tanθ = 0.41, θ ≈ 22°.

Common mistake

Don't resolve forces along the slope when the acceleration is horizontal; use horizontal and vertical.

Intermediate

Vehicle on flat circular turn

Friction provides centripetal force; maximum speed depends on μ.

Key relationv_max = √(μrg)

On a flat turn the centripetal force is static friction: μmg ≥ mv²/r, so vmax = √(μrg). Wet roads lower μ and the safe speed.

Worked example

μ = 0.7, r = 50 m: vmax ≈ 18.5 m/s.

Common mistake

The speed limit doesn't depend on the car's mass.

Intermediate

Conical pendulum

A bob moving in a horizontal circle with the string tracing a cone.

Key relationT = 2π√(L cosθ/g)

A bob on a string moves in a horizontal circle. T cosθ = mg and T sinθ = mv²/r give tanθ = v²/(rg) and period 2π√(L cosθ/g).

Worked example

L = 1 m, θ = 30°: T ≈ 1.87 s.

Common mistake

The tension is bigger than the weight.

Advanced

Vertical circular motion

Tension and speed vary around the loop; minimum speed at the top sets the loop condition.

Key relationv_top ≥ √(gr)

In a vertical circle the speed and tension vary: at the top T + mg = mv²/r, at the bottom T − mg = mv²/r. The string stays taut only if v_top ≥ √(gr).

Worked example

A ball on a 1 m string must have at least 3.13 m/s at the top.

Common mistake

The tension at the top is smaller than at the bottom.

Advanced

Loop-the-loop

Minimum speed at bottom to complete a loop is √(5gr).

Key relationv_bottom = √(5gr)

To complete a vertical loop of radius r, the minimum speed at the top is √(gr) and so, from energy conservation, the minimum at the bottom is √(5gr).

Worked example

r = 5 m: v_bottom ≥ 15.7 m/s.

Common mistake

At the top the normal force is zero at the critical speed; the object is not weightless in other senses.

Advanced

Centrifugal force

A fictitious outward force in a rotating frame.

Key relationF = mω²r

In a rotating frame a body appears pushed outward with fictitious force mω²r. In the inertial frame there is no such force; the body simply tends to go straight.

Worked example

Mud flies off a wheel because there's not enough force to keep it in a circle.

Common mistake

Don't draw centrifugal force in an inertial-frame diagram together with the centripetal force.

Advanced

Pseudo forces

In accelerating frames add −ma to apply Newton's laws.

Key relationF_pseudo = −ma

In an accelerating frame with acceleration a, every body feels an extra force −ma in that frame. Using it, Newton's laws apply as usual in the non-inertial frame.

Worked example

In a braking car (a = −3 m/s²) a 60 kg passenger feels +180 N forward.

Common mistake

Pseudo forces have no reaction partner.

Intermediate

Drag force

Fluid resistance grows with speed.

Key relationF = ½ρC_dAv²

Drag opposes motion through a fluid: F = ½ρC_dAv² at high speed (turbulent), growing with the square of speed. It depends on shape (C_d), frontal area A and density ρ.

Worked example

Doubling the speed of a cyclist increases drag fourfold and the power needed eightfold.

Common mistake

At low speeds drag is closer to linear in v.

Advanced

Stokes' law

Viscous drag on a small sphere at low speed.

Key relationF = 6πηrv

For a small sphere moving slowly through a viscous fluid, the drag is F = 6πηrv, proportional to speed. It governs the settling of fine particles and oil drops in Millikan's experiment.

Worked example

A 1 mm steel ball in glycerol quickly reaches a steady slow speed.

Common mistake

It holds only for low Reynolds number (smooth flow).

Advanced

Terminal velocity of a sphere

Reaches constant speed when weight, buoyancy and drag balance.

Key relationv_t = 2r²(ρ−σ)g/9η

At terminal speed the weight equals the buoyancy plus drag: v_t = 2r²(ρ − σ)g/(9η). The speed grows with the square of the radius, so large drops fall faster than small ones.

Worked example

Fog droplets with r = 10 μm fall a few millimetres per second.

Common mistake

This formula needs Stokes' law to be valid.

Beginner

Spring force

A spring pulls back in proportion to its stretch.

Key relationF = −kx

A spring exerts F = −kx, a restoring force proportional to its extension from the natural length, pointing opposite to the displacement. k (N/m) measures stiffness.

Worked example

k = 200 N/m stretched 5 cm: force 10 N.

Common mistake

x is the stretch from the natural length, not the total length.

Intermediate

Springs in series and parallel

Series reduces stiffness; parallel increases it.

Key relationk_p = k₁+k₂

Springs in parallel share the displacement: k = k₁ + k₂ (stiffer). Springs in series share the force: 1/k = 1/k₁ + 1/k₂ (softer).

Worked example

Two 100 N/m springs: parallel 200, series 50 N/m.

Common mistake

Series is the opposite of resistors in series for the same-looking formula; watch which quantity is shared.

Intermediate

Hooke's law limits

Beyond the elastic limit, deformation is permanent.

Hooke's law holds only up to the proportional limit. Beyond the elastic limit the material deforms permanently, and finally breaks. Real springs and wires follow F = kx only for small extensions.

Worked example

A paperclip pulled too far stays bent.

Common mistake

Never extrapolate F = kx for large stretches.

Intermediate

Equilibrium of concurrent forces

Lami's theorem links three forces in equilibrium.

Key relationF₁/sinα = F₂/sinβ = F₃/sinγ

Three forces in equilibrium on a point form a closed triangle, so Lami's theorem applies: F₁/sinα = F₂/sinβ = F₃/sinγ, with each angle opposite its force.

Worked example

A 100 N weight hanging from two ropes at 45° each: each tension ≈ 70.7 N.

Common mistake

Lami's angles are between the other two forces, opposite the one in the numerator.

Intermediate

Torque and equilibrium

Balance both forces and torques for static equilibrium.

Key relationΣτ = 0

A rigid body in equilibrium needs ΣF = 0 and Στ = 0 about any point. Choose a pivot that eliminates unknown forces to simplify.

Worked example

A 4 m plank, 20 kg, supported at its ends with a 10 kg load at 1 m: reaction at the far end by moments.

Common mistake

Moments must be taken about the same point for all forces.

Intermediate

Center of mass

The point where the system's mass behaves as concentrated.

Key relationx_cm = Σm x/Σm

The centre of mass is the mass-weighted average position: x_cm = Σm x/Σm. Net external force accelerates it as if all mass were there, no matter how complex the internal motion.

Worked example

Masses 1 kg at 0 and 3 kg at 4 m: x_cm = 3 m.

Common mistake

It may lie outside the body (a ring).

Intermediate

Motion of center of mass

The centre of mass moves as if all external forces act on it.

The centre of mass of a system moves as a point of mass M under the total external force: F_ext = M a_cm. Internal forces such as explosions don't change it.

Worked example

A thrown wrench spins complexly but its centre of mass follows a parabola.

Common mistake

An exploding firework's centre of mass keeps its projectile path until things hit the ground.

Intermediate

Normal force on curves

Normal force combines weight component and centripetal requirement.

On a curved path the normal force N adjusts to both gravity and the centripetal need. At the top of a hill N = mg − mv²/r (lighter); at the bottom of a dip N = mg + mv²/r (heavier).

Worked example

At 15 m/s over a 50 m radius crest, a 1000 kg car has N = 9800 − 4500 = 5300 N.

Common mistake

The car loses contact if v² ≥ gr at a crest.

Advanced

Constraint forces

Forces that enforce geometric restrictions such as strings and surfaces.

Constraint forces (normal, tension) arise to enforce a geometric restriction and adjust their size to whatever is needed. They have no fixed formula and are found from the equations of motion.

Worked example

A bead on a wire exerts whatever force is needed to keep it on the wire.

Common mistake

Constraint forces do no work if the constraint doesn't move.

Advanced

Newton's laws and inertial frames

The laws hold in inertial frames; failing that, add pseudo forces.

Newton's laws hold exactly only in inertial frames. Earth is nearly inertial for short experiments. To use Newton's laws in rotating or accelerating frames, add fictitious forces.

Worked example

A Foucault pendulum's rotating swing plane reveals Earth's rotation.

Common mistake

The laws don't claim that the frame is special, only that it is inertial.

Advanced

Dimensions and limits of Newtonian mechanics

Breaks down at high speeds and small scales, replaced by relativity and quantum mechanics.

Newtonian mechanics works for everyday sizes and speeds but fails at speeds near c (relativity), at atomic scales (quantum mechanics) and in strong gravity (general relativity). It's still an excellent approximation within its domain.

Worked example

GPS satellites need relativity but car engineering does not.

Common mistake

A theory's limits are a feature of science, not a failure.

Frequently asked questions

What is inertia?

Inertia is the tendency of an object to keep its current state of motion. The more mass an object has, the more inertia it has, and the harder it is to change its velocity.

Why does a heavier object not fall faster than a light one in a vacuum?

Gravity pulls harder on a heavier object, but it also has more inertia. The two effects cancel, so all objects fall with the same acceleration, g, when air resistance is negligible.

What is the difference between static and kinetic friction?

Static friction acts on objects that are not sliding and can take any value up to a maximum. Kinetic friction acts on sliding objects and is usually smaller and roughly constant.

What is the most common mistake in Newton's laws problems?

Drawing forces that the object does not experience, such as including the force it exerts on something else. Draw a free-body diagram showing only forces acting on the chosen object.

Is there a force that pushes an object forward when it moves?

No. Forward motion needs a net force in the direction of acceleration, but no separate forward force is needed to keep a body moving at constant velocity.

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