How to Draw a Free Body Diagram (Worked Examples)

Replace the object with a dot, then draw one labeled arrow for every force acting on that object alone, each arrow starting at the dot and pointing in the direction the force acts. Include only real pushes and pulls from identifiable sources; never draw velocity, acceleration, or ma as forces.

AP Physics: Unit 2 (topics 2.2 Forces and Free-Body Diagrams, 2.3 Newton's Third Law, 2.5 Newton's Second Law). Free-body diagrams are Topic 2.2 in AP Physics 1 Unit 2, Force and Translational Dynamics, worth 18 to 23 percent of the multiple-choice section. AP Physics C: Mechanics builds the same skill in its Unit 2, weighted 20 to 25 percent.

What a Free Body Diagram Shows

A free body diagram shows a single object, reduced to a dot, with one labeled arrow for every force acting on that object and nothing else. No velocity arrows, no acceleration arrows, no forces the object exerts on other things. Just the pushes and pulls the rest of the universe applies to your object.

The skill is Topic 2.2 in Unit 2: Force and Translational Dynamics, worth 18 to 23 percent of the AP Physics 1 multiple-choice section. It also shows up far beyond Unit 2: energy, momentum, and circular motion problems all start with a correct diagram. Get the diagram right and the algebra usually follows. Get it wrong and every equation after it inherits the mistake, which is why graders and teachers keep insisting on this one picture. It is also one of the highest-leverage skills for managing how hard AP Physics 1 actually is, since a clean diagram earns points across the whole exam.

The Four-Step Method

Every diagram follows the same routine.

Isolate the object. Decide exactly which object the diagram describes, then mentally cut every rope, surface, and hand away from it. Draw the object as a dot.
Inventory the forces acting on it. In AP Physics 1 mechanics you get one long-range force, gravity, plus contact forces from each thing touching the object. A rough surface contributes two: a normal force and friction.
Draw each force as an arrow that starts on the dot and points away from it, aimed in the direction the force acts. Use longer arrows for stronger forces.
Label every arrow with a subscripted symbol: $F_g$ for gravity, $F_N$ for the normal force, $F_f$ for friction, $F_T$ for tension.

If you cannot name the object exerting a force, the arrow does not belong. Try the routine in the free body diagram builder, which checks your force inventory as you build each diagram.

A Force Inventory Checklist

Run down this list for every problem and you will rarely miss an arrow or invent one.

Force	Symbol	Direction
Gravity	$F_g = mg$	Straight down, always
Normal	$F_N$	Perpendicular to the surface, pushing away from it
Friction	$F_f$	Parallel to the surface, opposing sliding or attempted sliding
Tension	$F_T$	Along the rope, pulling the object away from itself toward the point of attachment
Spring	$F_s$	Toward the spring's relaxed length
Applied push or pull	$F_A$	Whatever direction the push or pull acts

The AP equation sheet writes friction as one inequality, $|F_f| \leq |\mu F_N|$ , covering both static and kinetic cases; how to find the coefficient of friction works through it. The normal force has no formula of its own. You solve for it from the other forces on the diagram, which is why how to find normal force is worth reading before test day.

Inclined Planes: Where the Normal Force Tilts

On an incline, the two mistakes to avoid are drawing the normal force straight up and inventing a force that pushes the block down the ramp. Draw exactly what acts: gravity $F_g = mg$ points straight down regardless of the ramp angle, the normal force $F_N$ points perpendicular to the ramp surface, and friction (if any) points along the surface. For a block sliding down, kinetic friction points up the ramp.

The dot-and-arrows picture comes first; components come second. Tilt your axes so x runs along the ramp and y runs perpendicular to it, then split gravity into $mg\sin\theta$ along the ramp and $mg\cos\theta$ into it. On a plain ramp with nothing else pressing on the block, $F_N = mg\cos\theta$ , which is less than $mg$ . Drag the angle around in the inclined plane simulator to watch both components change, then work the full setups in inclined plane problems.

Common Errors That Cost Points

Drawing velocity or momentum as a force. An object moving right does not need a rightward arrow. A ball at the top of its arc has exactly one force on it, gravity, even though it is moving.
**Drawing $ma$ or the net force as its own arrow.** $ma$ is what the forces already on the diagram add up to, not an extra force. Adding it double counts everything.
Normal force drawn vertically on an incline. $F_N$ is perpendicular to the surface. It points straight up only when the surface is horizontal.
Both halves of a Newton's third law pair on one diagram. The two forces in a pair act on different objects (Topic 2.3). Your diagram shows only forces on your object; the reactions belong on the other objects' diagrams.
A centrifugal force pointing outward in circular motion. Circular motion requires a net inward force supplied by real forces such as tension, gravity, or friction; see centripetal force.
Arrows floating or chained tip-to-tail. Every arrow's tail sits on the dot.

From Diagram to Net Force

The diagram is step one of every dynamics problem; the payoff comes when you turn it into equations. Pick axes (tilted on inclines, horizontal and vertical otherwise), break any off-axis forces into components, then add each axis separately to get the net force. Newton's second law connects that sum to motion: the acceleration points in the same direction as the net force, with magnitude $a = F_{net}/m$ .

Two sanity checks catch most mistakes. First, count arrows: one for gravity plus one or two for each thing actually touching the object. Second, compare the net force direction with the motion you expect. An object speeding up has a net force along its velocity, one slowing down has a net force against it, and one moving at constant velocity has zero net force. The full procedure, with sign conventions, is in how to find net force, and you can check your arithmetic with the net force calculator.

Box pulled across a rough floor

You pull a 4.0 kg box across a horizontal floor with a 30.0 N horizontal force. The coefficient of kinetic friction between the box and floor is 0.25. Draw the free body diagram and find the box's acceleration.

Isolate the box as a dot. Four forces act on it: gravity $F_g$ (down, from Earth), the normal force $F_N$ (up, from the floor), the applied pull $F_A = 30.0 \text{ N}$ (horizontal, from you), and kinetic friction $F_f$ (horizontal, from the floor, opposing the sliding).
Gravity: $F_g = mg = (4.0 \text{ kg})(9.8 \text{ m/s}^2) = 39.2 \text{ N}$ , drawn downward from the dot.
Vertical direction: the box does not accelerate up or down, so the vertical arrows must cancel: $F_N = F_g = 39.2 \text{ N}$ .
Friction: $F_f = \mu_k F_N = (0.25)(39.2 \text{ N}) = 9.8 \text{ N}$ , drawn opposite the pull.
Horizontal net force: $F_{net} = 30.0 \text{ N} - 9.8 \text{ N} = 20.2 \text{ N}$ in the direction of the pull.
Newton's second law: $a = F_{net}/m = 20.2 \text{ N} / 4.0 \text{ kg} = 5.05 \text{ m/s}^2$ .

The diagram has four arrows leaving the dot: $F_N$ up and $F_g$ down with equal lengths, $F_A$ forward as the longest horizontal arrow, and $F_f$ backward, shorter. The box accelerates at about $5.1 \text{ m/s}^2$ in the direction of the pull.

Block on a frictionless 30 degree incline

A 2.0 kg block is released on a frictionless ramp inclined at 30 degrees above the horizontal. Draw the free body diagram, then find the normal force and the block's acceleration along the ramp.

Only two forces act: gravity $F_g$ straight down and the normal force $F_N$ perpendicular to the ramp surface. Frictionless means no third arrow, and nothing pushes the block down the ramp; the along-ramp pull is a component of gravity.
Gravity: $F_g = mg = (2.0 \text{ kg})(9.8 \text{ m/s}^2) = 19.6 \text{ N}$ .
Tilt the axes: x along the ramp, y perpendicular to it. Gravity splits into $mg\sin\theta$ along the ramp and $mg\cos\theta$ perpendicular to it.
Along the ramp: $mg\sin 30^\circ = (19.6 \text{ N})(0.500) = 9.80 \text{ N}$ .
Perpendicular to the ramp: $mg\cos 30^\circ = (19.6 \text{ N})(0.866) = 17.0 \text{ N}$ . The block never leaves the surface, so $F_N = 17.0 \text{ N}$ .
Acceleration: $a = F_{net}/m = 9.80 \text{ N} / 2.0 \text{ kg} = 4.9 \text{ m/s}^2$ directed down the ramp.

$F_N = 17.0 \text{ N}$ , noticeably less than the 19.6 N weight, as it is on a plain incline where gravity and the normal force are the only forces perpendicular to the surface, and $a = 4.9 \text{ m/s}^2$ down the ramp. On the diagram, $F_g$ points straight down and $F_N$ tilts with the surface.

Lamp hanging in an accelerating elevator

A 5.0 kg lamp hangs from a cord in an elevator that is accelerating upward at 2.0 m/s^2. Draw the free body diagram and find the tension in the cord.

Isolate the lamp. Two forces act: tension $F_T$ up (from the cord) and gravity $F_g$ down (from Earth). The elevator's acceleration is not a force and does not get an arrow.
Gravity: $F_g = mg = (5.0 \text{ kg})(9.8 \text{ m/s}^2) = 49 \text{ N}$ .
Take up as positive. Newton's second law along the vertical axis: $F_T - F_g = ma$ .
Solve: $F_T = F_g + ma = 49 \text{ N} + (5.0 \text{ kg})(2.0 \text{ m/s}^2) = 49 \text{ N} + 10 \text{ N} = 59 \text{ N}$ .

$F_T = 59 \text{ N}$ . The tension exceeds the 49 N weight because the net force must point upward for the lamp to accelerate upward, so the $F_T$ arrow is drawn longer than the $F_g$ arrow. For multi-rope setups, see how to find tension.

Frequently asked questions

Is ma a force that belongs on a free body diagram?

No. The product ma is what the forces already on the diagram add up to, not a separate push or pull, so drawing an ma arrow counts the same physics twice. Draw only forces with an identifiable source object, then set their sum equal to ma when you write Newton's second law.

Which way does the normal force point on an inclined plane?

Perpendicular to the ramp surface, pushing away from it. It points straight up only when the surface is horizontal. On a plain ramp at angle theta, its magnitude is mg cos theta, which is smaller than the object's weight.

Do I draw the forces my object exerts on other things?

No. A free body diagram shows only the forces acting on the chosen object. The force your box exerts on the floor is the Newton's third law partner of the normal force and belongs on the floor's diagram, not the box's. Mixing the two on one picture is among the most common errors on force problems.

Should the arrows be drawn to scale?

Relative lengths should be reasonable: equal-magnitude forces get equal-length arrows, and a larger force gets a visibly longer arrow. Nobody expects millimeter precision, but a book resting on a table with a normal arrow twice the length of the gravity arrow contradicts the physics, since the book is in equilibrium.

Do force components like mg sin theta go on the diagram?

Keep the diagram itself to whole forces. If you add components while solving, draw them clearly distinct from the forces (dashed lines are common) and never treat a force plus its own components as three separate forces. The safest workflow is to draw the clean diagram first, then do component work next to it.