Free-body diagrams begin with a boundary
Choose a mechanical system, distinguish external interactions from internal forces, and test a diagram against measured acceleration.
On this page
The question for today
What you’ll understand
- Draw forces acting on a selected system rather than every force nearby.
- Explain which interaction forces cancel when the boundary expands.
- Use components and an acceleration check to reject an incomplete diagram.
A force diagram is a compact physical claim. Every arrow names an interaction crossing the boundary of the object or collection of objects you chose to study. Before calculating, decide what lies inside that boundary and which observer measures acceleration. These choices turn an attractive drawing into a testable model. The same apparatus can support several correct diagrams, provided each diagram and its equation refer to the same system.
Select the system before naming forces
Imagine two carts joined by a light connector and pulled along a smooth horizontal track. A boundary around the rear cart excludes the connector and front cart. The connector's pull must therefore appear as an external force on that cart. A boundary around both carts makes the connector interaction internal. Its two opposing forces cancel in the sum for the combined system, although each cart still feels a force individually. An internal force has not ceased to exist; it has ceased to affect the total momentum of this chosen collection.
Represent a compact object by a point when its translational motion is the target. Draw weight from Earth, the track's normal reaction, and every other identified contact or field interaction. An arrow labeled motion or acceleration is not an additional force. Newton's law relates acceleration to the vector sum of interaction forces. If rotation matters, a point diagram loses essential information about application points and lever arms; retain the body's geometry and add a torque balance. Choose the model to answer a stated question rather than always drawing the simplest possible picture.
Components are bookkeeping, not extra interactions
Choose axes that make the constraints clear. On a straight incline, one axis along the surface and one perpendicular to it usually reduce the algebra. Weight remains one downward force even when you replace it by two components in the equations. Do not draw both the original weight and its components as three independent interactions. A normal force acts perpendicular to the local contact surface, so its magnitude is not automatically equal to weight. Resolve the complete equation perpendicular to the surface before deciding what the normal reaction must be.
Third-law partners act on different bodies. Earth's gravitational pull on a cart and the cart's gravitational pull on Earth therefore cannot cancel in the cart's own equation. Weight and a supporting normal force can balance on the same body, but they are not a third-law pair: they arise from different interactions. This distinction becomes practical when a lift accelerates. The passenger's acceleration requires unequal upward and downward forces even though every interaction still has an equal and opposite partner acting somewhere else. Balanced forces describe zero acceleration, not an absence of interactions.
Audit the diagram against a possible observation
After solving, check whether the assumed contacts remain physically possible. An ordinary floor can push upward but cannot pull a detached object downward. If the algebra demands a negative normal reaction, the maintained-contact assumption has failed and a free-flight model is needed. A slack string similarly cannot transmit compression. These conditions are part of the model, not optional corrections after the answer. An acceleration measured with video can distinguish a missing interaction from an incorrect mass estimate, especially when you repeat the experiment after changing the applied force.
A useful audit has three parts: inspect units, compare a limiting case, and identify an independent measurement. Doubling a horizontal pull on the same frictionless cart should double its horizontal acceleration. Joining another equal cart should halve the acceleration for the same total pull. If neither prediction holds, inspect track friction, connector mass, and the definition of the measured system. Real experiments can challenge several assumptions simultaneously, so change one control at a time. The diagram earns its value by explaining how a particular observation could show that its boundary or interaction list is inadequate.
Work it through
One apparatus, two useful boundaries
Two carts of masses and are joined by a massless connector on a horizontal frictionless track. A horizontal pull acts on the front cart. Find their acceleration and the connector tension.
Choose both carts as the first system. The connector forces are internal, while the horizontal external resultant is . Thus .
Choose only the rear cart. Its horizontal external force is the forward connector tension, so .
Check the front cart separately: its forward pull is and backward tension is . The resultant gives , agreeing with the common acceleration.
Both boundaries predict the same acceleration; the smaller boundary exposes an internal load that the combined-system equation cannot determine alone.
The living notebook
Explore the model
Interactive lab · vectors
Resolve one force consistently
Set angle to 30° and magnitude to 2; predict projections √3 and 1, then check them. Change the angle to 90° and predict which component vanishes before moving the control. This is a component illustration, not a full force-balance or multi-cart simulation.
- x component
- 1.393
- y component
- 0.975
- Magnitude
- 1.7
Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forComponents change with the axes or direction; they do not add new interactions.
Pause & reason
Check your understanding
Your turn
Try it yourself
A parcel sits on a horizontal lift floor accelerating upward at . Use . Find the normal force and name the third-law partner of that force.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Draw only forces on the parcel, take upward positive, and distinguish the floor's partner from Earth's pull.
Compare with the solution
The parcel obeys , giving . The third-law partner is the parcel's downward force on the floor. Its weight is a different interaction, so it need not equal N during upward acceleration.
Keep the idea
Take it with you
- A system boundary determines whether an interaction is internal or external.
- Acceleration and vector components are not additional interaction forces.
- Contact conditions and independent measurements test whether the diagram remains valid.
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