Chapter 04 · Newton's Laws · Full book

Newton’s Laws of Motion: The Causes Behind Change

Core intuition: Chapter Roadmap

Kinematics told us how things move; this chapter asks why. The whole subject rests on three laws and one diagram.

The three laws. The First defines when a frame is inertial and what “no net force” looks like; the Second, \(\vec{F}_{\text{net}} = m\vec{a}\), is the engine of all dynamics; the Third pairs every force with an equal and opposite partner on a different body.
The free-body diagram — the single most important skill in mechanics: isolate one object, draw every force acting on it, and resolve along well-chosen axes.
Constraints. Strings, pulleys, and contact surfaces lock the accelerations of different bodies together; reading the constraint correctly is half the battle.
Friction. Static (self-adjusting, up to \(\mu_s N\)) and kinetic (\(\mu_k N\)) — the force that makes the everyday world behave as it does.
Non-inertial frames. Solve a problem from an accelerating frame and a pseudo-force \(-m\vec{a}_{\text{frame}}\) restores \(\vec{F} = m\vec{a}\).
A problem-solving algorithm that ties it together: isolate, draw, resolve, apply the Second Law axis-by-axis, and use constraints to close the system.

Master the free-body diagram and the rest of mechanics — energy, momentum, rotation — becomes bookkeeping.

Perplexing Questions

Perplexing questions

Before we begin, sit with these questions. Do not reach for formulas. Each one is resolved explicitly at the relevant point in this chapter.

The table's missing force. A book rests on a table. The table pushes the book upward with a normal force. What pushes the table downward—is it the same force as the weight of the book, or a different force entirely?
The horse–cart paradox. A horse pulls a cart forward. By Newton's Third Law the cart pulls the horse backward with an equal and opposite force. If the forces are equal and opposite, why does the horse–cart system ever accelerate?
The lying scale. You stand on a bathroom scale inside a lift that accelerates upward. The reading increases. Have you gained mass?
The contact force trap. Two blocks \(A\) and \(B\) sit in contact on a frictionless floor. You push \(A\) into \(B\) with force \(F\). Is the contact force on \(B\) equal to \(F\)?
Tension in a string. A light string passes over a frictionless pulley and supports a mass at each end. Is the tension the same throughout? What changes if the string has mass?
Friction that pushes forward. A car accelerates from rest on a level road. The only horizontal force the road exerts on the tyres is static friction—but “friction opposes motion.” How can friction be the force that propels the car forward?
The wall that pushes back but goes nowhere. You push a wall with \(50\;\mathrm{N}\). The wall pushes you back with \(50\;\mathrm{N}\) (Newton's Third Law). Yet you do not fly backward. What other horizontal force acts on you, and why does the net force on you come out to zero?
The non-pair that looks like a pair. A book rests on a table. The weight of the book and the normal force on the book from the table are equal, opposite, and both act on the book. Are they a Newton's Third Law pair?
Unequal speeds, equal forces. Two ice skaters push off each other. The lighter skater moves away twice as fast. A bystander claims this violates Newton's Third Law because the effects are unequal. Where is the error?

Every question is answered where the relevant idea is developed, later in the chapter.

The Conceptual Shift: From How to Why

In kinematics we described motion: position, velocity, acceleration, and the equations connecting them. Not once did we ask why a ball accelerates or what keeps a planet curving around the Sun. We accepted acceleration as a given number and computed consequences.

Dynamics asks the deeper question: what determines acceleration? The answer, compressed into three laws by Newton, reorganises all of mechanics around a single concept—force.

Before Newton the dominant intuition was Aristotelian: a force is needed to sustain motion. Push a crate across the floor; stop pushing, and the crate stops. This seems self-evident, but it is profoundly wrong. What stops the crate is friction—another force—not the absence of your push. Galileo saw through the illusion centuries before Newton.

Core intuition: The Break With Aristotle

Galileo imagined a ball rolling down one smooth incline and up another, each time rising nearly to its original height. As the second incline is made progressively flatter, the ball must travel farther to regain that height. In the limit of a perfectly flat, frictionless surface the ball would never regain the height—and therefore would never stop. The conclusion is radical: the natural state of a body is not rest but uniform motion. Force does not cause velocity. Force causes change in velocity. This single insight is the dividing line between ancient and modern mechanics.

Before stating the laws we need an operational definition. A force is a push or pull—an interaction between two identifiable objects—that, when unbalanced, produces acceleration. Forces are vectors: they have magnitude and direction and obey superposition (they add as vectors). The quantity that governs motion is not any single force but the net (resultant) force: \[ \vec{F}_{\text{net}} = \sum_i \vec{F}_i. \] When the net force is zero the velocity does not change. When it is nonzero the velocity changes, and the rate of change is proportional to the net force. This is the heart of Newtonian mechanics.

Newton's First Law and Inertial Frames

Newton's First Law is often paraphrased as “a body at rest stays at rest; a body in motion stays in motion.” The paraphrase, while memorable, hides the law's deepest content. The First Law is not a special case of \(F = ma\) with \(F = 0\). It is a logically prior statement: it asserts the existence of a privileged class of reference frames in which the laws of mechanics take their simplest form.

Birth of the formula: What the First Law Actually Says

Statement. There exist reference frames—called inertial frames—in which a body subject to zero net external force moves with constant velocity (including the special case of remaining at rest). Three points that are routinely missed: 1. “Constant velocity” means constant in both magnitude and direction. A particle moving in a circle at constant speed is not in uniform motion: its velocity vector changes direction continuously, so a net force must be acting on it. 2. The First Law defines what an inertial frame is. An inertial frame is one in which an isolated particle (zero net force) moves in a straight line at constant speed. Without this definition, the Second Law \(\sum\vec{F}=m\vec{a}\) has no specified domain of validity. 3. N1L is not simply \(\sum\vec{F}=m\vec{a}\) with \(\vec{a}=\vec{0}\). The Second Law presupposes that you are working in an inertial frame. The First Law is what tells you such frames exist. Logically the First Law comes first; without it the Second has no foundation.

Inertial Frames

A reference frame in which Newton's First Law holds is called an inertial frame. Any frame moving at constant velocity relative to an inertial frame is also inertial—this follows from Galilean velocity addition: if the relative velocity is constant, then \(\vec{a}' = \vec{a}\), so zero acceleration in one frame implies zero acceleration in the other.

Is the Earth an inertial frame? Strictly, no. It rotates on its axis and orbits the Sun, so every point on the surface undergoes centripetal acceleration. At the equator this acceleration is roughly \(0.034\,\mathrm{m/s^2}\)—negligible compared with \(g\approx 9.8\,\mathrm{m/s^2}\). For most problems (time-scales of seconds to minutes, length-scales of metres to kilometres) we treat the ground as inertial. The approximation breaks down for large-scale or long-duration phenomena: the Foucault pendulum, Coriolis deflection of winds, and satellite orbits.

Frames that accelerate relative to an inertial frame are called non-inertial. In such frames, free particles appear to accelerate with no identifiable force acting on them. Handling this requires pseudo forces—discussed later in this chapter.

Misconception alert

Trap 1: “A moving body needs a force to keep moving.” This is Aristotle's ghost. A hockey puck gliding on smooth ice maintains constant velocity with zero net force. If it slows, friction is the culprit—not “the absence of a push.” JEE/NEET trap: “What force keeps a satellite moving in orbit?” No force keeps it moving; gravity changes its direction of motion. Trap 2: “N1L is just \(F=0\) in N2L.” It is not. N1L asserts the existence of inertial frames. N2L only works once you are in one. Their logical roles are different. Trap 3: “Inertia is a force.” Inertia is a property of mass—the tendency of a body to resist changes in velocity. It is not a force and it never appears on a free-body diagram. When a bus brakes suddenly and you lurch forward, no force pushes you forward; your body simply continues at the bus's original velocity (N1L) while the bus decelerates beneath you.

Equilibrium

◈ Concurrent forces — the hanging lamp

Two strings, one lamp. Drag the lamp or set the string angle φ — each strand pulls harder as the strings flatten, and the tension climbs steeply as φ shrinks.

drag the lamp ↕ or use sliders

Live readout

T =mg2 sin φ

φ (with ceiling)35°

weight mg19.6 N

tension (each)17.1 N

string angle φ 35° lamp mass m 2.0 kg

A body is in equilibrium when the net force on it vanishes: \[ \sum \vec{F} = \vec{0}. \] This does not require the body to be at rest. It requires that its velocity be constant—which could be zero (static equilibrium) or nonzero (dynamic equilibrium).

A car cruising at steady \(60\,\mathrm{km/h}\) on a level highway is in dynamic equilibrium: the engine's forward drive force exactly balances air drag and rolling resistance. A lamp hanging motionless from two strings is in static equilibrium: the vector sum of the two string tensions and gravity is zero.

In component form, equilibrium demands: \[ \sum F_x = 0, \qquad \sum F_y = 0. \] These two scalar equations, combined with a careful free-body diagram, are the starting point for every statics problem.

✓ Checkpoint · Newton's First Law

A car moves at constant \(60\,\mathrm{km/h}\) on a straight level road. A student argues: “The engine exerts a forward force, so the net force is nonzero, so the car accelerates.” Identify the flaw and state the net force on the car.
The student ignores resistive forces (air drag, rolling friction). At constant velocity these exactly balance the engine force; the net force is zero.
A puck slides freely on a smooth ice rink (neglect friction and air resistance). List all forces acting on the puck. Is it in equilibrium?
Weight \(mg\) downward and normal force \(N\) upward. These balance (\(\sum\vec{F}=\vec{0}\)), so the puck is in dynamic equilibrium, moving at constant velocity.
Two forces act on a particle: \(3\,\mathrm{N}\) due east and \(4\,\mathrm{N}\) due north. A third force \(\vec{F}_3\) is applied so that the particle is in equilibrium. Find the magnitude and direction of \(\vec{F}_3\).
\(\vec{F}_3\) must cancel the resultant of the first two. Magnitude \(=\sqrt{3^2+4^2}=5\,\mathrm{N}\). Direction: \(\arctan(4/3)\approx 53.1^\circ\) south of west (i.e. \(3\,\mathrm{N}\) westward \(+ 4\,\mathrm{N}\) southward).
A lamp of weight \(W\) hangs from two light strings that make angles \(\theta_1\) and \(\theta_2\) with the ceiling (measured from the vertical). Write the equilibrium equations for the junction point in terms of the tensions \(T_1\), \(T_2\).
Horizontal: \(T_1\sin\theta_1 = T_2\sin\theta_2\). Vertical: \(T_1\cos\theta_1 + T_2\cos\theta_2 = W\).

Newton's Second Law: \(\sum \vec{F} = m\vec{a}\)

The Second Law quantifies the relationship between force and motion. The net external force on a body equals the product of its mass and its acceleration: \[ \boxed{\sum \vec{F} = m\vec{a}.} \] This is a vector equation. The acceleration \(\vec{a}\) is always parallel to \(\sum\vec{F}\), not to the velocity \(\vec{v}\). A projectile at the top of its trajectory has \(\vec{v}\) horizontal but \(\vec{a}=\vec{g}\) pointing straight down.

The Second Law embodies the superposition principle: each force contributes independently to the acceleration, and the acceleration is determined by their vector sum. There is no interaction between forces themselves; the \(3\,\mathrm{N}\) push you apply to a block is unaffected by the \(5\,\mathrm{N}\) push applied by someone else.

Birth of the formula: From Observation to \(\sum F = ma\)

Two independent experimental facts underlie the Second Law: Observation 1 (fixed mass): Apply a net force \(F\) to a body of mass \(m\) and measure its acceleration. Double the force and the acceleration doubles. Conclusion: \(a \propto F\) at fixed \(m\). Observation 2 (fixed force): Apply the same net force to bodies of different mass. Double the mass and the acceleration halves. Conclusion: \(a \propto 1/m\) at fixed \(F\). Combining: \(a \propto F/m\), or \(F = k\,m\,a\). The unit of force is chosen so that \(k=1\): one newton (\(1\,\mathrm{N}\)) is the force that gives a \(1\,\mathrm{kg}\) mass an acceleration of \(1\,\mathrm{m/s^2}\). Dimensional check: \([\mathrm{N}] = [\mathrm{kg}][\mathrm{m/s^2}] = \mathrm{kg\cdot m\cdot s^{-2}}\). ✓ The momentum form (preview). Newton originally wrote the Second Law as \[ \sum\vec{F} = \frac{d\vec{p}}{dt}, \qquad \vec{p}=m\vec{v}. \] When \(m\) is constant this reduces to \(\sum\vec{F}=m\vec{a}\). The \(d\vec{p}/dt\) form is more fundamental: it handles situations where mass changes with time (a rocket expelling fuel, a chain piling on a surface). Variable-mass systems are not treated in this chapter. We will return to the momentum formulation when we study impulse and collisions.

Core intuition: The Vector Nature of N2L

Because \(\sum\vec{F}=m\vec{a}\) is a vector equation, it splits into independent scalar equations along each coordinate axis: \[ \sum F_x = ma_x, \qquad \sum F_y = ma_y. \] These are solved independently. Choose axes wisely—typically one axis along the expected direction of acceleration, the other perpendicular—and resolve every force onto both axes. A direct consequence: \(\vec{a}\) points along the net force, not along the velocity.

Projectile at the peak: \(\vec{v}\) is horizontal, \(\vec{a}=\vec{g}\) is vertically downward.
Car turning at constant speed: \(\vec{v}\) is tangential, \(\vec{a}\) is centripetal (toward the centre).
Block decelerating on a rough floor: \(\vec{v}\) is forward, \(\vec{a}\) is backward (friction opposes sliding).

Whenever your answer gives \(\vec{a}\) parallel to \(\vec{v}\) in a problem where the net force has a component perpendicular to \(\vec{v}\), revisit the calculation.

Misconception alert

Trap 1: “Force is in the direction of velocity.” At the top of a projectile's flight the velocity is horizontal, yet the only force (gravity) is vertical. Force determines the direction of acceleration, not velocity. JEE favourite: asking for the direction of net force on a projectile at various points in its trajectory. The answer is always downward (assuming no air resistance), regardless of where the projectile is. Trap 2: “Heavy objects fall slower because they need more force.” A \(10\,\mathrm{kg}\) mass does need ten times the force of a \(1\,\mathrm{kg}\) mass for the same acceleration—but gravity provides exactly that larger force (\(F=mg\)). The acceleration \(a = F/m = g\) is independent of mass. Galileo's insight, verified experimentally and tested on every major exam. Trap 3: “Zero net force means the body is at rest.” Zero net force means zero acceleration, which means constant velocity. That velocity may well be \(20\,\mathrm{m/s}\) northward. Rest is merely the special case \(\vec{v}=\vec{0}\).

Example 1 · Single block on a frictionless surface

A \(4\,\mathrm{kg}\) block sits on a smooth horizontal table. Two horizontal forces act on it: \(F_1 = 15\,\mathrm{N}\) to the right and \(F_2 = 7\,\mathrm{N}\) to the left. Find the acceleration of the block.

Solution. System: The block alone. FBD (vertical): Weight \(mg\) downward, normal force \(N\) upward. No vertical acceleration, so \(N = mg\)—these are not needed further. FBD (horizontal): Net force \(= F_1 - F_2 = 15 - 7 = 8\,\mathrm{N}\) to the right. N2L: \[ \sum F_x = ma_x \implies 8 = 4\,a_x \implies a_x = 2\,\mathrm{m/s^2}\;\text{(rightward).} \] Sign-convention note: We defined rightward as positive. Both \(\sum F_x\) and \(a_x\) are positive—consistent. Had \(F_2 > F_1\), we would have obtained \(a_x<0\) (leftward acceleration). The sign convention handles direction automatically; there is no need to “guess” which way the block moves before writing equations.

Example 2 · Two forces at an angle

A \(2\,\mathrm{kg}\) particle on a smooth horizontal surface is acted on by \(\vec{F}_1 = 6\,\mathrm{N}\) due east and \(\vec{F}_2 = 8\,\mathrm{N}\) due north. Find the magnitude and direction of the acceleration.

Solution. Component method: \[ \sum F_x = 6\,\mathrm{N}\;\text{(east)}, \qquad \sum F_y = 8\,\mathrm{N}\;\text{(north)}. \] \[ a_x = \frac{6}{2} = 3\,\mathrm{m/s^2}, \qquad a_y = \frac{8}{2} = 4\,\mathrm{m/s^2}. \] Magnitude: \[ a = \sqrt{a_x^2 + a_y^2} = \sqrt{9+16} = 5\,\mathrm{m/s^2}. \] Direction: \(\alpha = \arctan(a_y/a_x) = \arctan(4/3) \approx 53.1^\circ\) north of east. Key observation: The acceleration is in the direction of the resultant force, not in the direction of either individual force. Any time you see \(\vec{a}\) aligned with one particular force rather than the net force, re-examine the FBD.

✓ Checkpoint · Newton's Second Law

A \(5\,\mathrm{kg}\) block on a smooth surface is acted upon by three horizontal forces: \(10\,\mathrm{N}\) east, \(6\,\mathrm{N}\) west, and \(8\,\mathrm{N}\) north. Find the magnitude and direction of the acceleration.
Net east–west: \(10-6=4\,\mathrm{N}\) east. Net north: \(8\,\mathrm{N}\). Resultant \(=\sqrt{16+64}=4\sqrt{5}\,\mathrm{N}\). \(a=4\sqrt{5}/5 \approx 1.79\,\mathrm{m/s^2}\) at \(\arctan(8/4)=\arctan 2 \approx 63.4^\circ\) north of east.
A \(2\,\mathrm{kg}\) block accelerates at \(3\,\mathrm{m/s^2}\). (a) Find the net force. (b) If the mass is doubled but the net force stays the same, find the new acceleration.
(a) \(F_{\text{net}}=ma=6\,\mathrm{N}\). (b) \(a'=F/m'=6/4=1.5\,\mathrm{m/s^2}\).
A \(60\,\mathrm{kg}\) person stands on a scale inside a lift. Find the scale reading when the lift (a) accelerates upward at \(2\,\mathrm{m/s^2}\), (b) accelerates downward at \(2\,\mathrm{m/s^2}\), (c) moves at constant velocity. (\(g=10\,\mathrm{m/s^2}\).)
Core intuition: Resolution PQ 3 — The Lying Scale
The scale reads the normal force, not your weight. In an upward-accelerating lift, Newton's Second Law gives \(N - mg = ma\), so \(N = m(g+a) > mg\). The reading increases because the floor must push harder to give you upward acceleration. Your mass is unchanged; only the contact force changes.
(a) \(N=m(g+a)=60\times 12=720\,\mathrm{N}\). (b) \(N=m(g-a)=60\times 8=480\,\mathrm{N}\). (c) \(N=mg=600\,\mathrm{N}\).
Two blocks (\(m_1=3\,\mathrm{kg}\), \(m_2=2\,\mathrm{kg}\)) are in contact on a smooth horizontal surface. A horizontal force \(F=25\,\mathrm{N}\) is applied to \(m_1\). Find the common acceleration and the contact force between the blocks.
Core intuition: Resolution PQ 4 — The Contact Force Trap
The contact force on \(B\) is not \(F\). The applied force \(F\) accelerates the entire system; \(B\) need only be accelerated by its own mass \(m_B\), so the contact force is \(m_B a = m_B F/(m_A + m_B) < F\). To find any internal force, always use the system FBD first to get \(a\), then isolate the body of interest.
\(a=F/(m_1+m_2)=25/5=5\,\mathrm{m/s^2}\). Isolate \(m_2\): contact force \(=m_2 a=2\times 5=10\,\mathrm{N}\). Common trap: the contact force is \(10\,\mathrm{N}\), not \(25\,\mathrm{N}\).

Solved examples

Five fully-worked problems from this chapter, free — solution and answer shown in full. The complete set of worked examples is in the full book.

Example 1 · Block pushed across a rough floor

A \(5\,\mathrm{kg}\) block rests on a floor with \(\mu_s = 0.3\), \(\mu_k = 0.2\). A horizontal force \(F = 30\,\mathrm{N}\) is applied. Find the friction force and the acceleration. (\(g = 10\,\mathrm{m/s^2}\).)

Solution. Given: \(m = 5\,\mathrm{kg}\), \(\mu_s = 0.3\), \(\mu_k = 0.2\), \(F = 30\,\mathrm{N}\), \(g = 10\,\mathrm{m/s^2}\).
Find: the friction force and the acceleration.
Setup: First test motion by comparing \(F\) with the maximum static friction \(\mu_s mg\). If the block slides, kinetic friction \(\mu_k mg\) acts backward and Newton's second law gives the acceleration.
Solve: \(f_{s,\max} = \mu_s mg = 0.3 \times 50 = 15\,\mathrm{N}\). Since \(F = 30 > 15\), the block slides; \(f_k = \mu_k mg = 0.2 \times 50 = 10\,\mathrm{N}\) (backward). \[ a = \frac{F - f_k}{m} = \frac{30 - 10}{5} = 4\,\mathrm{m/s^2}. \] Answer: \(\boxed{f_k = 10\,\mathrm{N},\quad a = 4\,\mathrm{m/s^2}}\)
Check: Limiting case: just past breakaway (\(F \to 15^+\,\mathrm{N}\)) friction is already kinetic, so \(a \to (15-10)/5 = 1\,\mathrm{m/s^2}\), small and positive as expected. ✓ Dimensions: \(a = [\mathrm{N}]/[\mathrm{kg}] = \mathrm{m/s^2}\). ✓

Example 2 · Stacked blocks: do they move together?

Block \(B\) (\(2\,\mathrm{kg}\)) rests on block \(A\) (\(4\,\mathrm{kg}\)) on a frictionless floor; between them \(\mu_s = 0.5\). A horizontal force \(F = 12\,\mathrm{N}\) is applied to the lower block \(A\). Find the acceleration of each block. (\(g = 10\,\mathrm{m/s^2}\).)

Solution. Given: \(m_A = 4\,\mathrm{kg}\), \(m_B = 2\,\mathrm{kg}\), interface \(\mu_s = 0.5\), frictionless floor, \(F = 12\,\mathrm{N}\) on \(A\), \(g = 10\,\mathrm{m/s^2}\).
Find: the acceleration of each block.
Setup: Assume they move together, find the shared acceleration, then check whether the friction needed to drag \(B\) along stays within \(\mu_s m_B g\).
Solve: Together: \(a = F/(m_A+m_B) = 12/6 = 2\,\mathrm{m/s^2}\). Friction needed on \(B\): \(f = m_B a = 4\,\mathrm{N}\). Available: \(\mu_s m_B g = 0.5 \times 20 = 10\,\mathrm{N} > 4\,\mathrm{N}\), so they move together. Answer: \(\boxed{a_A = a_B = 2\,\mathrm{m/s^2}}\), with \(4\,\mathrm{N}\) of self-adjusting friction on \(B\).
Check: Threshold: slipping would need \(a = \mu_s g = 5\,\mathrm{m/s^2}\), i.e. \(F = (m_A+m_B)(5) = 30\,\mathrm{N}\); since \(12 < 30\,\mathrm{N}\), “together” is confirmed. ✓ Dimensions: \(\mathrm{N/kg = m/s^2}\). ✓

Example 3 · Galileo's falling bodies

Aristotle held that heavier bodies fall faster. Galileo refuted this with a thought experiment alone, no apparatus. Reconstruct his argument.

Solution. Tie a heavy stone \(H\) to a light stone \(L\) and drop the pair. On one hand, if heavier falls faster, the slow stone \(L\) should hold back the fast stone \(H\), so the combination falls slower than \(H\) alone. On the other hand, the combination \(H + L\) is heavier than \(H\), so by the very same rule it should fall faster than \(H\) alone. The premise yields two opposite conclusions, so it must be false: all bodies fall at the same rate in vacuum. Insight: Newton's second law makes this exact — gravity pulls with \(F = mg\), so \(a = F/m = g\), independent of mass; the mass that is pulled is precisely the mass that resists. Check: feather and hammer fall together in a vacuum, as famously shown on the Moon. ✓

Example 4 · How hard does a seatbelt pull?

Estimate the average force a seatbelt exerts on a \(70\,\mathrm{kg}\) passenger when a car crashes from \(54\,\mathrm{km/h}\) to rest, the body halting over about \(0.1\,\mathrm{s}\).

Solution. Setup: Average force from the impulse–momentum theorem, \(F \approx m\,\Delta v/\Delta t\).
Estimate: \(\Delta v = 54\,\mathrm{km/h} = 15\,\mathrm{m/s}\), \(\Delta t \approx 0.1\,\mathrm{s}\): \[ F \approx \frac{70 \times 15}{0.1} \approx 1.0 \times 10^4\,\mathrm{N}. \] That is about \(15\) times the passenger's weight (\(mg \approx 700\,\mathrm{N}\)), i.e. roughly \(15g\). Answer: \(\boxed{F \sim 10^4\,\mathrm{N} \approx 15\times\text{body weight}}\)
Check: Without a belt the body would stop against the dashboard in perhaps \(0.01\,\mathrm{s}\), ten times faster and so ten times the force (\(\sim 10^5\,\mathrm{N}\)) — which is why belts and airbags, by stretching \(\Delta t\), save lives. ✓

Example 5 · Reading a braking-distance table

A driving manual lists braking distances on dry road:

speed \(v\) (m/s)	\(10\)	\(20\)	\(30\)
distance \(d\) (m)	\(10\)	\(40\)	\(90\)

What law links \(d\) and \(v\), and what coefficient of friction does it imply? (\(g = 10\,\mathrm{m/s^2}\).)

Solution. Setup: Test \(d \propto v^2\) by forming \(d/v^2\), then compare with the braking model \(d = v^2/(2\mu g)\).
Solve: \(d/v^2 = 10/100 = 40/400 = 90/900 = 0.10\,\mathrm{s^2/m}\) — constant, so \(d = 0.10\,v^2\). Matching \(d = v^2/(2\mu g)\) gives \(1/(2\mu g) = 0.10\), hence \(\mu = 1/(2 \times 0.10 \times 10) = 0.5\). Answer: \(\boxed{d \propto v^2,\quad \mu = 0.5}\)
Check: Doubling the speed (\(10 \to 20\)) quadruples the distance (\(10 \to 40\)), the signature of a \(v^2\) law; \(\mu = 0.5\) is typical for rubber on dry road. ✓

Problem bank

Five questions from this chapter’s 50-question bank, free — attempt each one before you reveal the answer. The rest of the bank, and the timed test that draws on all of it, are in the full book.

✎ Free practice questions

Block on a Rough Floor
A \(6\,\mathrm{kg}\) block sits on a horizontal surface with \(\mu_s = 0.5\) and \(\mu_k = 0.4\). A horizontal force \(F = 40\,\mathrm{N}\) is applied. Find the friction force acting on the block and its acceleration. (\(g = 10\,\mathrm{m/s^2}\).)
\(f_{s,\max} = \mu_s mg = 0.5 \times 60 = 30\,\mathrm{N}\). Since \(F = 40 > 30\), the block slides. \(f_k = \mu_k mg = 0.4 \times 60 = 24\,\mathrm{N}\). \(a = (F - f_k)/m = (40 - 24)/6 = 16/6 \approx 2.67\,\mathrm{m/s^2}\).
Atwood Machine
Two masses \(m_1 = 7\,\mathrm{kg}\) and \(m_2 = 3\,\mathrm{kg}\) hang from a light inextensible string over a smooth massless pulley. Find the acceleration of the system and the tension in the string. (\(g = 10\,\mathrm{m/s^2}\).)
\(a = (m_1 - m_2)g/(m_1 + m_2) = 4 \times 10/10 = 4\,\mathrm{m/s^2}\). \(T = 2m_1 m_2 g/(m_1 + m_2) = 2(7)(3)(10)/10 = 42\,\mathrm{N}\).
Block on a Rough Incline at Rest
A \(5\,\mathrm{kg}\) block sits on a rough incline at \(\theta = 20^\circ\) with \(\mu_s = 0.6\). Find the friction force acting on the block. Is it equal to \(\mu_s N\)? (\(g = 10\,\mathrm{m/s^2}\).)
\(mg\sin\theta = 50\sin 20^\circ \approx 17.1\,\mathrm{N}\). \(f_{s,\max} = \mu_s mg\cos\theta = 0.6 \times 50\cos 20^\circ \approx 28.2\,\mathrm{N}\). Since \(17.1 < 28.2\), the block is stationary. Friction \(= mg\sin\theta \approx 17.1\,\mathrm{N}\) (self-adjusting), not \(28.2\,\mathrm{N}\).
Connected Blocks on a Rough Floor
Blocks \(A\) (\(4\,\mathrm{kg}\)) and \(B\) (\(6\,\mathrm{kg}\)) are connected by a light string and rest on a horizontal surface with \(\mu_k = 0.25\). A horizontal force \(F = 50\,\mathrm{N}\) is applied to block \(A\). Find the acceleration of the system and the tension in the string between \(A\) and \(B\). (\(g = 10\,\mathrm{m/s^2}\).)
Total friction \(= \mu_k (m_A + m_B)g = 0.25 \times 100 = 25\,\mathrm{N}\). System: \(a = (F - f_{\text{total}})/(m_A + m_B) = (50-25)/10 = 2.5\,\mathrm{m/s^2}\). Isolate \(B\): \(T - \mu_k m_B g = m_B a \implies T = m_B(a + \mu_k g) = 6(2.5 + 2.5) = 30\,\mathrm{N}\).
Block on the Front Wall of an Accelerating Cart
A block of mass \(m\) is held against the smooth-faced front vertical wall of a cart purely by the cart's horizontal acceleration \(a\). The coefficient of static friction between block and wall is \(\mu_s\). (a) Draw the forces on the block and write the horizontal and vertical equations. (b) Derive the minimum acceleration \(a_{\min}\) for which the block does not slide down. (c) Interpret the limits \(\mu_s \to 0\) and \(\mu_s\) large.
(a) Horizontal: the wall's normal force \(N\) supplies the block's acceleration, \(N = ma\). Vertical: static friction \(f\) (up) must support the weight, \(f = mg\), with \(f \le \mu_s N\). (b) \(mg \le \mu_s N = \mu_s m a \implies a \ge g/\mu_s\), so \(\boxed{a_{\min} = g/\mu_s}\). (Equivalently, for a given \(a\) the wall must be rough enough that \(\mu_s \ge g/a\).) (c) \(\mu_s \to 0\): \(a_{\min} \to \infty\) (a perfectly smooth wall can never hold the block). ✓ Large \(\mu_s\): \(a_{\min} \to\) small (a sticky wall holds the block with only gentle acceleration). ✓

Chapter test

A paper drawn at random from this chapter's bank. Choose the exam you are training for — the marking scheme, pace and difficulty mix follow the real pattern. Work on paper; when you finish (or the clock runs out), the answers are revealed and you mark yourself honestly.

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Questions

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Free practice questions

The Friction–Force Graph
A block of mass \(m\) rests on a horizontal floor with coefficients \(\mu_s\) and \(\mu_k\) (with \(\mu_s > \mu_k\)). A horizontal force \(F\) is increased slowly from zero. (a) Sketch the friction force \(f\) as a function of \(F\) for \(F\) from \(0\) to well beyond the value that starts the block moving. (b) Give \(f\) just before and just after the block begins to slide, and explain the discontinuity. (c) Write the block's acceleration once it is sliding.
(a) For \(0 \le F \le \mu_s mg\) the block is static and friction self-adjusts: \(f = F\) (a straight line of slope \(1\)). At \(F = \mu_s mg\), \(f\) reaches its ceiling \(f_{s,\max} = \mu_s mg\). For \(F > \mu_s mg\) the block slides and \(f\) drops to the constant value \(f_k = \mu_k mg\), independent of \(F\) (a horizontal line). (b) Just before sliding: \(f = \mu_s mg\). Just after: \(f = \mu_k mg < \mu_s mg\). The drop occurs because static friction can grow up to \(\mu_s N\), but once contact slips the operative coefficient falls to \(\mu_k\). (c) \(a = (F - \mu_k mg)/m\).
Incline–Pulley System (Symbolic)
Mass \(m_1\) hangs vertically from a light string that runs over a smooth pulley at the apex of a frictionless incline of angle \(\theta\); the other end is tied to mass \(m_2\) resting on the incline. (a) Taking \(m_1\) descending as positive, derive the acceleration \(a\) of the system. (b) State the condition for \(m_2\) to move up the incline. (c) Derive the tension and verify the limits \(\theta \to 0\) and \(\theta \to 90^\circ\).
(a) For \(m_1\) (down positive): \(m_1 g - T = m_1 a\). For \(m_2\) (up-slope positive): \(T - m_2 g\sin\theta = m_2 a\). Adding: \(a = \dfrac{(m_1 - m_2\sin\theta)\,g}{m_1 + m_2}\). (b) \(m_2\) moves up iff \(a > 0 \iff m_1 > m_2\sin\theta\). (c) \(T = m_1(g - a) = \dfrac{m_1 m_2 (1 + \sin\theta)\,g}{m_1 + m_2}\). Limit \(\theta \to 0\): \(a \to \dfrac{m_1 g}{m_1+m_2}\), \(T \to \dfrac{m_1 m_2 g}{m_1+m_2}\) (horizontal block pulled by a hanger). ✓ Limit \(\theta \to 90^\circ\): \(a \to \dfrac{(m_1-m_2)g}{m_1+m_2}\), \(T \to \dfrac{2 m_1 m_2 g}{m_1+m_2}\) (vertical Atwood). ✓