Free-Body Diagrams: Turning Real Situations into Solvable Models

What a Free-Body Diagram (FBD) Really Is

A free-body diagram is a simplified picture of a chosen system (often one object) showing only the external forces acting on that system. The purpose is not to make a pretty drawing; it is to turn a real situation into a set of equations you can solve. Once the forces are correctly identified and drawn, you translate the diagram into component equations using ΣF = m a.

An FBD answers two questions: (1) What forces act on the system? (2) In what directions do they act? Everything else (paths, scenery, other objects) is removed unless it helps identify a force.

(1) Step-by-Step Procedure for Building an FBD

Step 1: Choose the system (be explicit)

Decide what you are analyzing. Common choices: a single block, a person in an elevator, or a combined system (two blocks together). Write it down: “System = block A” or “System = (A + B) together.”

• If you choose a single object, forces from other objects appear as external forces on your diagram.
• If you choose a multi-object system, forces between objects inside the system become internal and usually do not appear (details in section 3).

Step 2: Draw the system as a simple shape

Use a dot or a box. Do not draw the whole environment. The diagram should be uncluttered.

Step 3: Add only the forces acting on the system

For each interaction with something outside the system, draw one force arrow on the system. Label each force with a clear symbol (and optionally its magnitude if known). Typical labels: W (weight), N (normal), T (tension), F_s (spring), F_app (applied).

Continue in our app.

• Listen to the audio with the screen off.
• Earn a certificate upon completion.
• Over 5000 courses for you to explore!

Or continue reading below...

Download the app

• Each force arrow has a tail on the object and points in the direction the force acts on the object.
• Do not include “motion arrows” as forces. Velocity and acceleration are not forces.

Step 4: Choose coordinate axes that simplify components

Pick axes that make the math easy. You are allowed to rotate axes.

• On an incline, choose x along the plane and y perpendicular to the plane.
• In elevator problems, choose y vertical (up positive is common).
• For a force at an angle, choose axes horizontal/vertical unless another choice reduces components.

Step 5: Resolve forces into components (only if needed)

Any force not aligned with your axes should be split into components. Use consistent signs based on your axis directions.

Example: if a force F makes angle θ above +x, then F_x = F cos θ, F_y = F sin θ.

Step 6: Write ΣF = m a in components

Write one equation per axis:

ΣF_x = m a_x  ΣF_y = m a_y

Then substitute expressions for each force component from your FBD. If an axis has no acceleration, set that component to zero (e.g., a_y = 0 for many contact problems where the object doesn’t leave the surface).

Step 7: Check your equations against the physical situation

• Do the signs match your chosen positive directions?
• If acceleration is zero, do the forces plausibly balance?
• If acceleration is upward, does the net force come out upward?

(2) Common Forces and How to Recognize Them

Weight (gravitational force)

Weight acts on the object due to gravity and points toward Earth’s center (downward near the surface).

• Magnitude: W = m g.
• It acts on the object itself, not on the surface.

Normal force (contact force perpendicular to a surface)

The normal force N is the force a surface exerts on an object, perpendicular to the surface. It adjusts to satisfy the constraint “no interpenetration” (and often “no acceleration through the surface”).

Key idea: N is not automatically equal to mg.

• Flat surface, no vertical acceleration, no other vertical forces: often N = mg.
• Incline: N is typically mg cos θ (if no other perpendicular forces).
• Elevator accelerating upward: the normal force on a person can be greater than mg.
• If you pull up on a block with an angled force, the normal force can be less than mg.

What sets N is the perpendicular component of ΣF = m a relative to the surface. You usually solve for N from that equation rather than guessing it.

Tension (force from a taut rope/cable)

Tension T acts along the rope, pulling away from the object. In idealized problems (massless rope, frictionless pulley), the tension is the same throughout a continuous rope segment.

• Tension is not “always equal to weight.” It depends on the motion/acceleration and the rest of the system.
• A rope can only pull, not push; tension direction is along the rope away from the object.

Spring force (Hooke’s law)

A spring exerts a force proportional to its extension/compression from equilibrium.

• Magnitude: F_s = k x, where x is displacement from the spring’s natural length.
• Direction: opposite the displacement (restoring).

Applied forces (pushes/pulls by a person or device)

An applied force F_app is any external push/pull not already categorized (not weight, not normal, not tension, not spring). If it is at an angle, it must be resolved into components.

Be careful: “applied force” is not a special physics force; it is just a label for “some external force.” Its direction and point of application come from the problem statement.

(3) Multi-Object Systems and Connected Masses

Why system choice matters

In connected-mass problems, you can draw an FBD for each object or for a combined system. The best choice depends on what you want to eliminate.

Internal vs external forces

Internal forces are forces between parts of your chosen system. External forces come from outside the system.

• If system = block A alone, the rope tension on A is external (it comes from the rope, which is outside the system).
• If system = (block A + block B) together, the tension between them becomes internal and does not appear on the combined-system FBD (it cancels when summing forces on the whole system).

This is powerful: choosing a combined system can remove unknown internal forces (like tension) from the net-force equation, leaving fewer unknowns.

Connected masses: shared acceleration constraint

For an ideal rope over an ideal pulley, connected objects often share the same acceleration magnitude (directions may differ). Your FBD gives force equations; the rope constraint links accelerations.

Typical workflow:

• Draw separate FBDs for each mass to write two ΣF = m a equations.
• Add a constraint like “both have acceleration magnitude a.”
• Solve simultaneously for a and T.

(4) Practice Set (with FBD Guidance and Equation Setup)

Problem A: Block on a frictionless incline

Situation: A block of mass m rests on a frictionless incline at angle θ. Find the acceleration down the plane and the normal force.

FBD (system = block):

• Weight W = mg straight down.
• Normal force N perpendicular to the plane.

Choose axes: x along plane (positive down-slope), y perpendicular outward from plane.

Resolve weight:

• Component along plane: W_x = mg sin θ (down the plane).
• Component perpendicular: W_y = mg cos θ (into the plane).

Write component equations:

ΣF_x: mg sin θ = m a  →  a = g sin θ  ΣF_y: N - mg cos θ = m a_y

Since the block does not accelerate through the surface, a_y = 0, so:

N = mg cos θ

Checkpoint: N is smaller than mg unless θ = 0.

Problem B: Person standing on a scale in an accelerating elevator

Situation: A person of mass m stands on a scale in an elevator accelerating upward with acceleration a. What does the scale read?

FBD (system = person):

• Weight mg downward.
• Normal force N upward from the scale (this equals the scale reading).

Choose axis: vertical y, positive upward.

Equation:

ΣF_y: N - mg = m a  →  N = m(g + a)

Variations to test understanding:

• If elevator accelerates downward with magnitude a, then N = m(g - a).
• If elevator moves upward at constant speed, a = 0 and N = mg.

Problem C: Two blocks connected by a rope over an ideal pulley (table + hanging mass)

Situation: Block m1 sits on a frictionless table and is connected by a massless rope over a frictionless pulley to a hanging block m2. Find the acceleration and the tension.

FBDs:

System 1 = block on table (m1):

• Tension T pulling horizontally toward the pulley.
• Normal N up and weight m1 g down (these often balance if no vertical acceleration).

System 2 = hanging block (m2):

• Weight m2 g downward.
• Tension T upward.

Choose axes:

• For m1: +x toward pulley.
• For m2: +y downward (so its acceleration is +a if it moves down).

Write equations:

For m1: ΣF_x: T = m1 a  For m2: ΣF_y: m2 g - T = m2 a

Solve: Substitute T = m1 a into the second equation:

m2 g - m1 a = m2 a  →  m2 g = (m1 + m2) a  →  a = (m2 g)/(m1 + m2)

Then:

T = m1 a = (m1 m2 g)/(m1 + m2)

Checkpoint: If m2 is very small, a is small; if m2 is very large compared to m1, a approaches g (but never exceeds it in this ideal model).

Problem D: Pulling a crate with a force at an angle (normal force is not mg)

Situation: A crate of mass m is pulled across a horizontal floor by a force F at angle θ above the horizontal. Assume no friction for this setup. Find the acceleration and the normal force.

FBD (system = crate):

• Weight mg downward.
• Normal force N upward.
• Applied force F at angle θ above horizontal.

Choose axes: +x horizontal to the right, +y upward.

Resolve applied force: F_x = F cos θ, F_y = F sin θ.

Write equations:

ΣF_x: F cos θ = m a  →  a = (F cos θ)/m  ΣF_y: N + F sin θ - mg = m a_y

The crate stays on the floor, so a_y = 0:

N = mg - F sin θ

Checkpoint: Pulling upward reduces the normal force. If F sin θ becomes larger than mg, the model “crate stays on floor” breaks (it would lift off), signaling you must revisit assumptions.

Error-Checklist: Common FBD Mistakes to Catch Early

• Including forces that do not act on the system: If the system is the block, do not draw forces the block exerts on other objects (those are reaction forces on other bodies).
• Drawing both action and reaction on the same FBD: Newton’s third-law pairs act on different objects, so they belong on different diagrams.
• Forgetting a force: Common omissions are the normal force, tension, or an applied force at an angle.
• Adding “motion forces”: Do not draw velocity, acceleration, or “centrifugal force” in an inertial-frame introductory setup. Only real interactions.
• Assuming N = mg automatically: Check the vertical/perpendicular equation. Any vertical acceleration or angled/applied forces change N.
• Wrong tension direction: Tension pulls along the rope away from the object; it does not push.
• Mixing axes and components: After choosing axes, every term in ΣF_x must be an x-component, and every term in ΣF_y must be a y-component.
• Sign confusion: Decide positive directions first, then assign signs consistently. If you change your axis choice, update signs everywhere.
• Using the wrong system for the question: If you need tension, analyze one mass; if you want acceleration without tension, consider combining masses to eliminate internal forces.
• Not checking constraints: For connected masses, ensure accelerations are linked by the rope constraint (same magnitude for an ideal rope).