In short
A free body diagram (FBD) isolates a single object, removes everything it touches, and replaces each contact with the force it exerts on that object. You draw the object as a dot or simple shape, then add arrows for weight (downward), normal force (perpendicular to each contact surface), friction (along the surface, opposing relative motion), tension (along the string, pulling away), and any applied force. Once all forces are drawn, you resolve them into components along your chosen axes and apply Newton's second law: \sum F_x = ma_x, \sum F_y = ma_y.
A 5 kg wooden block sits on a tilted plank in a physics lab. The plank makes a 30° angle with the horizontal. The block does not slide — friction holds it in place. Your physics teacher asks: what forces act on the block, and how large is the friction?
You know Newton's second law: \vec{F}_{\text{net}} = m\vec{a}. You know the block is stationary, so \vec{a} = 0, which means the net force is zero. But the block is touching a surface, gravity is pulling it down, friction is pushing it up the slope, and the surface is pushing it outward. That is at least three forces in different directions. How do you keep track of all of them without losing your mind?
You draw a free body diagram.
What a free body diagram actually is
A free body diagram is not a picture of the physical situation. It is a model — a stripped-down sketch of a single object with arrows showing every external force acting on it. You remove the table, the string, the ramp, the floor, the wall — everything the object touches — and replace each point of contact with the force it exerts.
The word "free" means you have freed the body from its surroundings. The body floats alone in space, and the only things left are the forces. This is the single most important problem-solving tool in mechanics. Every problem in dynamics — from a book resting on a table to a satellite orbiting Earth — starts with an FBD.
The five forces you will draw most often
Before learning how to draw an FBD, you need to know what to draw. Five forces appear again and again in mechanics problems.
1. Weight (\vec{W} or m\vec{g}) Gravity pulls every object toward the centre of the Earth. The weight acts at the centre of gravity of the object and always points straight down, regardless of how the object is tilted or oriented.
2. Normal force (\vec{N}) Any surface in contact with the object pushes it perpendicular to the surface. A table pushes a book upward. A ramp pushes a block outward, perpendicular to the ramp's face. "Normal" means perpendicular — the force is always at right angles to the contact surface, never along it.
3. Friction (\vec{f}) A rough surface resists the object's tendency to slide. Friction acts along the surface, opposing the direction the object would move if there were no friction. If the block on the ramp would slide down without friction, friction points up the ramp.
4. Tension (\vec{T}) A string, rope, or wire pulls the object along the line of the string, away from the object. If a lamp hangs from a wire, the wire pulls the lamp upward and toward the attachment point.
5. Applied force (\vec{F}_{\text{app}}) Any push or pull that a person, engine, or mechanism exerts on the object. A porter pushing a trolley at a railway station, a crane lifting a crate — these are applied forces. You draw them in the direction the force is actually applied.
The five-step FBD construction process
Here is the method. Follow these five steps in order, every time, and you will never miss a force or add a fictitious one.
Step 1: Choose your body
Decide which single object you are analysing. Draw a boundary around it — real or imaginary. Everything inside the boundary is "the system." Everything outside is "the environment." Forces from the environment on the system go into the FBD. Forces internal to the system do not.
Why: the system boundary is the most important decision in the entire problem. Change the boundary and you change which forces are internal and which are external. If you analyse a block on a ramp, the ramp's push on the block is external. If you analyse the block-plus-ramp as a single system, that push becomes an internal force and disappears from the FBD.
Step 2: Draw the body as a dot or simple shape
Replace the real object with a point (for particle problems) or a simple rectangle (if shape matters for torque). Do not draw a detailed picture — that defeats the purpose. The FBD is a force diagram, not an engineering drawing.
Step 3: Draw weight first — always
Weight is the one force that acts on every object near Earth. It points straight down from the centre of the object. Start with this because you will never forget it if it is always first.
Why: starting with gravity is a habit that prevents the most common FBD mistake — forgetting weight. Every object has weight. No exceptions (unless the problem explicitly says "in outer space, far from any massive body").
Step 4: Identify every contact and draw the contact forces
Go around the boundary you drew in Step 1. At every point where the object touches something — a surface, a string, a spring, a hinge, another object — ask two questions:
- Is there a normal force? (Yes, if there is a surface pushing the object.)
- Is there a friction force? (Yes, if the surface is rough and the object has a tendency to slide.)
For every string or rope: draw a tension force along the string, pulling away from the object.
For every spring: draw a spring force along the spring's axis.
Why: this systematic sweep ensures you do not miss any contact force. The most common source of error in FBDs is forgetting a contact — especially when two surfaces touch at different angles.
Step 5: Draw a coordinate system
Choose your axes. For flat surfaces, use horizontal and vertical. For inclined planes, tilt your axes so that one axis runs along the slope and the other is perpendicular to it. Label the axes with x and y (or whatever symbols the problem uses) and mark the positive directions.
Why: the coordinate system determines how you decompose forces into components. A smart choice of axes minimises the number of forces you need to resolve. On an incline, tilting the axes to match the slope means the normal force has no component to resolve — it sits entirely along one axis.
The following diagram shows a complete FBD for a block on a 30° incline, built by following all five steps.
Notice what is not in the diagram: no ramp, no floor, no lab bench. The block floats alone with its forces. That is the whole point.
What NOT to include — the most common FBD mistakes
Students make five predictable mistakes when drawing FBDs. Knowing them in advance will save you marks on every exam.
Mistake 1: Forgetting a force. The most common omission is weight. The second most common is the normal force from a surface. If the object touches a surface, there is a normal force — always.
Mistake 2: Adding a force that does not exist. A block sliding on a frictionless surface has no friction. A block in free fall has no normal force. Do not draw forces that the problem has eliminated.
Mistake 3: Drawing reaction forces on the wrong body. Newton's third law says the block pushes on the table and the table pushes on the block. But the FBD of the block shows only the table's push on the block (the normal force). The block's push on the table goes in the table's FBD, not the block's.
Mistake 4: Adding "the force of motion." There is no such thing. If a block is sliding to the right, there is no rightward "force of motion" keeping it going. Newton's first law says objects keep moving without any force. Do not invent a force to explain existing velocity.
Mistake 5: Drawing weight at an angle. Weight always points straight down — toward the centre of the Earth — regardless of whether the object sits on a flat surface, a ramp, or is flying through the air. On an incline, weight still points straight down. You resolve it into components along and perpendicular to the slope, but the arrow itself is vertical.
Resolving forces into components
Once you have drawn the FBD, you need to turn the arrows into numbers. This means resolving each force into components along your chosen coordinate axes.
On a flat surface, this is straightforward: weight is purely in the -y direction, normal is purely in the +y direction, friction is along x. But on an inclined plane, the geometry demands more care.
Force resolution on an inclined plane
Consider the block on the 30° incline from earlier. You have tilted your axes so that x runs along the slope (positive up the slope) and y is perpendicular to the slope (positive away from the surface).
Weight \vec{W} = mg points straight down. You need to decompose it into two components: one along the slope (x-direction) and one perpendicular to the slope (y-direction).
The angle between the weight vector and the -y axis (the direction into the surface) is the same as the incline angle \theta. This is a geometric fact you can verify by drawing the angles.
Why: the incline makes angle \theta with the horizontal. The y-axis is perpendicular to the incline, so it makes angle \theta with the vertical. Since the weight vector is vertical and the y-axis is tilted \theta from vertical, the angle between them is \theta.
The components of weight in the tilted frame are:
Why: the component along the slope is the "opposite" side of the triangle formed by the weight vector and the tilted axes, so it uses \sin\theta. The component perpendicular to the slope is the "adjacent" side, so it uses \cos\theta. A useful sanity check: at \theta = 0° (flat surface), W_x = 0 and W_y = mg — all of the weight is perpendicular to the surface, which is correct.
The normal force N is entirely along the +y axis (perpendicular to the surface, pushing outward): N_x = 0, N_y = N.
The friction force f is entirely along the x-axis. If friction prevents the block from sliding down, it points up the slope: f_x = +f, f_y = 0.
Writing Newton's equations from the FBD
Once you have the components, applying Newton's second law is mechanical. For each axis, the sum of force components equals mass times the acceleration component along that axis.
Along the slope (x-direction):
Why: friction f points up the slope (positive x), and the weight component mg\sin\theta points down the slope (negative x). If the block is stationary, a_x = 0, so f = mg\sin\theta.
Perpendicular to the slope (y-direction):
Why: the normal force N points away from the surface (positive y), and the weight component mg\cos\theta points into the surface (negative y). The block does not accelerate perpendicular to the slope, so a_y = 0, giving N = mg\cos\theta.
These two equations — one per axis — are all you need to solve for the unknowns. For the 5 kg block on a 30° incline at rest:
The friction required to hold the block is 24.5 N, and the surface pushes back with a normal force of 42.4 N. This answers the question from the opening paragraph.
Worked examples
Example 1: Block on a 30° incline with friction
A 5 kg block sits on a rough incline tilted at 30° to the horizontal. The coefficient of static friction between the block and the surface is \mu_s = 0.4. Determine whether the block stays stationary or begins to slide. Find all forces acting on the block and write Newton's equations.
Step 1. Draw the FBD. The block touches only the incline surface, so there are three forces: weight mg (straight down), normal force N (perpendicular to the incline, outward), and static friction f (along the incline, up the slope — opposing the tendency to slide down).
Why: the block would slide down without friction, so friction points up the slope. There is no applied force, no string, no spring — just gravity and the surface contact.
Step 2. Compute the weight.
Step 3. Resolve weight into components along the tilted axes.
Why: the x-component (along the slope) is the force trying to pull the block downward along the incline. The y-component (into the surface) is the force pressing the block against the ramp, which determines how hard the surface pushes back.
Step 4. Apply Newton's second law along each axis. Assume the block is stationary (a_x = 0, a_y = 0).
Perpendicular to slope (y):
Along the slope (x):
Why: if the block is stationary, the forces must balance in both directions. The normal force balances the perpendicular weight component, and friction balances the component trying to pull the block down the slope.
Step 5. Check whether static friction can actually provide 24.5 N.
Why: the maximum static friction is \mu_s N. The block needs 24.5 N of friction to stay put, but the surface can only provide up to 16.96 N.
Since the required friction (24.5 N) exceeds the maximum available friction (16.96 N), the block slides down the incline. Static friction cannot hold it.
Result: The block does not stay stationary. It slides down because the gravitational component along the slope (mg\sin 30° = 24.5 N) exceeds the maximum static friction (\mu_s mg\cos 30° = 16.96 N). The FBD reveals this by showing that the force balance along the slope is impossible with the given coefficient of friction.
What this shows: An FBD does not just label forces — it tells you whether equilibrium is possible. By resolving forces into components and checking whether the available friction can match the demand, you answered a qualitative question ("does it slide?") with a quantitative calculation.
Example 2: A lamp hanging from two wires at different angles
A lamp of mass 4 kg hangs from a ceiling hook by two wires. Wire A makes an angle of 40° with the horizontal ceiling, and Wire B makes an angle of 60° with the horizontal ceiling. Find the tension in each wire.
Step 1. Choose the system. The system is the junction point where the two wires meet and the lamp hangs. This point is in equilibrium (it does not accelerate), so the net force on it is zero.
Why: choosing the junction as the system simplifies the problem to three concurrent forces at a single point. You do not need to worry about the shape of the lamp or where on the lamp gravity acts.
Step 2. Identify the forces. Three forces act on the junction:
- Tension T_A along Wire A, pulling up and to the left at 40° above horizontal.
- Tension T_B along Wire B, pulling up and to the right at 60° above horizontal.
- Weight mg = 4 \times 9.8 = 39.2 N, pulling straight down.
Step 3. Set up horizontal and vertical axes. Since the junction is not on a slope, use standard x (horizontal, positive right) and y (vertical, positive up) axes.
Step 4. Resolve each force into components.
For T_A (40° above horizontal, pointing up-left):
Why: the x-component of T_A points to the left (negative), and the y-component points upward (positive). Use cosine for the horizontal component and sine for the vertical component because the angle is measured from the horizontal.
For T_B (60° above horizontal, pointing up-right):
For weight:
Step 5. Apply Newton's second law. Since the junction is in equilibrium, \sum F_x = 0 and \sum F_y = 0.
Horizontal (x):
Why: there is no horizontal acceleration, so the leftward pull from Wire A must exactly balance the rightward pull from Wire B.
Vertical (y):
Why: there is no vertical acceleration, so the upward components of both tensions must together support the weight of the lamp.
Step 6. Solve the system. From equation (1):
Why: isolating T_B in terms of T_A reduces the system to one unknown.
Substitute into equation (2):
And from the relation T_B = 1.532 \, T_A:
Step 7. Verify. Check the vertical equilibrium:
Result: The tension in Wire A is approximately 19.9 N, and the tension in Wire B is approximately 30.5 N. Wire B, which makes the steeper angle with the ceiling, carries more of the load.
What this shows: When two wires support a lamp at different angles, the steeper wire bears more weight. The FBD of the junction point turns a geometrically confusing three-dimensional setup into two simple algebraic equations — one for horizontal balance, one for vertical balance. The angles determine how each tension splits between horizontal and vertical support.
Common confusions
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"The normal force always equals mg." Only on a flat, horizontal surface with no other vertical forces. On an incline, N = mg\cos\theta, which is less than mg. If someone pushes down on the block, N increases. The normal force is whatever the surface needs to exert to prevent the object from passing through it — it adjusts to the situation.
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"Friction always opposes motion." More precisely, friction opposes relative motion (or the tendency of relative motion) between the surfaces in contact. If you push a heavy box to the right but it does not move, static friction points to the left. But if you place a box on a moving truck that suddenly brakes, friction on the box points backward (in the direction the truck is decelerating) — it is friction that was carrying the box forward in the first place, and now it decelerates the box along with the truck.
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"I need to add a centripetal force arrow for circular motion." No. The centripetal force is not a new force — it is the name for whatever combination of existing forces (gravity, normal, tension, friction) points toward the centre of the circle. If a ball swings in a vertical circle on a string, the centripetal force is provided by the tension minus the weight component. You draw tension and weight in the FBD, not a separate "centripetal force" arrow.
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"The FBD should show the forces the object exerts on other things." No — that is Newton's third law, and those reaction forces go on the other object's FBD. The block pushes the table down with force mg, but in the block's FBD, you draw only the table's upward push N on the block.
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"Weight can point along the slope on an incline." Weight always points straight down. The components of weight along and perpendicular to the slope are mathematical decompositions you perform after drawing the FBD. The arrow for weight in the diagram is vertical.
If you can draw an FBD, resolve forces, and write Newton's equations for a single body, you have the core skill. What follows is for readers who want to tackle multi-body systems and subtler system-boundary choices.
System boundary and internal forces
The choice of system boundary is not always obvious, and different choices lead to different FBDs that are all correct — but some are more convenient than others.
Consider two blocks stacked on top of each other, sitting on a frictionless floor. Block A (3 kg) sits on top of Block B (5 kg). You push Block B to the right with a force F.
Choice 1: Analyse Block A alone. The forces on A are: weight m_A g (down), normal from B on A, N_{BA} (up), and friction from B on A (to the right, if A accelerates with B). You get one FBD with these three forces.
Choice 2: Analyse Block B alone. The forces on B are: weight m_B g (down), normal from floor on B, N_{\text{floor}} (up), normal from A on B, N_{AB} (down — Newton's third law partner of N_{BA}), friction from A on B (to the left — Newton's third law partner of friction from B on A), and applied force F (to the right). More forces, but you get the information about F directly.
Choice 3: Analyse the A+B system together. The internal forces (N_{BA}, N_{AB}, friction between A and B) cancel in pairs. The external forces are: total weight (m_A + m_B)g (down), N_{\text{floor}} (up), and F (to the right). The system accelerates as a = F/(m_A + m_B). This is the simplest FBD — but it hides the contact forces between A and B.
Why: internal forces always cancel when you analyse the combined system. This is Newton's third law at work — every internal force has an equal and opposite partner within the system. Only external forces survive in the combined FBD.
The strategy: use the combined system to find the acceleration, then use an individual block's FBD to find the internal contact forces.
When forces are not concurrent — torque considerations
All the FBDs in this article treat the object as a particle — a point where all forces meet. In reality, forces act at different points on an extended body, and if they do not all pass through the same point, they can create a torque that causes rotation.
For a rigid body in complete equilibrium (no translation and no rotation), you need three conditions:
The third equation — torques must balance — is essential for problems like a ladder leaning against a wall, a beam supported at two points, or a see-saw. The FBD is drawn the same way, but you must mark where each force acts on the body, because the torque depends on the distance from the pivot.
This leads to the study of rotational equilibrium, where the free body diagram becomes even more critical.
The FBD as a constraint equation generator
Every FBD produces exactly as many equations as there are axes. A two-dimensional FBD gives you two equations (x and y). A three-dimensional FBD gives three. If the problem has more unknowns than equations, you need more FBDs — one per body, plus constraint equations that relate the motions (for example, "the two blocks have the same acceleration because they are connected by a rigid string").
This is the foundation of the connected bodies technique: draw a separate FBD for each body, write Newton's second law for each, and add constraint equations. The FBD is not the end of the solution — it is the beginning. But it is the step that, if done wrong, makes every subsequent calculation wrong.
Where this leads next
- Applications: Connected Bodies and Systems — using multiple FBDs for pulleys, Atwood machines, and blocks connected by strings.
- Static and Kinetic Friction — the force that appears in almost every FBD, with the full theory of \mu_s and \mu_k.
- Motion on Inclined Planes — applying the FBD and force resolution techniques to blocks sliding, rolling, or being pushed up and down slopes.
- Newton's Second Law — the law that turns your FBD into equations of motion.
- Newton's Third Law — the law that governs which forces appear on which body's FBD.