A free body diagram (FBD) is a sketch that shows every force acting on a single object as an arrow. It is the most important problem-solving skill in all of mechanics — drawing one correctly is the difference between getting the right answer and getting lost.
Every physics teacher says the same thing: “Start with a free body diagram.” There is a reason. Once you see all the forces laid out clearly, applying Newton’s Second Law (F = ma) becomes straightforward algebra.
What Is a Free Body Diagram?
A free body diagram isolates one object from everything else and shows only the forces acting on that object. The object is drawn as a simple dot or box. Each force is drawn as an arrow starting from the object, pointing in the direction the force acts, with a label identifying the force.
The word “free” means the object is drawn free from its surroundings. You remove the table, the rope, the ramp, the floor — everything — and replace each with the force it exerts. The table becomes a normal force arrow. The rope becomes a tension arrow. Gravity is always present, pointing straight down.
Why Free Body Diagrams Matter (They Solve Every Problem)
Every mechanics problem — whether it involves a box on a floor, a car on a hill, or a satellite in orbit — follows the same workflow:
- Draw the FBD.
- Choose coordinate axes.
- Write ΣF = ma along each axis.
- Solve.
Without the FBD, you are guessing which forces to include and which direction they point. With the FBD, the math writes itself. A correct free body diagram turns a confusing physical situation into a clean set of equations.
Step-by-Step: How to Draw a Free Body Diagram
Follow this three-step method every time.
Step 1: Isolate the object. Pick the single object you want to analyze. Draw it as a dot or a simple rectangle. Erase everything else from your mental picture — the surface it sits on, the ropes attached to it, the person pushing it. They all disappear.
Step 2: Identify every force. Ask two questions:
- What is touching the object? Every contact point produces a force. A surface gives a normal force (perpendicular to the surface) and possibly friction (parallel to the surface). A rope gives tension. A hand gives an applied force.
- What field forces act at a distance? Gravity acts on everything with mass (W = mg, pointing straight down). In some problems, electric or magnetic forces act at a distance too, but in classical mechanics, gravity is usually the only one.
Step 3: Draw arrows. For each force, draw an arrow from the center of the object pointing in the correct direction. Make the arrow length roughly proportional to the force magnitude. Label every arrow: W (or mg) for weight, N for normal force, f for friction, T for tension, F_app for applied force.
That is it. Three steps, and you have the foundation for solving the problem.
The 5 Most Common Forces in Free Body Diagrams
1. Weight (W or mg): Gravity pulls every object toward the center of the Earth. Always points straight down. Magnitude: W = mg, where g = 9.8 m/s².
2. Normal Force (N): A contact surface pushes back perpendicular to itself. A table pushes up on a book. A ramp pushes perpendicular to the ramp’s surface (not straight up). The normal force adjusts to prevent the object from passing through the surface.
3. Friction (f): Acts parallel to the contact surface, opposing relative motion or the tendency of motion. A box pushed to the right on a floor has friction pointing to the left. A block about to slide down a ramp has friction pointing up the ramp. See the full guide at Friction in Physics.
4. Tension (T): The pulling force transmitted through a rope, string, chain, or cable. Tension always pulls — ropes cannot push. It acts along the direction of the rope.
5. Applied Force (F_app): Any external push or pull from a hand, motor, engine, or other mechanism. Direction and magnitude are given in the problem.
Free Body Diagram Examples (8 Situations)
Situation 1: Book on a Table
Forces: Weight (mg, down) and normal force (N, up). These are equal and opposite. Net force is zero — the book stays at rest. (Newton’s First Law in action.)
Situation 2: Box Being Pushed Across a Floor
Forces: Weight (down), normal force (up), applied force (horizontal, in the direction of the push), kinetic friction (horizontal, opposite to the push). The net horizontal force determines horizontal acceleration.
Situation 3: Block on an Inclined Plane
Forces: Weight (straight down — NOT along the slope), normal force (perpendicular to the ramp surface), friction (parallel to ramp, up the slope if the block tends to slide down). Weight must be decomposed into components parallel and perpendicular to the ramp. See Inclined Plane Problems.
Situation 4: Object Hanging From Two Ropes at Angles
Forces: Weight (down), tension T₁ (along rope 1, angled), tension T₂ (along rope 2, angled). Each tension must be broken into horizontal and vertical components. The object is in equilibrium, so ΣF_x = 0 and ΣF_y = 0.
Situation 5: Block Pulled by a Rope at an Angle
Forces: Weight (down), normal force (up), tension (along the rope, at an angle above horizontal), friction (horizontal, opposing motion). The upward component of tension reduces the normal force (and therefore reduces friction).
Situation 6: Object in Free Fall
Forces: Weight (down). That is the only force. No normal force, no friction, no tension. The FBD has a single arrow. Acceleration = g = 9.8 m/s² downward.
Situation 7: Car Braking on a Flat Road
Forces: Weight (down), normal force (up), friction from brakes (horizontal, opposing the car’s motion). Air resistance may also act horizontally. The net horizontal force (friction + air resistance) decelerates the car.
Situation 8: Object on an Incline With Friction and Applied Force Pushing Up the Ramp
Forces: Weight (down), normal force (perpendicular to ramp), applied force (up the ramp), kinetic friction (down the ramp, opposing the upward motion). This is the most complex common FBD — and the method handles it exactly the same way.
The #1 Mistake: Drawing “ma” as a Force
This is the single most common error students make on physics exams, and it is critical to get right.
📌 Common Misconception: Students draw “ma” as a force arrow on the free body diagram.
ma is NOT a force. It is the result of all forces combined. Think of it this way: forces are the players. Acceleration is the score. You do not draw the score on the field — it is the outcome of what the players do.
When you write ΣF = ma, the left side (ΣF) is the sum of all real forces that appear on your FBD. The right side (ma) is what those forces produce. If you add “ma” to the FBD, you are double-counting — like adding the score as an extra player.
Here is the test: Can you identify what object or surface is exerting the force? Weight comes from Earth’s gravity. Normal force comes from the surface. Tension comes from the rope. Friction comes from the surface contact. What exerts “ma”? Nothing — because it is not a force.
Free Body Diagrams for Inclined Planes
Inclined plane FBDs are the most common source of confusion, so they deserve special attention.
The key insight: weight always points straight down — even on a ramp. Do not draw weight along the slope. Draw it straight down, then decompose it into two components:
- mg sin θ parallel to the ramp surface (down the slope)
- mg cos θ perpendicular to the ramp surface (into the ramp)
The normal force equals mg cos θ (not mg!) because it only has to balance the perpendicular component of weight.
Memory trick: “Sin slides.” The component with sin θ is the one that slides the object along the surface. The component with cos θ is the one that presses the object into the surface.
Full worked problems with step-by-step FBDs are in our Inclined Plane Problems guide.
Free Body Diagrams with Multiple Objects
When a problem involves two blocks connected by a rope (or stacked on top of each other), draw a separate FBD for each object. Each object gets its own diagram with only the forces acting on that object.
For example, if block A sits on a table and a rope over a pulley connects it to block B hanging off the edge:
- Block A’s FBD: weight A (down), normal force (up), tension (horizontal, toward the pulley), friction (horizontal, opposing motion).
- Block B’s FBD: weight B (down), tension (up).
The tension is the same in both diagrams (same rope). Write F = ma for each block. You now have two equations and two unknowns (acceleration and tension). Solve simultaneously.
This technique works for any connected system — Atwood machines, stacked blocks, objects on pulleys over ramps.
Practice — Draw These FBDs Yourself
Test your skills with these scenarios. Draw each FBD before reading on.
Challenge 1: A 10 kg box on a flat floor is pushed with a 30 N force at 25° below the horizontal. There is kinetic friction (μ_k = 0.2). Draw the FBD, find the normal force, and find the acceleration.
Solution: Forces on the box — weight (98 N, down), normal force (N, up), applied force (30 N, at 25° below horizontal), kinetic friction (to the left, opposing motion).
The applied force has components: horizontal = 30 cos 25° = 27.2 N, vertical = 30 sin 25° = 12.7 N (downward, because the push angle is below horizontal).
Vertical equilibrium: N = mg + F sin 25° = 98 + 12.7 = 110.7 N. (The downward push increases the normal force.)
Friction: f_k = μ_k × N = 0.2 × 110.7 = 22.1 N.
Horizontal: F cos 25° − f_k = ma → 27.2 − 22.1 = 10 × a → a = 0.51 m/s².
Challenge 2: A 5 kg block on a 35° incline with μ_k = 0.3. Find the acceleration.
Solution: Weight components: mg sin 35° = 28.1 N (down slope), mg cos 35° = 40.1 N (into surface). Normal force: N = 40.1 N. Friction: f = 0.3 × 40.1 = 12.0 N (up slope).
Net force down slope: 28.1 − 12.0 = 16.1 N. Acceleration: a = 16.1 / 5 = 3.22 m/s² down the slope.
Challenge 3: Two blocks connected by a rope over a frictionless pulley. Block A (5 kg) on a frictionless table. Block B (3 kg) hanging vertically. Find acceleration and tension.
Solution:
- Block A FBD: T (horizontal, toward pulley). Apply F = ma: T = 5a.
- Block B FBD: Weight (3 × 9.8 = 29.4 N, down), T (up). Apply F = ma: 29.4 − T = 3a.
Substitute T = 5a into the second equation: 29.4 − 5a = 3a → 29.4 = 8a → a = 3.68 m/s². Then T = 5 × 3.68 = 18.4 N.
These problems use exactly the same method described in Newton’s Second Law (F = ma) and Newton’s Third Law (the tension force is a Third-Law pair between the rope and each block).
Use the Free Body Diagram Builder tool to check your diagrams visually.
All concepts connect back to the Classical Mechanics.
Frequently Asked Questions
What is a free body diagram?
A free body diagram is a simplified drawing that shows a single object isolated from its surroundings, with every force acting on it represented as a labeled arrow. It is the first and most essential step in solving any mechanics problem.
How do you draw a free body diagram step by step?
Isolate the object (draw it as a dot or box). Identify every force by asking what touches it (normal, friction, tension, applied) and what field forces act on it (gravity). Draw each force as an arrow from the center, label it, and make arrow lengths reflect relative magnitudes.
What forces do you include in a free body diagram?
Include only forces acting on the chosen object. Common forces are weight (mg, always down), normal force (perpendicular to surface), friction (parallel to surface), tension (along rope), and any applied push or pull. Never include forces the object exerts on other things.
Is ma a force on a free body diagram?
No. Acceleration (ma) is not a force — it is the result of all forces combined. Drawing ma on a free body diagram is the most common student mistake. Forces cause acceleration; acceleration does not cause itself. Only include real, identifiable forces with a physical source.
What is the purpose of a free body diagram?
A free body diagram lets you see every force clearly so you can apply Newton’s Second Law (ΣF = ma) correctly. Without it, you risk missing forces, getting directions wrong, or confusing forces on the object with forces the object exerts on others. It turns a messy physical situation into solvable math.
How do you draw a free body diagram for an inclined plane?
Draw weight (mg) pointing straight down — not along the slope. Draw the normal force perpendicular to the ramp surface. Draw friction parallel to the ramp surface. Then decompose weight into two components: mg sin θ along the ramp and mg cos θ perpendicular to the ramp. The normal force equals mg cos θ.