Free-Body Diagrams

Every bridge you have ever crossed was certified by an act of imagination. Somewhere, an engineer mentally sawed the structure apart — one girder, one bolt, one cable at a time — lifted each piece onto a clean page, drew every force the rest of the world exerts on it, and demanded that the piece obey Newton's laws all by itself. If every fragment passes that interrogation, the whole bridge stands. This ritual of isolation has a name: the free-body diagram, and it is the single most consequential picture in mechanics. The idea is deceptively simple. Pick one body. Cut it free from everything it touches. Replace each touch — and each long-range pull like gravity — with an explicit force arrow. The unbreakable rule is that only forces exerted on the chosen body appear, and every arrow must answer the question "who, physically, is exerting you?" Forces the body exerts on other things belong on their diagrams; the body's own motion is not a force at all and never gets an arrow. Why care so much about a sketch? Because the diagram is where physics problems are actually won or lost. Physics-education research finds that drawing a correct free-body diagram is among the strongest single predictors of solving the problem itself — and the skill scales without modification from a block on an incline to spacecraft docking, robot grasping, and the million-node force ledgers inside every finite-element simulation. Master this one habit and Newton's laws stop being three sentences and become a working machine.

Key takeaways

A free-body diagram (FBD) isolates ONE body and shows every external force acting on it as a labeled vector — gravity, normal forces, tension, friction, applied pushes — and nothing else.
The admission test for an arrow: name the physical agent exerting it. "The table pushes the book up" qualifies; "the motion carries it forward" does not — motion is not a force, and $m\vec{a}$ is the response to forces, never an arrow on the diagram.
Newton's third law partners act on different bodies, so they never appear together on one diagram; weight and normal force on the same body are not a third-law pair.
Once the diagram is drawn, choose axes (tilt them along an incline!) and project: each axis turns $\sum\vec{F}=m\vec{a}$ into one scalar equation.
For connected bodies, draw one FBD per body; the string or contact supplies equal-and-opposite internal forces across diagrams, and a constraint (inextensible string, no interpenetration) links the accelerations.
Key idea: cut one body free, draw every external force acting on it and nothing else, then project $\sum\vec{F}=m\vec{a}$ along well-chosen axes — the diagram is the complete interface between a physical situation and Newton's mathematics.

The core insight

Cut one body free, draw every external force acting on it and nothing else, then project $\sum\vec{F}=m\vec{a}$ along well-chosen axes — the free-body diagram is the complete interface between a physical situation and Newton's mathematics.

Where this lives today

The free-body diagram never retired; it multiplied. Every finite-element simulation — the software that certifies aircraft wings, crash-tests cars before they exist, and checks that a skyscraper sways but does not snap — is at bottom millions of free-body diagrams solved simultaneously: the structure is meshed into tiny elements, and each node's equation is exactly $\sum\vec{F} = m\vec{a}$ for the forces its neighbors exert on it. The engineer's mental saw became a discretization algorithm, but the physics per element is this article's recipe unchanged. Robotics runs the method in real time. A robot planning to grasp a mug or a legged robot deciding where to plant a foot must solve contact problems — which normal forces and friction forces, within their $f_s \le \mu_s N$ cones, keep the object or the body in the intended motion — and game-physics engines resolve thousands of such per-body force ledgers every frame at 60 Hz. Meanwhile the Fundamentals of Engineering exam, the licensure gate for practicing engineers, still opens its mechanics sections the way your next problem set does: draw the free-body diagram first.

The physical picture

Think of a free-body diagram as an interrogation. You grab one object, lift it out of its environment, and ask: who is pushing or pulling on you, and in what direction? Each truthful answer becomes an arrow. The Earth pulls down on the block — arrow. The incline presses on its underside, perpendicular to the surface — arrow. The surface grips it against sliding — arrow. And that is the complete testimony: anything the block does to the incline is the incline's problem, recorded on the incline's diagram, not here.

Figure

The cut — isolate the block and replace every touch with a force arrow

The admission test for every arrow is brutally simple: name the source. "The Earth pulls the block down" — legitimate, that is weight. "The table pushes the book up" — legitimate, that is the normal force. "The block's speed carries it forward" — rejected: motion is a state, not an interaction, and no object in the universe is exerting that push. Forces come in exactly two kinds, and the inventory is short. Contact forces (normal force, friction, tension, a hand's push) act only where something physically touches the body, so you find them by tracing the body's boundary with a finger. Long-range forces — for our purposes, gravity — act on the whole body at once and are drawn as a single arrow from its center of mass. Trace the boundary, add gravity, and you are provably done: there is nothing else. Two illegal arrows account for most wrong diagrams ever drawn, and both are worth staring at once in their natural habitat:

Figure

The two illegal arrows — motion is not a force and ma is not a force

The first illegal arrow is the "force of motion" — the medieval impetus, the feeling that a moving thing must be carrying a push. Newton's first law killed it in 1687: uniform motion needs no force, only changes of motion do. The second is drawing $m\vec{a}$ as if it were a force. The product $m\vec{a}$ is the output of the diagram — the response the forces produce — and writing it among the inputs double-counts the physics. A useful slogan: the left side of $\sum\vec{F}=m\vec{a}$ lives on the diagram; the right side is what the diagram predicts. One more habit completes the picture. When two bodies interact — block stacked on block, masses tied by a string — imagine slicing through the connection. The slice exposes an equal-and-opposite pair of forces, one arrow plunging into each body. Draw separate diagrams and both arrows become visible, one per diagram; fuse the bodies into a single system and the pair hides inside, cancelling. Choosing where to cut is choosing which forces you can see, and skilled problem-solvers cut deliberately: the whole system to find the acceleration in one line, then a single body to expose the internal tension they actually want.

From picture to equations

The method is a fixed five-move recipe, and it is worth naming the moves because every mechanics problem you will ever solve runs through them in order.

Choose the body. One object, stated explicitly. "The block", not "the block and rope, sort of". For extended objects treat the body as a particle at its center of mass — legitimate whenever rotation is not in play.

Inventory the forces. Trace the body's boundary for contacts (each touch contributes a normal force, possibly friction, or a tension if it is a string), then add gravity. Every arrow must have a nameable agent.

Draw and label. Arrows start on the body and point in the force's direction, each labeled with a symbol ( $W$ , $N$ , $T$ , $f$ , $F_{\mathrm{app}}$ ) — never a number yet; algebra first, arithmetic last.

Choose axes. Any orthogonal pair works, but the algebra collapses when one axis lies along the acceleration. On an incline, tilt the axes with the slope; in circular motion, point one axis at the center.

Project and solve. Each axis converts the vector law into one scalar equation; count unknowns, add constraint equations as needed, and solve. The law the diagram feeds is Newton's second, and it is worth writing with the subscript that carries the whole philosophy. Only forces exerted by other objects on this body enter the sum — that is exactly what the diagram displays:

\sum \vec{F}_{\mathrm{ext}} = m\vec{a} \qquad (1)

where: $\vec{F}_{\mathrm{ext}}$ — each force drawn on the diagram; $m$ — the body's mass; $\vec{a}$ — the acceleration of its center of mass. The equation is a vector statement, which is precisely why the diagram matters: it is the bookkeeping device that guarantees no force is forgotten, none is invented, and every one enters with the correct direction. A vector equation is solved one axis at a time. After choosing axes, take components:

\sum F_x = m a_x, \qquad \sum F_y = m a_y \qquad (2)

where: $F_x, F_y$ — signed components of each drawn force along the chosen axes; $a_x, a_y$ — the acceleration components along the same axes. Each axis yields one equation, so a two-dimensional diagram is worth exactly two scalar equations per body — knowing that count in advance tells you immediately how many bodies you must isolate to pin down all your unknowns.

The force inventory

Nearly every arrow you will ever draw comes from this short menu. The direction rules are absolute; the magnitude rules tell you which forces are known and which are unknowns the equations must determine.

force	direction rule	magnitude rule
weight $W$	vertically down, from the center of mass	$W = mg$ , always known
normal $N$	perpendicular to the surface, pushing away	unknown — whatever prevents interpenetration; $N \ge 0$
tension $T$	along the string, always pulling the body toward the rope	unknown — set by the dynamics; uniform in an ideal rope
static friction $f_s$	along the surface, opposing the tendency to slide	unknown, self-adjusting up to $f_s \le \mu_s N$
kinetic friction $f_k$	along the surface, opposing the actual sliding	known once $N$ is: $f_k = \mu_k N$
spring force $F_{\mathrm{sp}}$	along the spring, toward its natural length	$F_{\mathrm{sp}} = k\,\vert x\vert$ for stretch or compression $x$
applied push or pull	as given	as given

Three of these deserve their own displays, because their magnitude rules are where beginners stumble. Gravity first — the one long-range force on the menu, and the only arrow that appears on literally every diagram near Earth's surface:

\vec{W} = m\vec{g}, \qquad g \approx 9.81\ \mathrm{m/s^2} \qquad (3)

where: $\vec{W}$ — the gravitational force, drawn straight down from the center of mass regardless of how the surface is tilted; $g$ — the near-surface gravitational field strength. On an incline the weight does not tilt; its components along tilted axes do — a distinction the first derivation below exploits. The normal force is the great self-adjusting unknown. A surface pushes with exactly the force required to stop the body sinking into it — no more, no less:

\vec{N} \perp \text{surface}, \qquad N \ge 0 \qquad (4)

where: $\vec{N}$ — the contact force perpendicular to the touching surfaces, always a push, never a pull. Its magnitude is not a formula but an output of the equations: $mg$ for a book on a level table, $mg\cos\theta$ on an incline, $m(g+a)$ in an accelerating elevator, and zero at the instant contact is lost. Treating $N = mg$ as a law rather than a special case is among the most reliable ways to get mechanics wrong. Friction lives in the surface plane and comes in two regimes with different logical status:

f_s \le \mu_s N, \qquad f_k = \mu_k N \qquad (5)

where: $f_s$ — static friction, an unknown that adjusts to whatever value prevents sliding, up to the ceiling $\mu_s N$ ; $f_k$ — kinetic friction, a known magnitude once sliding begins, opposing the relative motion; $\mu_s, \mu_k$ — dimensionless coefficients with $\mu_k \lesssim \mu_s$ . On the diagram the only decision is the direction: opposite the tendency to slide (static) or opposite the actual sliding (kinetic). An ideal string adds the last standard entry. Massless and inextensible, it can only pull, and Newton's second law applied to any piece of it — zero mass times any finite acceleration — forces the tension to be uniform along its length:

T_{\text{same string}} = \text{const} \qquad (6)

where: $T$ — the tension magnitude, identical at both ends of a massless string even when the string accelerates, and preserved around an ideal (massless, frictionless) pulley, which merely redirects it. This is why one symbol $T$ serves both blocks in every pulley problem — and why a string appearing in two diagrams silently links their equations.

Derivation ABlock on a frictionless incline with tilted axes

The goal is the acceleration of a block released on a frictionless plane inclined at angle $\theta$ — the simplest problem whose answer genuinely requires a free-body diagram, and the standard demonstration of why axes should be tilted. The strategy: isolate the block, find exactly two forces (weight and normal), then choose axes along and perpendicular to the slope so that the unknown acceleration lies entirely along one axis. The payoff for that choice is that only one force needs decomposing — the weight — while $\vec{N}$ and $\vec{a}$ each live cleanly on a single axis. Watch how the normal force comes out as $mg\cos\theta$ , not $mg$ : the diagram plus the perpendicular equation derives its value, illustrating that $N$ is an output, never an assumption.

Step 1. Draw the diagram: weight $mg$ straight down, normal $N$ perpendicular to the incline surface. Nothing else touches the block. Choose the $x$ -axis pointing down the slope and the $y$ -axis along the outward normal. Why tilt the axes: the block accelerates along the slope, so tilted axes make $a_y = 0$ exactly, and the vector unknown $\vec{a}$ collapses to a single scalar $a$ along $x$ . With horizontal-vertical axes both equations would contain both unknowns $N$ and $a$ — solvable, but twice the algebra.

Step 2. Decompose the one force that lies along neither axis. The weight makes angle $\theta$ with the $-y$ axis, so its components are $mg\sin\theta$ along $+x$ (down the slope) and $mg\cos\theta$ along $-y$ (into the surface). Why it is $\sin\theta$ on the slope: tip $\theta$ to zero and the along-slope pull must vanish — $\sin\theta$ does, $\cos\theta$ does not. Checking limits on the decomposition catches sine-cosine swaps before they propagate.

Step 3. Project Newton's second law (2) onto the $y$ -axis, where the block never leaves the surface, so $a_y = 0$ :

N - mg\cos\theta = 0 \quad\Rightarrow\quad N = mg\cos\theta

Why $N$ is less than $mg$ : the surface only has to prevent motion into itself, and only the perpendicular slice $mg\cos\theta$ of the weight is trying to move the block that way. The rest of the weight acts along the slope, where the surface's perpendicular push is powerless against it.

Step 4. Project onto the $x$ -axis, where the normal force has no component:

mg\sin\theta = ma \quad\Rightarrow\quad a = g\sin\theta \qquad (7)

where: $a$ — the acceleration down the slope. The mass has cancelled: heavy and light blocks slide identically, Galileo's inclined-plane discovery falling out of two lines of projection.

Step 5. Interrogate the limits. At $\theta \to 0$ : $a \to 0$ , $N \to mg$ — a block on a level floor. At $\theta \to 90°$ : $a \to g$ , $N \to 0$ — free fall beside a vertical wall it barely touches. Both endpoints are exactly right, and $N \to 0$ at the vertical limit re-confirms that the normal force is a response, fading away as the surface has less and less to resist.

Third-law bookkeeping across diagrams

Newton's third law is the rule that connects one diagram to another. Whenever body A exerts a force on body B, body B exerts the reverse on A:

\vec{F}_{AB} = -\vec{F}_{BA} \qquad (8)

where: $\vec{F}_{AB}$ — the force on A from B, which lives on A's diagram; $\vec{F}_{BA}$ — the force on B from A, which lives on B's. To find a partner, swap the two labels and nothing else. The partner of "Earth pulls block down" is "block pulls Earth up" — a force on the planet, absent from the block's diagram entirely. This is why third-law partners can never cancel: they act on different bodies, and no single diagram ever contains both. The weight and the normal force on a resting book — the pair everyone wants to call action-reaction — are forces from two different agents (Earth, table) on the same body, equal only because the book happens to have $\vec{a} = 0$ . Put the book in a descending elevator and they part company immediately, while every true third-law pair in the problem stays exactly matched. The cut principle now pays off in full. Here is the same machinery — one diagram per body, third-law pairs across diagrams, a constraint linking accelerations — laid out for the two workhorse systems:

Figure

One system, two ledgers — each body gets its own diagram and the string ties them together

Go deeper · Derivation B · 2 minTwo blocks in contact and the force between them

Derivation BTwo blocks in contact and the force between them

A hand pushes with force $F$ on block 1 (mass $m_1$ ), which in turn pushes block 2 (mass $m_2$ ) ahead of it across a frictionless floor. The goal is the contact force $N_{12}$ between the blocks — a quantity that is invisible on any diagram of the combined system and becomes visible only when we cut between the blocks. The strategy is the two-cut method: treat the pair as one system to get the acceleration cheaply, then isolate one block to expose the internal force. The result, $N_{12} = F\,m_2/(m_1+m_2)$ , carries a satisfying moral: the contact force is exactly the fraction of $F$ needed to accelerate the mass downstream of the cut.

Step 1. Cut around both blocks together (total mass $m_1 + m_2$ ). The pushes between the blocks are now internal — a third-law pair buried inside the system — so the only horizontal external force is $F$ :

F = (m_1 + m_2)\,a \quad\Rightarrow\quad a = \frac{F}{m_1 + m_2}

Why the internal pair drops out: by (8) the two contact forces are equal and opposite, and both act inside the cut, so they cancel in the external sum. That cancellation is the entire advantage of the system view.

Step 2. Now cut around block 2 alone. The only horizontal force on it is the push from block 1, $N_{12}$ — the hand touches only block 1 and cannot act on block 2 directly:

N_{12} = m_2\,a

Why block 2 is the smarter choice: isolating block 1 instead gives $F - N_{21} = m_1 a$ — two forces instead of one. Both routes agree, but the cleanest diagram is always the body with the fewest arrows.

Step 3. Substitute the system acceleration:

N_{12} = \frac{m_2}{m_1 + m_2}\,F \qquad (9)

where: $N_{12}$ — the force on block 2 from block 1; by Newton's third law block 2 pushes back on block 1 with the same magnitude. With $F = 12\ \mathrm{N}$ , $m_1 = 2.0\ \mathrm{kg}$ , $m_2 = 1.0\ \mathrm{kg}$ : $a = 4.0\ \mathrm{m/s^2}$ and $N_{12} = 4.0\ \mathrm{N}$ — one third of the applied force, because the cut leaves one third of the mass downstream.

Step 4. Test it by reversing the push. Apply the same $F$ to block 2 instead, so block 2 drives block 1: the downstream mass is now $m_1$ and the contact force becomes $F\,m_1/(m_1+m_2) = 8.0\ \mathrm{N}$ . Same blocks, same applied force, double the internal squeeze — because the cut now has twice as much mass to feed. As a check: $N_{12}$ plus the net force needed to accelerate block 1 equals $4.0 + 2.0\times4.0 = 12\ \mathrm{N} = F$ , so the ledger balances. The limits also behave: $m_2 \to 0$ makes the contact force vanish, and $m_1 \to 0$ makes the contact force carry all of $F$ .

Go deeper · Derivation C · 2 minThe Atwood machine from two diagrams and one constraint

Derivation CThe Atwood machine from two diagrams and one constraint

Two masses $m_1 > m_2$ hang from a string over an ideal pulley. The goal is their common acceleration and the string tension — the archetype of every connected-body problem, and the cleanest showcase of the full method: one diagram per body, tension uniform by (6), and an inextensibility constraint standing in as the extra equation that two unknowns demand. The strategy is to write Newton's second law once per diagram with signs chosen along each body's direction of motion, then add. Watch what addition does: the tension — internal to the string-and-masses system — cancels in the sum, exactly as an internal force should.

Step 1. State the constraint. The string cannot stretch, so when $m_1$ descends a distance $d$ , $m_2$ rises the same $d$ : their accelerations share one magnitude $a$ . This kinematic fact is the second equation the two unknowns $(a, T)$ require — diagrams alone would leave the system underdetermined.

Step 2. Diagram of $m_1$ (weight down, tension up), taking down as positive since $m_1$ descends:

m_1 g - T = m_1 a

Why the sign convention is per-body: each diagram gets its own axis aligned with that body's motion, so "positive" means "the way this body actually accelerates" and both equations carry the same symbol $a$ with no minus-sign gymnastics.

Step 3. Diagram of $m_2$ (weight down, tension up), taking up as positive since $m_2$ rises:

T - m_2 g = m_2 a

Why the same $T$ appears: the string is massless and the pulley ideal, so by (6) the tension is uniform along the entire string — one symbol serves both diagrams. A real pulley with mass would break this, needing a torque equation of its own.

Step 4. Add the two equations — the tension cancels:

(m_1 - m_2)\,g = (m_1 + m_2)\,a \quad\Rightarrow\quad a = \frac{m_1 - m_2}{m_1 + m_2}\,g \qquad (10)

where: the numerator is the unbalanced weight driving the system and the denominator is the total inertia resisting — gravity acts only on the mass difference, but must drag all the mass along.

Step 5. Back-substitute for the tension:

T = \frac{2\,m_1 m_2}{m_1 + m_2}\,g

where: $T$ — always strictly between the two weights: less than $m_1 g$ (so $m_1$ can accelerate down) and greater than $m_2 g$ (so $m_2$ can accelerate up). For $m_1 = 1.2\ \mathrm{kg}$ , $m_2 = 1.0\ \mathrm{kg}$ : $a = 0.89\ \mathrm{m/s^2}$ and $T = 10.7\ \mathrm{N}$ , sitting between $9.8\ \mathrm{N}$ and $11.8\ \mathrm{N}$ as promised.

Step 6. Limits: $m_1 = m_2$ gives $a = 0$ , $T = m g$ — static balance. $m_2 \to 0$ gives $a \to g$ , $T \to 0$ — $m_1$ free-falls trailing a slack string. Both endpoints are the right physics. The numeric values can be confirmed directly: $a = 0.2\times9.81/2.2 = 0.892\ \mathrm{m/s^2}$ and $T = 2\times1.2\times1.0\times9.81/2.2 = 10.70\ \mathrm{N}$ .

Limiting cases

case	result	meaning
statics, $\vec{a} = 0$	$\sum\vec{F} = 0$ (and $\sum\vec{\tau} = 0$ for extended bodies)	the diagram encodes pure balance; all of structural engineering lives here
whole-system cut	internal pairs cancel; only external forces remain	fastest route to the acceleration — but internal forces become invisible
single-body cut	internal forces of the system become external to the body	the only way to see a tension or contact force
massless connector	$T$ uniform along the string, preserved over ideal pulleys	one tension symbol per string; a massless body obeys $\sum\vec{F} = 0$ even while accelerating
circular motion	net inward component $= mv^2/r$	no new arrow — "centripetal force" is a role played by the forces already drawn
non-inertial frame	add fictitious force $-m\vec{a}_{\mathrm{frame}}$	the one legal extra arrow, and only if you insist on working in an accelerating frame

Common trapThe block is moving, so there must be a forward force on it

This is the impetus instinct, and it is the most common wrong arrow in the world's homework. Uniform motion requires no force — only changes of motion do. A coasting puck's diagram holds exactly two arrows, weight and normal, no matter how fast it slides. Its cousins are equally illegal: an " $ma$ " arrow (that is the response, not a force), a "centripetal force" arrow added alongside tension and gravity in circular motion (centripetal is a role the drawn forces fill, not a new agent), and a third-law "reaction" placed on the same body as its partner (partners live on different bodies, by definition). Finally, never write $N = mg$ by reflex: the normal force is an unknown that the perpendicular equation determines, and it equals $mg$ only on a level surface with no vertical acceleration and no other vertical forces.

Worked examples

1 · Apparent weight in an elevator (confidence builder)

A 70 kg passenger stands on a bathroom scale in an elevator. Find the scale reading when the elevator accelerates upward at $2.0\ \mathrm{m/s^2}$ , and when it accelerates downward at the same rate.

Step 1. Choose the body: the passenger. Two forces act: weight $W = mg$ down, and the scale's normal force $N$ up. The scale reads $N$ — a scale measures the contact force pressed into it, not gravity.

Step 2. Take up as positive and project equation (1): $N - mg = ma$ , so $N = m(g + a)$ with $a$ signed.

Step 3. Accelerating upward, $a = +2.0\ \mathrm{m/s^2}$ : $N = 70 \times (9.81 + 2.0) = \boxed{826.7\ \mathrm{N}}$ — about 20% over the at-rest reading of $70 \times 9.81 = 686.7\ \mathrm{N}$ . The passenger "feels heavy" because the floor genuinely pushes harder.

Step 4. Accelerating downward, $a = -2.0\ \mathrm{m/s^2}$ : $N = 70 \times (9.81 - 2.0) = \boxed{546.7\ \mathrm{N}}$ . And the limit is instructive: at $a = -g$ the normal force hits zero — free fall, weightlessness, the scale reading nothing while gravity is fully present. Weight never changed in any of these cases; only the normal force did, and the diagram keeps the two cleanly apart.

2 · Incline, pulley, and friction — the full method (exam level)

A block of mass $m_1 = 4.0\ \mathrm{kg}$ sits on a $30°$ incline with kinetic friction coefficient $\mu_k = 0.20$ . A string over an ideal pulley at the top connects it to a hanging mass $m_2 = 3.0\ \mathrm{kg}$ . The system is released and $m_2$ descends. Find the acceleration and the tension.

Step 1. Two bodies, two diagrams. On $m_1$ (moving up the slope): tension $T$ up-slope, weight component $m_1 g\sin\theta$ down-slope, friction $f_k$ down-slope (kinetic friction opposes $m_1$ 's sliding, which is up-slope), normal $N$ perpendicular. On $m_2$ : weight $m_2 g$ down, tension $T$ up.

Step 2. Perpendicular equation for $m_1$ : $N = m_1 g\cos\theta = 4.0 \times 9.81 \times 0.866 = 33.98\ \mathrm{N}$ , so $f_k = \mu_k N = 0.20 \times 33.98 = 6.80\ \mathrm{N}$ .

Step 3. Check the assumed direction before trusting it: the drive is $m_2 g = 29.43\ \mathrm{N}$ against the hold-back $m_1 g\sin\theta = 19.62\ \mathrm{N}$ — an imbalance of $9.81\ \mathrm{N}$ , comfortably above what static friction could supply at these coefficients. The system indeed moves with $m_2$ descending, so the friction arrow points down-slope as drawn.

Step 4. Along-motion equations, each body's positive axis along its own motion — for $m_1$ : $T - m_1 g\sin\theta - f_k = m_1 a$ ; for $m_2$ : $m_2 g - T = m_2 a$ . The constraint (inextensible string) has already been used: one $a$ , one $T$ .

Step 5. Add the equations — $T$ cancels: $m_2 g - m_1 g \sin\theta - f_k = (m_1 + m_2)\,a$ , so

a = \frac{29.43 - 19.62 - 6.80}{7.0} = \boxed{0.43\ \mathrm{m/s^2}}

Step 6. Back-substitute into $m_2$ 's equation: $T = m_2(g - a) = 3.0 \times (9.81 - 0.43) = \boxed{28.1\ \mathrm{N}}$ .

Step 7. Audit the answer with the other diagram: $T - m_1 g\sin\theta - f_k = 28.1 - 19.62 - 6.80 = 1.72\ \mathrm{N}$ , and $m_1 a = 4.0 \times 0.43 = 1.72\ \mathrm{N}$ — the two ledgers agree to the last digit. Cross-checking the unused equation is the professional's five-second insurance policy, and note $T$ sits between the naive extremes ( $19.6$ and $29.4\ \mathrm{N}$ ), exactly as a connecting tension must.

Origins · 1 minIsaac Newton and how this was discovered

Isaac Newton

1643–1727 · Trinity College, Cambridge

Newton never drew a free-body diagram — the Principia of 1687 argues in the austere language of Euclidean geometry, with forces appearing as line segments in elaborate geometrical constructions. But the license for the method is entirely his. The second law is stated for a body under the forces "impressed upon it", and the third law guarantees the bookkeeping stays consistent when you cut a system apart: whatever force the cut exposes on one side, its exact opposite appears on the other. What Newton supplied as axioms, later mechanicians — above all Euler, who recast the laws analytically and applied them to any body "cut free" from its surroundings — turned into the diagrammatic craft every engineer now learns first. The free-body diagram is Newton's three sentences, weaponized as a picture.

History

1687

— Newton's Principia states the three laws of motion — the license for isolating any body and equating impressed forces to its change of motion.

1736

— Euler's Mechanica recasts Newton's geometry as differential equations for a single isolated particle, the analytic ancestor of every FBD calculation.

1775

— Euler states the linear and angular momentum principles for any body cut free from its surroundings — the "cut principle" that underwrites the modern diagram.

1937

— Timoshenko and Young's Engineering Mechanics makes the free-body diagram the compulsory opening move of statics education worldwide.

2009

— Physics-education research (Rosengrant, Van Heuvelen and Etkina) documents that students who draw correct free-body diagrams dramatically outperform those who skip them.

Going further

Texts. Kleppner & Kolenkow, An Introduction to Mechanics, ch. 2 (Newton's laws — the classic force-diagram bootcamp) · Morin, Introduction to Classical Mechanics, ch. 3 (F = ma and its problem set) · Taylor, Classical Mechanics, ch. 1 · Goldstein, Poole & Safko, Classical Mechanics, ch. 1 (survey of the elementary principles). The key results derived in this article are: the frictionless incline gives $a = g\sin\theta$ and $N = mg\cos\theta$ (with correct $\theta \to 0$ and $90°$ limits); the contact force between two pushed blocks is $N_{12} = F\,m_2/(m_1+m_2)$ , giving $a = 4.0\ \mathrm{m/s^2}$ and $N_{12} = 4.0\ \mathrm{N}$ for $F = 12\ \mathrm{N}$ , $m_1 = 2.0\ \mathrm{kg}$ , $m_2 = 1.0\ \mathrm{kg}$ (and $8.0\ \mathrm{N}$ with the push reversed); and the Atwood machine gives $a = (m_1-m_2)g/(m_1+m_2)$ and $T = 2m_1m_2g/(m_1+m_2)$ , numerically $0.89\ \mathrm{m/s^2}$ and $10.7\ \mathrm{N}$ for $m_1 = 1.2\ \mathrm{kg}$ and $m_2 = 1.0\ \mathrm{kg}$ . The worked-example numbers (elevator readings $826.7 / 686.7 / 546.7\ \mathrm{N}$ ; incline-pulley $N = 33.98\ \mathrm{N}$ , $f_k = 6.80\ \mathrm{N}$ , $a = 0.43\ \mathrm{m/s^2}$ , $T = 28.1\ \mathrm{N}$ ) follow from the boxed formulas with $g = 9.81\ \mathrm{m/s^2}$ .

Builds on: Newton's Second Law · Newton's Third Law · Vectors, Coordinate Systems and Kinematics · Newton's First Law and Inertial Frames. Leads to: Normal Force · Gravitational Force Near Earth's Surface · Circular Motion and Centripetal Acceleration · Projectile Motion.

Why doesn't the force I exert on the wall appear on my own free-body diagram?

Because your diagram lists only forces exerted on you, and that force acts on the wall — it belongs on the wall's diagram. What appears on yours is its third-law partner: the wall's push back on you. The two are equal and opposite but live on different bodies, which is exactly why they can never cancel each other. This one sorting rule — every force filed under the body it acts on — is the entire discipline of the method. When you push a wall while standing on ice and slide backward, your diagram makes it obvious: the only horizontal force on you is the wall's push, so backward you go.

Is centripetal force a separate arrow on the diagram?

No — and adding one is a classic double-count. "Centripetal" names a role, not an agent: it is whatever net inward component the already-drawn forces provide. A ball whirling on a string has exactly two arrows, tension and weight; the tension (plus a slice of gravity, depending on geometry) is the centripetal force. The right procedure is to draw the ordinary forces, point one axis at the circle's center, and set the net inward component equal to $mv^2/r$ . If your diagram has an arrow literally labeled "centripetal force" with no nameable agent behind it, delete it — it fails the name-the-source test.

Where does ma go if not on the diagram?

On the other side of the equals sign. The diagram supplies the left side of $\sum\vec{F} = m\vec{a}$ — the inputs; $m\vec{a}$ is the output, the response those forces produce. Drawing it as a force double-counts the physics and wrecks the signs. (There is one honest exception: in an accelerating reference frame you may add a fictitious force $-m\vec{a}_{\mathrm{frame}}$ — note, the frame's acceleration, not the body's — to keep using Newton's law inside that frame. That is a deliberate, labeled bookkeeping device, not the body's own $ma$ sneaking in.) A clean habit: write the force sum from the diagram, an equals sign, then $ma$ — the geography of the equation mirrors the logic.

Weight and normal force on a resting book are equal and opposite — aren't they a third-law pair?

No, and the distinction is the most-tested subtlety in the topic. Third-law partners are the same interaction seen from both ends: swap the labels in "Earth pulls book" and you get "book pulls Earth" — a force on the planet. Weight (from the Earth) and normal (from the table) are two different interactions that happen to balance because the book has $\vec{a} = 0$ . Put the book in an elevator accelerating downward and $N$ drops below $mg$ immediately — they were never partners, just temporarily equal. True third-law pairs remain exactly equal and opposite under all circumstances, no equilibrium required.

How do I decide which way friction points?

Ask what the surfaces are doing, or trying to do, relative to each other. Kinetic friction opposes the actual relative sliding of the surfaces in contact. Static friction opposes the tendency to slide — the motion that would happen if friction vanished. Run that thought experiment: no friction, which way would the contact patch slip? Static friction points the other way. This produces answers that surprise beginners: friction points forward on an accelerating car's drive wheels (the tire pushes the road backward, so the road pushes the tire forward), and forward on a crate in an accelerating truck bed — friction there is the only thing accelerating the crate. Friction is not "always against the motion"; it is against relative slipping at the contact.

Why is the normal force not always mg? When can I trust that shortcut?

Because $N$ is not a law — it is an unknown, determined afresh by each problem's perpendicular equation. The surface pushes exactly hard enough to prevent interpenetration, no more. $N = mg$ survives only when the surface is horizontal, the perpendicular acceleration is zero, and nothing else has a vertical component. Tilt the surface: $N = mg\cos\theta$ . Accelerate the elevator: $N = m(g \pm a)$ . Add a rope pulling up at an angle, or press down on the block: $N$ shifts again. The reliable procedure is always the same — write the perpendicular component equation and let it tell you $N$ . Reflexively substituting $mg$ is the single most common quiet error in friction problems, because it silently corrupts $f = \mu N$ downstream.

When am I allowed to lump several bodies into one system, and when must I separate them?

Cut wherever you like — the method is valid for any choice of body, including composite ones, provided every body inside the cut shares the same acceleration. Lumping is the fast lane when you want only the overall acceleration: internal forces cancel by the third law and drop out, as in the two-block derivation where $a = F/(m_1+m_2)$ came in one line. But the price of lumping is blindness: internal forces are invisible from outside the cut. The moment the question asks for a tension, a contact force, or anything between the parts, you must cut through that connection so the force becomes external to some body. Professionals routinely do both in one problem: system first for $a$ , then one strategic single-body cut to expose the internal force asked for.

Do I draw each force where it actually acts, or all from the center?

For particle mechanics — everything in this article — slide all arrows to the center of mass and think of the body as a point. Only the vector sum enters $\sum\vec{F} = m\vec{a}$ , so the point of application changes nothing. The moment rotation matters, the point of application becomes physics: torque is $\vec{r}\times\vec{F}$ , and where the force acts determines its lever arm. Then the convention is: gravity from the center of mass (that is what "center of gravity" means), normal and friction at the contact patch, tension at the attachment point. A ladder leaning against a wall is the classic case where sliding forces to the center would destroy the problem — the torque balance is the whole content. Rigid-body articles take that up properly.

Free-Body Diagrams

Where this lives today

The physical picture

From picture to equations

The force inventory

Third-law bookkeeping across diagrams

Limiting cases

Worked examples

1 · Apparent weight in an elevator (confidence builder)

2 · Incline, pulley, and friction — the full method (exam level)

Going further

Before this

Where this leads