Free Body Diagram Generator
Build, visualize, and analyze forces with interactive free body diagrams and step-by-step explanations.
Background
A free body diagram isolates one object and shows every external force acting on it. This generator helps you identify forces, draw clear arrows, break angled forces into components, write ΣF equations, and decide whether the object is balanced or accelerating.
How to use this generator
- Choose the physical situation: flat surface, inclined plane, hanging object, pulley, elevator, or custom.
- Enter mass, gravity, angle, friction, applied force, tension, or drag values as needed.
- Use auto forces for common diagrams, then add custom forces if your problem includes extra pushes or pulls.
- Click Generate diagram to see the force arrows, components, ΣF equations, acceleration, and diagram feedback.
How this generator works
A free body diagram starts by isolating one object and drawing only the external forces acting on that object. The generator converts each force into x- and y-components, adds them to find ΣFx and ΣFy, and then uses Newton’s second law to estimate acceleration when mass is provided.
For inclined planes, the generator rotates the thinking to the slope: weight is split into mg sinθ down the ramp and mg cosθ into the ramp. The normal force is perpendicular to the surface, not always equal to the full weight.
Formula & Equation Used
Newton’s second law: ΣF = ma
Components: Fx = F cosθ, Fy = F sinθ
Weight: W = mg
Friction: Ff = μN
Incline components: mg sinθ parallel to the ramp and mg cosθ perpendicular to the ramp.
Example Problems & Step-by-Step Solutions
Example 1: Box resting on a table
Problem: A 10 kg box rests on a horizontal table. Draw the free body diagram and decide whether it accelerates.
- Identify forces on the box: weight downward and normal force upward.
- Calculate weight: W = mg = (10)(9.8) = 98 N downward.
- Because the box is not accelerating vertically, the table's normal force balances the weight: N = 98 N upward.
- Write the force sums: ΣFx = 0 and ΣFy = N − W = 98 − 98 = 0 N.
- Conclusion: Net force is 0 N, so the box is in equilibrium.
Common mistake: Do not include the force the box exerts on the table. The diagram shows only forces acting on the box.
Example 2: Block sliding tendency on an incline
Problem: A 12 kg block sits on a 30° incline with μ = 0.15. Build the force diagram and estimate the down-slope net force.
- Start with weight: W = mg = (12)(9.8) = 117.6 N vertically downward.
- Split weight into incline components: mg sinθ = 117.6 sin(30°) = 58.8 N down the ramp.
- Find the perpendicular component: mg cosθ = 117.6 cos(30°) ≈ 101.8 N into the ramp.
- The normal force balances the perpendicular component: N ≈ 101.8 N.
- Estimate friction: Ff = μN = (0.15)(101.8) ≈ 15.3 N up the ramp.
- Find the net parallel force: ΣFparallel = 58.8 − 15.3 = 43.5 N down the ramp.
- Conclusion: The block tends to accelerate down the incline.
Common mistake: Normal force equals mg cosθ on this simple incline, not the full mg.
Example 3: Hanging lamp
Problem: A 5 kg lamp hangs motionless from a cable. Draw the FBD and find the cable tension.
- Identify forces on the lamp: tension upward and weight downward.
- Calculate weight: W = mg = (5)(9.8) = 49 N downward.
- The lamp is motionless, so ΣFy = 0.
- Set the equation: T − W = 0, so T = W.
- Substitute: T = 49 N.
- Conclusion: The FBD has an upward 49 N tension arrow and a downward 49 N weight arrow.
Common mistake: Tension acts along the cable. It is not automatically equal to weight if the object accelerates or if multiple cables share the load.
Example 4: Pushing a crate with friction
Problem: A 20 kg crate is pushed right with 80 N on a flat floor. If μ = 0.25, find the net horizontal force.
- Identify vertical forces: W = mg = (20)(9.8) = 196 N downward and N = 196 N upward.
- Identify horizontal forces: applied force 80 N right and friction left.
- Estimate friction: Ff = μN = (0.25)(196) = 49 N left.
- Sum horizontal forces: ΣFx = 80 − 49 = 31 N right.
- Sum vertical forces: ΣFy = 196 − 196 = 0 N.
- Use Newton's second law: a = ΣF/m = 31/20 = 1.55 m/s² right.
- Conclusion: The crate accelerates to the right.
Common mistake: Friction opposes motion or impending motion; it should point left in this example.
Example 5: Elevator accelerating upward
Problem: A 70 kg person stands in an elevator accelerating upward at 2 m/s². Find the normal force.
- Identify forces on the person: normal force upward and weight downward.
- Calculate weight: W = mg = (70)(9.8) = 686 N downward.
- Choose upward as positive and apply Newton's second law: ΣFy = ma.
- Set the equation: N − mg = ma.
- Solve: N = m(g + a) = 70(9.8 + 2) = 826 N.
- Conclusion: The normal force is larger than weight, so the person feels heavier.
Common mistake: The normal force is not always equal to weight. It changes when the object accelerates vertically.
Frequently Asked Questions
What is a free body diagram?
A free body diagram is a simplified drawing of one object with all external forces acting on it shown as arrows.
Which forces should I include?
Include only forces acting on the isolated object: weight, normal, friction, tension, applied force, spring force, drag, or other contact forces.
Why don't action-reaction pairs appear together?
Action-reaction pairs act on different objects. A free body diagram shows forces acting on only one selected object.
When is normal force equal to weight?
Normal force equals weight only in some flat-surface equilibrium cases. On inclines or accelerating systems, it can be different.
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