Free-body diagrams — AP Physics C: Mechanics Study Guide

For: AP Physics C: Mechanics candidates sitting AP Physics C: Mechanics.

Covers: Identification of contact vs non-contact forces, correct vector notation, system isolation, internal vs external force distinction, drawing diagrams for connected objects and inclined planes, and exam-specific problem-solving techniques.

You should already know: Vector addition and resolution, Newton’s three laws of motion, how to define a coordinate system.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics C: Mechanics style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Free-body diagrams?

A free-body diagram (often abbreviated FBD, sometimes called a force diagram) is a simplified vector diagram that isolates a single object or defined system of objects, showing only the net external forces acting on that system while omitting internal forces and the object's surrounding environment. Per the AP Physics C: Mechanics Course and Exam Description (CED), this topic is a core foundational skill for all Newton’s laws of motion problems. While it directly accounts for roughly 2-4% of total exam weight, it is required to earn points on nearly 30% of all exam points, since almost every FRQ and roughly one quarter of MCQ require drawing or interpreting an FBD to reach a correct solution. FBDs appear in both MCQ and FRQ sections of the exam; on FRQs, you will often be explicitly asked to draw an FBD, and incorrect force labels or extra, unbalanced forces will lose you points even if your final numerical answer is correct. The standard AP convention is to represent the object as a point (or simple box, if needed for clarity) with all force vectors originating at the object’s center of mass, pointing outward in the direction the force acts, and each force labeled with a clear, standard notation (e.g., $n$ for normal force, $T$ for tension, $f_s$ for static friction).

2. Contact vs Non-Contact Forces and System Isolation

The first step to drawing a correct FBD is defining your system (the object(s) you are analyzing) and classifying all forces acting on it as either external (originating outside the system) or internal (originating from another part of the system, which cancel out per Newton’s third law and are omitted). Next, you must separate forces into contact and non-contact: contact forces require physical contact between the system and another object, and include normal force, tension, friction, drag, and applied push/pull. Non-contact forces act at a distance, and the only non-contact force you will encounter in AP Physics C Mechanics is gravitational force (weight). A reliable starting routine that eliminates most missed or extra forces is: 1) Draw weight first: it always acts straight down toward the center of the Earth, with magnitude $mg$, unless the problem specifies another gravitational acceleration. 2) Then trace around the system outline: every place the system touches another object, there is at least one contact force. For a smooth surface, that’s just normal force perpendicular to the surface; for a rough surface, add friction parallel to the surface. For a rope/string attached, that’s tension pulling along the string away from the system.

Worked Example

A 2 kg block is being pulled up a rough inclined plane by a rope attached to its top edge. Draw the full, 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 $\vec{W}$ pointing straight down (not perpendicular to the incline) with magnitude $W=mg=19.6\text{ N}$.
3. Identify all contact points: the block touches the incline along its bottom edge, and touches the rope at the top edge, so there are two sets of contact forces.
4. From contact with the incline: draw normal force $\vec{n}$ perpendicular to the incline surface pointing outward toward the block, and kinetic friction ${\vec{f}}_k$ 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 $\vec{T}$ 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, do not draw net force or acceleration vectors on an FBD you are asked to draw. Only draw individual forces acting on the system; net force is a result, not an actual force, and including it will cost you a point.

3. Internal vs External Forces for Connected Systems

When analyzing multiple connected objects (e.g., two blocks stacked, two blocks connected by a string over a pulley), you can either draw a separate FBD for each individual object, or draw a single FBD for the entire combined system. The key rule here is that internal forces (forces between objects within the system) are always equal in magnitude and opposite in direction per Newton’s third law, so they sum to zero net force and can be omitted from the system-level FBD. Only external forces (forces from objects outside the combined system) are included. This is a huge time-saver for problems asking for the acceleration of the entire system: you don’t need to solve for tension first if you treat the whole system as one. But if the question asks for tension (the force between the two objects), you must isolate the individual object to get that internal force into your FBD as an external force for that individual system.

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 is 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 to the right. No other external forces are present, 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 $\vec{T}$ from the string pointing right. This is the complete individual FBD.

Exam tip: If the problem asks for the force between two connected objects, never use the combined system FBD alone to solve for it; always isolate the individual object to get that force as an external force in your FBD.

4. Coordinate System Alignment for Inclined Planes

One of the most common contexts for FBDs on the AP exam is an object on an inclined plane, and the biggest mistake students make here is misaligning forces with the coordinate system after drawing the FBD. 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 you only need to resolve the weight vector into components; all other forces (normal, tension, friction) already lie along one of the axes. The weight vector points straight down, so its components for an incline at angle $\theta$ from the horizontal are: $$W_x=mg\sin \theta (\text{parallel to the incline, pointing down the incline})$$ $$W_y=mg\cos \theta (\text{perpendicular to the incline, pointing into the incline})$$ A quick check to confirm you did not swap sine and cosine is to test the edge case: if $\theta =0^{\circ }$ (flat ground), $W_x=0$ which makes sense (no component parallel to the ground), and $W_y=mg$, which matches the full weight on the flat surface. If $\theta =90^{\circ }$ (vertical wall), $W_x=mg$ 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 $mg$ straight down, normal force $n$ along positive y-axis, static friction $f_s$ along positive x-axis (opposes the tendency to slide down the incline).
3. Resolve weight into components: $W_x=-mg\sin \theta =-(5)(9.8)(\sin 30^{\circ })=-24.5\text{ N}$ (negative because it points down the incline, opposite the positive x direction).
4. $W_y=-mg\cos \theta =-(5)(9.8)(\cos 30^{\circ })\approx -42.4\text{ N}$ (negative because it points into the incline, opposite the positive y direction).
5. All other forces are already aligned with the axes: $n=+n\hat{y}$, $f_s=+f_s\hat{x}$. For a stationary block, the sum of forces in the y-direction is $n-mg\cos \theta =0$, which matches our expectation.

Exam tip: Always confirm your weight components with the $\theta =0^{\circ }$ edge case if you forget which trig function goes with which component; this check takes 2 seconds and eliminates 50% of common errors here.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Drawing a "force of motion" or "inertia force" pointing in the direction the object is moving. Why: Students confuse momentum/kinetic energy with an actual force acting on the object, especially for moving objects that are 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 the normal force straight up for an object on an incline. Why: Students default to the 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: Including tension as a pushing force on an object attached to a rope. Why: Students mix up the direction of tension, drawing it pointing toward the rope instead of away. 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: Keeping internal tension between two connected blocks in a combined system FBD. Why: Students forget that internal forces cancel, so they end up with an extra force in their net force calculation that gives the wrong acceleration. Correct move: When drawing a system-level FBD for multiple connected objects, cross out any forces between objects inside the system boundary before calculating net force.
• Wrong move: Drawing weight as a component perpendicular to the incline instead of straight down on the original FBD. Why: Students confuse the resolved components of weight with the original force vector. Correct move: On the original FBD, always draw the full weight vector straight down; add the components as separate vectors only after drawing the original full force.
• Wrong move: Adding both the total weight of a system and the weights of individual parts when drawing a combined system FBD. Why: Students double-count gravitational force when combining multiple objects. Correct move: For a combined system, the total gravitational force is just $Mg$ where $M$ is the sum of the masses of all parts; do not add individual weights separately.

6. Practice Questions (AP Physics C: Mechanics Style)

Question 1 (Multiple Choice)

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

Worked Solution: First, draw the correct FBD, which has three forces: total weight $mg$, normal force $n=mg\cos \theta$, and friction $f=mg\sin \theta$. Constant speed means net force is zero, so these relationships hold. For $\theta =30^{\circ }$, $\cos 30^{\circ }\approx 0.866$ and $\sin 30^{\circ }=0.5$, so $mg>mg\cos 30^{\circ }>mg\sin 30^{\circ }$. Option B is incorrect because it adds forces as scalars, not vectors. The other options have incorrect ranking. The correct answer is A.

Question 2 (Free Response)

Two blocks, of mass $m_1=2\text{ kg}$ and $m_2=5\text{ kg}$, are connected by a massless inextensible string over a massless frictionless pulley. Block $m_1$ rests on a horizontal rough table, coefficient of kinetic friction ${\mu }_k=0.2$, and block $m_2$ hangs vertically off the edge of the table. The system is released from rest. (a) Draw a correctly labeled free-body diagram for each block. (b) Write Newton's second law for each block, using your FBDs. (c) Calculate the magnitude of the acceleration of the system and the tension in the string.

Worked Solution: (a) For $m_1$ (on the table): Forces are: 1) Weight $m_{1}g$ straight down, 2) Normal force $n$ straight up from the table, 3) Tension $T$ from the string pointing horizontally toward the pulley, 4) Kinetic friction $f_k$ pointing horizontally opposite the direction of motion. For $m_2$ (hanging): Forces are: 1) Weight $m_{2}g$ straight down, 2) Tension $T$ from the string pointing straight up. (b) For $m_1$, take positive $x$ right, positive $y$ up. No acceleration in $y$: $$\sum F_y=n-m_{1}g=0⟹n=m_{1}g$$ $$\sum F_x=T-{\mu }_k{m}_{1}g=m_{1}a$$ For $m_2$, take positive $y$ downward (aligned with acceleration direction): $$\sum F_y=m_{2}g-T=m_{2}a$$ (c) Add the two force equations to eliminate $T$: $$m_{2}g-{\mu }_k{m}_{1}g=a(m_1+m_2)$$ $$a=9.8\frac{5-(0.2)(2)}{2+5}\approx 6.44{\text{ m/s}}^2$$ Substitute back to find $T$: $T=m_2(g-a)\approx 16.8\text{ N}$

Question 3 (Application / Real-World Style)

A 1500 kg elevator accelerates upward at $1.2{\text{ m/s}}^2$ when starting from rest on the ground floor of a building. The elevator is supported by a single steel cable. Draw the FBD for the elevator and calculate the tension in the cable, then explain what your answer means for the cable's required strength.

Worked Solution: The correctly drawn FBD for the isolated elevator has two forces: weight $mg$ pointing straight down, and tension $T$ from the cable pointing straight up. Taking upward as the positive direction, Newton's second law gives: $$\sum F=T-mg=ma$$ Solve for $T$: $$T=m(g+a)=1500(9.8+1.2)=16500\text{ N}=16.5\text{ kN}$$ This result means the support cable must be engineered to withstand a minimum tension of 16.5 kN when accelerating the loaded elevator upward from rest; a cable weaker than this will snap under the load.

7. Quick Reference Cheatsheet

8. 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 second law to solve for acceleration, force, or motion parameters. Even if your final calculation is correct, an incorrectly drawn FBD on an AP FRQ will cost you points. Immediately after mastering FBDs, you will apply your skill to solving Newton’s second law problems for single and connected systems, including inclined planes and pulley systems. FBDs also remain critical for later topics across the course: you need them to identify which forces do work in energy problems, to find the net centripetal force for circular motion, and to calculate the net restoring force for mass-spring oscillations.

• Newton's Second Law
• Connected Systems and Pulley Problems
• Uniform Circular Motion
• Work and Kinetic Energy