Free-body diagrams

AP Physics C: Mechanics· AP Physics C: Mechanics CED — Newton's Laws of Motion· 14 min read

1. Core Definition and AP Exam Conventions★★☆☆☆⏱ 3 min

A free-body diagram (abbreviated FBD) is a simplified vector diagram that isolates a single object or defined system of objects, showing only net external forces acting on the system, omitting internal forces and surrounding environment. FBDs are a foundational skill for all Newton's laws problems on the AP exam: while directly worth 2-4% of total points, they are required for nearly 30% of all exam points. On FRQs, incorrect FBDs cost points even if your final numerical answer is correct.

📘 Definition

Free-Body Diagram (FBD)

FBD

A standardized simplified vector diagram of an isolated system that shows only external forces acting on the system, omitting all internal forces and surrounding objects.

2. Force Classification and System Isolation★★☆☆☆⏱ 4 min

The first step to drawing a correct FBD is defining your system boundary, then classifying all forces as external (originating from outside the system) or internal (originating from inside the system). Internal forces cancel per Newton's third law and are always omitted from the FBD. Next, separate forces into contact and non-contact categories:

Non-contact forces: Act at a distance; only gravitational weight ( $m g$ ) appears in AP Physics C: Mechanics, and it always acts straight down toward the center of the Earth.
Contact forces: Require physical contact with another object; include normal force, tension, friction, drag, and applied pushes/pulls. Every contact point between the system and another object produces at least one contact force.

A reliable routine to avoid missed or extra forces is: 1) Draw weight first, 2) Trace the system outline and add one contact force for every point of contact with another object.

📐 Worked Example

A 2 kg block is pulled up a rough inclined plane by a rope attached to its top edge. Draw a correctly labeled FBD for the block.

1
Define the system as the isolated block, represented as a point at its center of mass per AP convention.
2
Add the only non-contact force: gravitational weight $W$ pointing straight down (not perpendicular to the incline), with magnitude $W = m g = 19.6 N$ .
3
Identify all contact points: the block touches the incline along its bottom edge, and touches the rope at its top edge, so two sets of contact forces exist.
4
From contact with the incline: draw normal force $n$ perpendicular to the incline pointing toward the block, and kinetic friction $f_{k}$ parallel to the incline pointing opposite the direction of motion (down the incline, since the block moves up).
5
From contact with the rope: draw tension $T$ along the rope pointing away from the block (up the incline). No other forces are present, so the FBD is complete.

Exam tip:

On AP FRQs, never draw net force or acceleration vectors on an FBD you are explicitly asked to draw. Only include individual actual forces acting on the system.

3. Internal vs External Forces for Connected Systems★★★☆☆⏱ 3 min

When analyzing multiple connected objects (e.g., two blocks connected by a string over a pulley), you can draw a separate FBD for each object, or a single FBD for the entire combined system. The key rule is: internal forces (forces between objects inside the system boundary) cancel out and are omitted from the combined system FBD. Only external forces (from objects outside the system) are included.

Combined system FBDs save time when calculating the acceleration of the entire system, but if you need to find the force between two connected objects (e.g., tension in the connecting string), you must isolate the individual object to treat that internal force as an external force in its FBD.

📐 Worked Example

A 1 kg block and a 3 kg block are connected by a massless string, resting on a frictionless horizontal table. A 10 N horizontal push is applied to the 1 kg block from the left, toward the 3 kg block. Draw the system-level FBD for the combined two-block system, and the individual FBD for the 3 kg block.

1
For the combined system: define the system boundary to include both blocks, so the tension between them is internal and omitted from the FBD.
2
Add the non-contact force: total weight $(m_{1} + m_{2}) g$ pointing straight down.
3
Add contact forces: total normal force $n_{1} + n_{2} = (m_{1} + m_{2}) g$ pointing straight up from the table, and the 10 N applied external push pointing right. No other external forces exist, so the system FBD is complete.
4
For the individual 3 kg block: redefine the system as just the 3 kg block, so tension from the string is now an external force. Add weight $m_{2} g$ pointing down, normal force $n_{2}$ pointing up, and tension $T$ pointing right. This is the complete individual FBD.

Exam tip:

If a problem asks for the force between two connected objects, never use only the combined system FBD to solve for it; always isolate the individual object.

4. Coordinate Alignment for Inclined Plane FBDs★★★☆☆⏱ 4 min

Inclined planes are one of the most common FBD contexts on the AP exam. The standard convention to simplify calculations is to align the x-axis parallel to the incline surface, and the y-axis perpendicular to the incline. This means only the weight vector needs to be resolved into components, as all other forces (normal, tension, friction) already lie along one of the axes.

W_x = mg \sin\theta, \quad W_y = mg \cos\theta

where $h e t a$ is the angle of the incline measured from the horizontal. A quick check to confirm you did not swap sine and cosine is to test the edge case: if $h e t a = 0 h e t a$ (flat ground), $W_{x} = 0$ (no parallel component) and $W_{y} = m g$ (full weight perpendicular to the ground), which is correct. If $h e t a = 9 0^{\circ}$ (vertical wall), $W_{x} = m g$ and $W_{y} = 0$ , which is also correct.

📐 Worked Example

A 5 kg block rests on a 30° incline, held stationary by static friction. Draw the FBD and resolve all forces into components aligned with the standard inclined-plane coordinate system.

1
Draw the block as a point at its center of mass, set the x-axis parallel to the incline (positive up the incline) and y-axis perpendicular to the incline (positive outward from the incline).
2
Draw all original force vectors: full weight $m g$ straight down, normal force $n$ along the positive y-axis, static friction $f_{s}$ along the positive x-axis (opposes the tendency to slide down the incline).
3
Resolve weight into components: $W_{x} = - m g sin θ = - (5) (9.8) (sin 3 0^{\circ}) = - 24.5 N$ . The negative sign indicates it points down the incline, opposite the positive x direction.
4
$W_{y} = - m g cos θ = - (5) (9.8) (cos 3 0^{\circ}) \approx - 42.4 N$ . The negative sign indicates it points into the incline, opposite the positive y direction.
5
All other forces are already aligned with the axes: $n = + n \overset{y}{^}$ , $f_{s} = + f_{s} \overset{x}{^}$ . For a stationary block, $\sum F_{y} = n - m g cos θ = 0$ , which matches our expectation.

Exam tip:

If you forget which trig function matches which component, confirm with the $h e t a = 0 h e t a$ edge case; this check takes 2 seconds and eliminates half of all common errors here.

5. AP-Style Concept Check★★★☆☆⏱ 2 min

✓ Quick check

Test your understanding of FBD force magnitudes with this AP-style multiple choice question:

A 10 kg block slides down a 30° rough inclined plane at constant speed. Which of the following correctly ranks the magnitudes of the forces acting on the block, as they would appear on a correctly drawn FBD?
- A) Weight > Normal force > Friction
- B) Weight = Normal force + Friction
- C) Weight > Normal force = Friction
- D) Friction > Normal force > Weight
Reveal answer
A) Weight > Normal force > Friction —
For constant speed, net force is zero, so $n = m g cos 3 0^{\circ} \approx 0.866 m g$ , $f = m g sin 3 0^{\circ} = 0.5 m g$ , so $m g > n > f$ . Option B incorrectly adds force magnitudes as scalars instead of treating them as vectors.

6. Common Pitfalls

Wrong move:

Drawing a 'force of motion' or 'inertia force' pointing in the direction of motion

Why:

Students confuse momentum/kinetic energy with an actual force acting on the object, especially for moving objects no longer being pushed.

Correct move:

Only add a force if there is a physical interaction (contact or gravitational) acting on the object; motion itself does not create a force.

Wrong move:

Pointing normal force straight up for an object on an incline

Why:

Students default to flat-surface normal force direction instead of remembering normal force is always relative to the contact surface.

Correct move:

Always draw normal force perpendicular to the contact surface, regardless of the surface's orientation.

Wrong move:

Drawing tension as a pushing force pointing toward the rope for an object attached to a rope

Why:

Students mix up tension direction, drawing it toward the rope instead of away from the system.

Correct move:

Tension in a flexible rope always pulls on the object, so draw tension pointing along the rope away from the system.

Wrong move:

Including internal tension between connected blocks in a combined system FBD

Why:

Students forget internal forces cancel, leading to an extra force in net force calculations that gives the wrong acceleration.

Correct move:

When drawing a system-level FBD for multiple connected objects, omit all forces between objects inside the system boundary.

Wrong move:

Drawing only weight components on an incline FBD instead of the full original weight vector straight down

Why:

Students confuse resolved components of weight with the original actual force vector.

Correct move:

On the original FBD, always draw the full weight vector straight down; add components as separate vectors only after drawing the original full force.

Wrong move:

Adding both total system weight and individual weights of parts in a combined system FBD

Why:

Students double-count gravitational force when combining multiple objects.

Correct move:

For a combined system, total gravitational force is $M g$ where $M$ is the sum of individual masses; do not add individual weights separately.

7. Quick Reference Cheatsheet

Category	Formula / Rule	Notes
Force classification	Contact: $n, T, f, F_{applied}$ ; Non-contact: $W = m g$	Only gravity is non-contact in AP C Mechanics; every contact force comes from a touch point
Internal force rule	$\sum F_{internal} = 0$	Omit internal forces from all FBDs; only include external forces from outside the system
Inclined plane weight components ( $θ$ from horizontal)	$W_{x} = m g sin θ$ , $W_{y} = m g cos θ$	$W_{x}$ parallel to incline, $W_{y}$ perpendicular; check with $h e t a = 0^{\circ}$ to confirm no swapped trig functions
Normal force direction	Always perpendicular to contact surface	Never draw normal straight up for inclined surfaces
Tension direction for flexible ropes	Always pulls away from the system	Ropes cannot push, so tension never points toward the system
Connected system acceleration	$a = \frac{\sum F _{external, net}}{M _{total}}$	Valid for combined system FBD only; cannot solve for internal forces (tension) with this alone
AP FBD convention	All force vectors originate at center of mass	Do not draw net force or acceleration on the FBD; only draw individual forces

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · 1
Draw FBD for block on rough incline
2022 · 2
FBD for connected Atwood system

What's Next

Free-body diagrams are the non-negotiable foundation for every problem involving Newton’s laws of motion, and without correctly drawing and interpreting FBDs, you will not be able to correctly apply Newton's laws to solve dynamics problems. This skill transfers to every subsequent topic in AP Physics C: Mechanics, including circular motion, work and energy, rotational dynamics, and oscillations. Mastering FBD conventions and avoiding common pitfalls will pay off across the entire exam, as nearly every FRQ requires at least one correctly drawn FBD to earn full credit.