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SCIENCE · INVESTIGATE AND EXPLAIN

Draw forces on one object without double counting

A free-body diagram isolates one object or an explicitly stated system. Each arrow names an interaction acting on that object: Earth on cart, track on cart, or string on cart. Forces the cart exerts on something else belong on a different diagram. Velocity and acceleration can be shown nearby but are not additional force arrows.

Learning outcome: Compare pull angles 0°,30° and 60° at 20 N. Draw the three actual interaction arrows before using the ledger, then resolve components in equations. Check the vertical sum and the horizontal sum separately. Predict what would break if an unbounded pull lifted the cart off the track.

Starting recap

Review signed vectors, components, acceleration, graph slopes and system boundaries. Every force must name its agent and recipient.

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Try

For m=4 kg, g=10 m/s² and a 20 N pull at 30°, the horizontal pull is 20 cos 30°≈17.32 N and the upward part is 10 N. Vertical balance gives N=40−10=30 N. Horizontal acceleration is 17.32/4≈4.33 m/s². Do not add a separate “net force” arrow to these interactions.

Open the experiment

Predict

At fixed pull magnitude, does a steeper upward angle increase or decrease horizontal acceleration and normal force?

Experiment

Compare pull angles 0°,30° and 60° at 20 N. Draw the three actual interaction arrows before using the ledger, then resolve components in equations. Check the vertical sum and the horizontal sum separately. Predict what would break if an unbounded pull lifted the cart off the track.

Resolve an angled pull on a supported cart

Worked teaching model · synthetic values · A 4 kg cart on a horizontal frictionless track; g=10 m/s². No vertical acceleration while normal force is positive. Pull has horizontal and upward components. All supported settings maintain contact.

Reset keeps your prediction, captures and practice history.

Interaction arrows are grouped by recipient. For horizontal-only cart/block diagrams, balanced vertical interactions are omitted explicitly; this is a component diagram, not a claim that gravity vanished. For the spring case only the selected spring interaction is drawn; the opposite external holding force is tabulated but omitted from that diagram. For circular cases, the arrow is the required tension, not a claim it is available when the feasibility check fails. All arrows in one view share a scale; scale changes with controls. A zero force has no visible arrow. The force table supplies exact signs and units.

CartF1: (0, -40) NF2: (0, 30) NF3: (17.32051, 10) N
Proposed-model interaction components — right and up are positive
RecipientInteractionFx (N)Fy (N)
CartEarth on cart: weight0-40
CartTrack on cart: normal030
CartPulling agent on cart17.3205110
Bounded synthetic force-model results
QuantityValueUnit
Horizontal pull component17.32051N
Upward pull component10N
Normal force30N
Horizontal acceleration4.33013m/s²

The normal force is not automatically equal to weight: the upward pull supports part of the load.

Vertical balance gives N+F sinθ−mg=0; horizontal dynamics gives F cosθ=ma. Draw the actual pull once, not both the pull and its components as additional forces.

The normal force is not automatically equal to weight: the upward pull supports part of the load.

Vertical balance gives N+F sinθ−mg=0; horizontal dynamics gives F cosθ=ma. Draw the actual pull once, not both the pull and its components as additional forces.

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

1. Orient: Draw forces on one object without double counting

A free-body diagram isolates one object or an explicitly stated system. Each arrow names an interaction acting on that object: Earth on cart, track on cart, or string on cart. Forces the cart exerts on something else belong on a different diagram. Velocity and acceleration can be shown nearby but are not additional force arrows.

2. Explain and derive

Choose axes that make constraints simple. For a cart on a horizontal track with no vertical acceleration, vertical force components sum to zero. If a pull F makes angle θ above horizontal, its components are F cosθ to the right and F sinθ upward. The normal force is then N=mg−F sinθ, not automatically mg.

In the horizontal direction of the frictionless model, F cosθ=maₓ. Increasing the pull angle while holding its magnitude fixed decreases the horizontal component and decreases the normal force. The pull is a single force: drawing it and also counting both of its components would count the same interaction twice.

A normal contact force cannot pull the cart down onto the surface. If the calculation required N<0, the assumed contact would fail and a new motion model would be needed. The bounded settings here keep N positive. On an incline, rotating axes along and perpendicular to the surface makes the gravitational components mg sinθ and mg cosθ; the normal still follows the perpendicular constraint, not a memorized equality.

3. Work an example

For m=4 kg, g=10 m/s² and a 20 N pull at 30°, the horizontal pull is 20 cos 30°≈17.32 N and the upward part is 10 N. Vertical balance gives N=40−10=30 N. Horizontal acceleration is 17.32/4≈4.33 m/s². Do not add a separate “net force” arrow to these interactions.

4. Predict and investigate

Predict first: At fixed pull magnitude, does a steeper upward angle increase or decrease horizontal acceleration and normal force?

Compare pull angles 0°,30° and 60° at 20 N. Draw the three actual interaction arrows before using the ledger, then resolve components in equations. Check the vertical sum and the horizontal sum separately. Predict what would break if an unbounded pull lifted the cart off the track. Use the lesson controls and record the settings with your evidence.

5. Transfer and explain

Try a new practice case

A 5 kg cart is pulled horizontally by 24 N on a frictionless track. Find horizontal acceleration.

This extra practice is not saved and does not award mastery. Record your first answer, explanation and whether you revealed help in the lesson notebook. Use the lesson’s saved question separately.

6. Reflect and retain evidence

For a frictionless inclined plane with angle 30°, derive acceleration from mg sinθ=ma. Explain why changing mass does not change this ideal acceleration.

Retain your first answer, force diagram, equations, units, assumptions and a changed-case prediction. Public worked checks are practice, not independent mastery evidence. Supplied synthetic outputs do not certify actual laboratory skill.

Contact mechanism and an exam-style dot diagram

Normal and friction forces are macroscopic descriptions of microscopic electromagnetic interactions at contacting surfaces; gravity acts without contact. A force must still name an external agent and a recipient. A body cannot produce a net force on itself by adding its own internal interactions. In the force analysis here, electrical microscopic detail is explanatory context, not an additional electric-force calculation.

Represent the selected cart by a dot and draw each actual force as a straight arrow originating there. Do not draw component arrows in addition to their parent force, or an extra net-force arrow. Two distinct forces pointing in the same direction must remain distinguishable side by side rather than overlap. Choose an axis along an incline when helpful, while the gravity arrow stays vertical. Changed representation: two strings pull a dot rightward with 3 and 5 N while a third pulls leftward with 4 N. Draw three arrows with a declared common scale, then calculate net force +4 N. For a 2 kg cart the acceleration is +2 m/s². Replacing the two rightward strings by one 8 N string preserves the net acceleration but changes the interaction inventory.

Framework and source notes

Original teaching prose and numerical cases. College Board AP Physics 1 CED, Unit 2 topic mapping: 2.2; 2.5. Checked 30 October 2026. This is a content cross-reference, not AP authorization.

Apply

A 4 kg cart has weight 40 N and an applied upward force component of 12 N, with zero vertical acceleration. What is the normal force?

Public practice checks a specific claim. It does not establish independent mastery.

For a frictionless inclined plane with angle 30°, derive acceleration from mg sinθ=ma. Explain why changing mass does not change this ideal acceleration.

Continue

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