Draw forces on one selected object

Decide what belongs in the diagram

A cart moves along a horizontal track while an agent applies a horizontal force. Earth pulls downward on the cart; the track pushes upward. Before adding arrows, name the object being analyzed: the cart alone. A free-body diagram lists external forces acting on that selected object. It does not collect every force somewhere in the scene.

The OpenStax guide to free-body diagrams explains this object-selection method. Here the dot represents the cart as a particle. The original diagram uses a common force scale for all arrows, with direction and numerical values also supplied in a table. It does not show the shape of the track or the path traveled.

Name the interaction, not just the arrow

The model includes three interactions on the cart: agent on cart, track on cart, and Earth on cart. Choose right as positive horizontal direction and up as positive vertical direction. Each interaction has a horizontal component and a vertical component, which are added separately.

The agent applies a constant horizontal force, labeled Fₓ; the subscript x identifies its horizontal component. Positive Fₓ points right and negative Fₓ points left. The track’s normal force points upward, perpendicular to the horizontal contact surface. Earth’s gravitational force, the cart’s weight, points downward.

A velocity arrow is not a force. “The cart is moving right” does not justify drawing an additional rightward force. Likewise, the net force is the sum of the existing forces, not a fourth interaction to add to them again. The diagram shows interactions; the ledger separately reports their sum.

Distinguish kilograms from newtons

Mass m is measured in kilograms. Weight W is a force measured in newtons. In this teaching model, W = mg, where g = 10 m/s² is an assigned gravitational acceleration magnitude. A 2 kg cart therefore has weight magnitude 2 × 10 = 20 N. The downward component is −20 N under our positive-up convention.

The unit newton is defined so that 1 N = 1 kg·m/s². The letter N beside a numerical value is the unit symbol. In the diagram, the variable name N is also used for the normal-force magnitude; context distinguishes that quantity from the unit. The table names the interaction explicitly to avoid relying on the letter alone.

The cart has no vertical acceleration and the applied force is purely horizontal. Thus the upward normal force must equal the downward weight magnitude: N − mg = 0, giving N = 20 N. This equality comes from the specific vertical balance. It is not a universal statement that every normal force always equals weight. An inclined surface, a vertical applied force or vertical acceleration would require another calculation.

Your action: double mass to 4 kg while holding the horizontal force fixed. Weight and normal force both become 40 N, and their arrows grow equally. Vertical net force remains zero. The larger individual forces have not vanished merely because their vector sum is zero.

Equal and opposite is not enough to identify a third-law pair

Newton’s third law relates the two forces in the same interaction between two bodies. Track-on-cart pairs with cart-on-track. Earth-on-cart pairs with cart-on-Earth. Agent-on-cart pairs with cart-on-agent. Each pair has equal magnitude and opposite direction, but its two forces act on different bodies.

The upward normal force and downward weight both act on the cart, so they are not a third-law pair. They happen to balance because of the model’s vertical constraint. Their actual partners act on the track and Earth, not on the cart. The diagram excludes those partners rather than making them cancel forces on the wrong object.

For a horizontal applied force of +4 N, the cart exerts −4 N horizontally on the agent. That reaction is real within the interaction model, but belongs on the agent’s free-body diagram. Adding it to the cart’s +4 N would incorrectly predict zero horizontal net force for the cart. The agent’s own motion would require its mass and other forces, which are not supplied here.

Test the zero-force claim precisely

Set the applied force to zero. Its horizontal arrow disappears, while normal force and weight remain. The horizontal net force is zero, and the vertical forces still sum to zero. This means zero total net force, not “no forces exist.” The cart already has initial horizontal velocity +2 m/s; with no net force to change it, that velocity remains constant in this ideal inertial frame.

This frictionless track is a teaching idealization. A real cart might experience rolling resistance, air drag or track irregularities. If observations disagree, inspect the interactions and assumptions before inserting an unexplained “motion force.” A useful diagram must represent the selected system and relevant evidence, not just produce a convenient cancellation.

Transfer: A 3 kg object rests on a horizontal support under teaching g=10 m/s². A separate agent pulls vertically upward with 6 N while the object remains in contact and has no vertical acceleration. What normal force is needed? Identify the third-law partner of the upward 6 N force.

Check the force-diagram transfer

Weight is 30 N downward. Vertical balance requires N+6−30=0, so the support’s normal force is 24 N upward. The partner of agent-on-object upward 6 N is object-on-agent downward 6 N; it does not belong in the object’s force sum. This changed vertical-force case is outside the horizontal-only widget and shows why N=mg is not universal.