Exploring Mechanics / 2. Newton's Laws
This section explains why motion changes.
Make a prediction first: a magician yanks a tablecloth out from under a full table setting, and the plates barely move. A driver slams the brakes, and an unbelted passenger keeps sliding forward even though the car has already stopped. What do these two tricks have in common? Both are Newton's First Law doing exactly what it says: an object keeps doing whatever it was already doing — staying still, or moving at constant velocity — until a net force makes it do otherwise.
ΣF = 0 ⟹ a = 0 ⟹ constant velocity (including velocity = 0, i.e. staying at rest)
Inertia isn't a force — it's a property. It's the reason a net force is required to change motion at all, and the reason two objects of different mass respond differently to the same force. Every widget below is really the same law, staged three different ways: something continues moving (or sitting still) simply because nothing has managed to force it to do otherwise yet.
Newton's First Law is often mis-stated as "an object at rest stays at rest, and a moving object keeps moving, unless a force acts on it." The precise statement uses net force: a book sitting on a table has two real forces on it (gravity pulling down, the table pushing up) that are individually nonzero but sum to ΣF = 0 — that's why it doesn't accelerate. A hockey puck gliding at constant velocity on ice isn't force-free either (gravity and the ice's support still act on it); it's simply that those vertical forces balance and the horizontal friction force is small enough to ignore over a short glide. "Constant velocity" is the general case; "at rest" is just the special case where that constant velocity happens to be zero.
This is the tablecloth trick shrunk down to a coin balanced on a playing card over a cup's opening. Flick the card away fast enough and the coin barely moves before the card is gone — then gravity (not the flick) drops it straight into the cup. Flick too gently, or use too high a friction coefficient, and the coin gets dragged along with the card instead.
While the card is still under the coin, kinetic friction is the only horizontal force on the coin, giving it acceleration a = μg — fixed, independent of how fast you flick. What changes with flick speed is time: the faster the card moves, the sooner its trailing edge clears the coin's position, so the coin only gets to accelerate for a shorter contact time tcontact. Its sideways drift during contact is ½at² — squared in a quantity that's shrinking — so a faster flick reduces drift dramatically, not just a little. That's the whole trick: outrun the friction, not overpower it.
A car brakes hard. Its unbelted passenger doesn't get pushed forward by any force — nothing is pushing them — they simply keep moving at the speed they already had, per Newton's First Law, while the car (which the brakes are acting on) slows down underneath them. From inside the car, that looks exactly like being "thrown forward." Toggle the seatbelt to see what force it takes to make the passenger decelerate along with the car instead.
Without a belt, the passenger has zero net horizontal force on them (ignoring the small friction of a car seat), so by the First Law their position is just xpax(t) = v₀t — unchanged constant-velocity motion. The car, acted on by the brakes, follows xcar(t) = v₀t − ½at². The passenger's drift relative to the car's cabin is the difference: Δx = xpax − xcar = ½at², growing the whole time the car is still decelerating. With the seatbelt on, the belt supplies exactly the extra backward force needed to give the passenger the same deceleration a as the car, so Δx stays at essentially zero — the belt is what turns "passenger" and "car" into one connected system.
Newton's First Law describes an idealized world with zero net horizontal force. Real surfaces always have some friction, but you can watch motion approach that ideal as friction shrinks: push two identical pucks with the same speed, one on a low-friction surface (ice) and one on a higher-friction one (carpet), and compare how far each coasts before stopping.
Kinetic friction gives each puck a constant deceleration a = μg (friction force μmg divided by mass m — the mass itself cancels out, which is why heavier pucks don't inherently coast farther or shorter on the same surface). Using v² = v₀² − 2as with final velocity v = 0: s = v₀² / (2μg). Push twice as hard (double v₀) and the coasting distance quadruples, not doubles — a fact drivers should find uncomfortably relevant to braking distance, covered properly in the next tabs.
Make a prediction first: you push an empty shopping cart and a fully loaded one with the exact same effort. Which one speeds up faster? Now push the empty cart twice as hard. Does it speed up twice as fast, or by some other factor? Newton's Second Law answers both questions with one equation.
F = ma
Rearranged as a = F/m, the law says acceleration is directly proportional to net force and inversely proportional to mass. Double the force on the same object and it accelerates twice as fast. Double the mass under the same force and it accelerates half as fast. Neither relationship is obvious until you've felt both — which is exactly what Widget 1 lets you compare directly.
F here is always the net (vector-summed) force — if three forces pull an object in different directions, you add them as vectors first, then apply F = ma to that single resultant. Also note the equation uses m (mass, an intrinsic property, the same on the Moon as on Earth), not weight (mg, which depends on local gravity). A given net force produces the exact same acceleration on the Moon as on Earth, even though that object weighs six times less there — mass is what actually resists acceleration, not weight.
Push a cart with a chosen force and watch it accelerate — then switch which quantity you vary. In "vary mass" mode the applied force stays fixed while you change the load; in "vary force" mode the mass stays fixed while you change how hard you push. Either way, the live graph plots exactly how a responds.
Holding F fixed, a(m) = F/m is an inverse relationship — its graph is a hyperbola that drops steeply at first and then flattens, because doubling an already-large mass matters less in absolute terms than doubling a small one. Holding m fixed, a(F) = F/m is directly proportional — a straight line through the origin with slope 1/m. Same equation, two very differently shaped graphs, purely because of which variable is doing the varying.
Real pushes and pulls also fight friction. Increase the pull force past the sled's friction resistance and it accelerates; stay below that threshold and it doesn't move at all — friction isn't a "slowing-down tax," it's a real force that can cancel your entire push.
Drag the tips of the two force vectors below to change their size and direction. The dashed green arrow is their vector sum — the single net force that actually determines how the object accelerates. Press release to see the object move in that direction.
Force vectors add component-by-component just like the displacement and velocity vectors from the Toolkit: ΣF = (F₁ₓ + F₂ₓ, F₁ᵥ + F₂ᵥ). The block only cares about that single resultant — it has no way to "know" it's being pulled by two separate forces instead of one equivalent force, which is exactly why free-body diagrams are drawn as one summed arrow when you only need the resulting motion.
Make a prediction first: two ice skaters of very different sizes push off from each other, each using exactly the same push. Does the lighter skater feel a bigger shove than the heavier one? Or do they experience the same force, and just respond to it differently?
For every force one object exerts on a second object, the second exerts a force equal in magnitude and opposite in direction back on the first.
Common misconception: action-reaction pairs do not cancel out, because they act on two different objects. A force only cancels another force acting on the same object. The push you feel and the push you give are a real pair of equal-and-opposite forces that never cancel — they just act on you and on whatever you pushed, separately.
The forces are always equal (FA on B = −FB on A), but Newton's Second Law (a = F/m) still applies separately to each object with its own mass. Equal force, different mass, therefore different acceleration — the lighter skater always flies off faster than the heavier one, even though the push between them was perfectly equal in both directions the entire time.
Pick a scenario, set the two masses, and push off. Both objects feel the exact same force (shown in the readout) for the exact same instant — watch how differently they end up moving anyway.
Swap "skater B" for "the entire Earth" (via the wall you're pushing, bolted to the ground) and the force pair is still exactly equal: you push the wall with some force, the wall pushes back on you with the same force, which is why you feel resistance at all. But Earth's mass is on the order of 10²⁴ kg, so its resulting acceleration a = F/m is immeasurably tiny — not zero, just far too small to ever notice. The law doesn't stop applying; the effect just becomes invisible at extreme mass ratios.
Try to make the two forces unequal. Slide "how hard you push" as far as you like — the sensor reading for "how hard it pushes back" won't budge from matching it. That's not a limitation of this widget; it's the law itself.
Make a prediction first: you push a heavy box that doesn't budge. You push a little harder — still nothing. You push just a little harder still, and suddenly it lurches forward far more easily than you expected. Why does it take a bigger push to start the box moving than it does to keep it moving?
Static friction: fs ≤ μsN (matches the applied push, up to a maximum) · Kinetic friction: fk = μkN (roughly constant while sliding)
Friction isn't a fixed drag that always opposes you by some constant amount — static friction is more like a rubber band that stretches to match whatever push you apply, up to a maximum of μsN. Push harder than that maximum and the object breaks free; friction switches to kinetic friction, which is usually a bit smaller. That drop is exactly why a stuck box suddenly "lurches" the instant it starts moving.
Both friction laws are empirical (not derived from something deeper) but they fit real surfaces surprisingly well: friction force is proportional to how hard the two surfaces are pressed together (N), essentially independent of the visible contact area and, for kinetic friction, roughly independent of sliding speed too. A heavier box needs a bigger push to start sliding, not because it has more surface area, but purely because its weight increases N.
Increase the push force gradually and watch the friction force track it exactly — until it hits the static maximum and the box breaks free. The graph shows this exact handoff.
The same push, the same box, four different surfaces. Coefficients below are typical real-world values — notice how much of the difference between "impossible to budge" and "slides at the slightest touch" comes down to one number.
Friction resists your push on this box — but it's also the only reason walking, climbing a slope, or a car's tires gripping the road work at all. Your foot pushes backward on the ground; without friction, the ground couldn't push forward on you in return (that's Newton's Third Law again), and you'd simply slip in place. "Friction is bad" and "friction is necessary" are both true depending entirely on whether you're trying to prevent sliding or trying to cause it.
Braking distance and how sharply you can turn are governed by exactly the same number: how much friction the tires can supply. (Reaction time and raw stopping distance for a chosen deceleration were covered back in Kinematics → Constant Acceleration; here, the deceleration itself comes directly from road friction.)
A tire can only supply so much total friction force before it slides, no matter which direction that force points — braking, turning, or (realistically) some mix of both at once. That's why racing drivers brake in a straight line before a turn rather than mid-corner: braking friction and cornering friction draw from the same limited budget, roughly μmg total, in any direction. This "budget" idea is developed further, with real circular motion, in Circular Motion → Banked Curves.
Make a prediction first: a box sits on a ramp. As you slowly tilt the ramp steeper, at some angle it suddenly starts sliding. Does that critical angle depend on how heavy the box is?
On a slope, gravity splits into two components: mg sinθ (pulling the box down the slope) and mg cosθ (pressing the box into the slope, which is exactly what the normal force N has to balance).
Gravity itself never changes — it's always mg straight down. What changes is which directions are convenient to measure it in. Tilting your axes to match the slope (one axis along the surface, one perpendicular to it) turns a single vector into two components, mg sinθ and mg cosθ, using the same right-triangle trigonometry as the vector components in the Toolkit — just with gravity playing the role of the vector being split, and the slope's tilt playing the role of the angle.
Tilt the ramp and watch gravity's two components change — sinθ grows as the ramp steepens while cosθ shrinks, and they always trade off so that the two components' squares still add up to mg.
The box starts sliding exactly when the down-slope pull outgrows the maximum static friction available — that crossing point defines the critical angle, and it has a surprisingly clean formula.
Sliding begins exactly when mgsinθ = μₛN = μₛmgcosθ. The mass m and g cancel from both sides — which is the answer to the prediction question: the critical angle never depends on the box's weight. What's left is sinθ = μₛcosθ, or tanθ = μₛ, so θ꜀ = atan(μₛ). A rougher surface (larger μₛ) tolerates a steeper ramp before sliding starts.
A ladder leans against a frictionless wall, held in place purely by floor friction. Move a person up the ladder and watch how much harder the floor has to grip to keep it from slipping — this is the same balance-of-turning-effects idea ("torque") that gets its own full section later, in Rotational Mechanics → Torque.
Taking torques about the ladder's foot (the point where it touches the floor) cancels out both the unknown floor normal force and the unknown floor friction, leaving one equation in the wall's push-back force Nwall: NwallLsinθ = gcosθ(MladderL/2 + m·d), where d is the person's distance up the ladder. Horizontal equilibrium then forces the floor friction to exactly equal Nwall. If that required friction ever exceeds μfloor times the floor's normal force, the ladder slips — which is exactly why climbing higher on a leaning ladder (increasing d) makes it progressively less stable, even though your weight itself hasn't changed.
Make a prediction first: a heavy block and a light block hang from opposite ends of a rope over a pulley. Does the heavier one accelerate downward as fast as it would falling freely, slower, or not at all?
A rope is treated as massless (its own weight is negligible) and inextensible (it doesn't stretch), which forces two objects tied together to share exactly the same speed and the same magnitude of acceleration at every instant — they move as a single connected system, even though the rope pulls on each one from a different direction.
The general strategy: draw a separate free-body diagram for each object, write F = ma for each one with the same unknown a and the same unknown T appearing in both equations (pulling in opposite, connected directions), then solve the two equations together.
A common guess is that the tension in a hanging system just equals one of the weights, or their average. It doesn't, in general — tension is whatever value makes both connected objects share the exact same acceleration, and that value comes out of solving the system, not from either mass alone. Widget 2 below lets you check this directly: set the two hanging masses equal and the system sits in perfect balance with T = mg exactly; make them unequal and T lands somewhere between the two weights, but never exactly at their average.
One more thing a rope can't do: push. It only ever pulls, along its own length. (If you swapped the rope in Widget 1 for a spring instead, pulling it and letting go wouldn't give one clean acceleration — it would set up a back-and-forth oscillation, the subject of its own full section later: Oscillations → Simple Harmonic Motion.)
Pull the front block with force F; the rope drags the back block along at the exact same acceleration. Notice that the tension is not equal to F — it's only the share of the pull that block 2 actually needs.
Two masses hang from a rope over a pulley. If they're equal, nothing moves. Make one heavier and the whole system accelerates — but (answering the prediction above) always slower than free fall, because the rope is also busy lifting the lighter mass.
Taking "A descending" as positive: for A, mAg − T = mAa; for B (moving up), T − mBg = mBa. Adding the two equations cancels T entirely, giving a = (mA−mB)g / (mA+mB). Substituting back gives T = 2mAmBg / (mA+mB) — notice this is always less than either individual weight, which is exactly why the heavier side never falls as fast as it would in free fall: the rope is holding it back precisely as much as it's lifting the other side.
A block on a ramp is tied over a pulley to a hanging mass. Whether the hanging mass wins (pulling the block up the slope) depends on a three-way tug between its own weight, the block's down-slope pull, and friction.