Interaction pairs and ideal tension
You will be able to: Separate third-law pairs from forces on one object and state ideal-string assumptions.
Why do equal and opposite forces not always cancel?
Two skaters push apart. Each feels the same force magnitude, yet the lighter skater accelerates more. Equal forces do not require equal accelerations.
A useful starting point: Draw the forces on one chosen object →
Words and symbols before equations
- Third-law pair
- Forces exchanged by two interacting objects; equal magnitude and opposite direction.
- Tension T
- A pull transmitted by a stretched string or cable, in N.
- Ideal string
- Negligible mass and no stretch.
- Ideal pulley
- Negligible rotational inertia and friction in this unit’s simple pulley models.
What this picture assumes
Left skater mass is 40 kg. Horizontal external forces are neglected. Interaction arrows act on different objects; vertical weight and support cancel for each skater.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- Forces are ±120 N. Left acceleration -3 m/s²; right 2 m/s². Combined center-of-mass acceleration is zero.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
Label a pair as F_A on B = −F_B on A. The arrows act on different objects, so they do not cancel in either object’s individual equation. They cancel when both objects are inside a combined system boundary.
For two isolated skaters, m_Aa_A = −m_Ba_B. The lighter mass has the larger acceleration magnitude while the system’s center of mass has zero acceleration. Contact durations and force histories match for the pair.
A massless string needs equal opposing end forces along each straight segment, so ideal tension is uniform; an ideal pulley redirects it without changing its magnitude. A massive rope can have unequal tensions because its own mass needs a net force to accelerate or support its weight.
| Question | Third-law pair | Balanced forces |
|---|---|---|
| Where do forces act? | On two different objects | On the same chosen object |
| Must magnitudes match? | Always for the interaction pair | When their vector sum vanishes |
| Implication | Equal interaction forces | Zero acceleration for constant mass |
A worked example, step by step
A 40 kg skater and a 60 kg skater push with a 120 N interaction force. Find both acceleration magnitudes.
- The interaction forces both have magnitude 120 N and opposite directions.
- For the 40 kg skater, a = 120/40 = 3 m/s².
- For the 60 kg skater, a = 120/60 = 2 m/s², opposite the first acceleration.
- The internal force pair sums to zero for the two-skater system, so its center of mass does not accelerate if external horizontal forces vanish.
Weight and normal force on a resting object are not a third-law pair; both act on the same object.
Does the heavier skater exert the larger interaction force?
Compare with an explanation
No. The pair is equal in magnitude; the masses determine their different accelerations.
Predict. Change one thing. Explain.
Hold the interaction force fixed. Increase only the right skater’s mass. Predict the two force magnitudes and both acceleration magnitudes.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Forces are ±120 N. Left acceleration -3 m/s²; right 2 m/s². Combined center-of-mass acceleration is zero.
Left skater mass is 40 kg. Horizontal external forces are neglected. Interaction arrows act on different objects; vertical weight and support cancel for each skater.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant force, system boundary, acceleration or calculus relationship to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionTwo carts of 2 kg and 5 kg push apart with 10 N each on a frictionless track. State the pair, calculate accelerations and describe the system center-of-mass acceleration.
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Compare with the answer and four-point rubric
- 1 point: Forces are +10 N on one and −10 N on the other.
- 1 point: The 2 kg cart has acceleration magnitude 5 m/s².
- 1 point: The 5 kg cart has acceleration magnitude 2 m/s², in the opposite direction.
- 1 point: Combined external horizontal force is zero, so a_cm = 0.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Why do third-law forces not cancel on one cart?
The partner force acts on the other cart.
RECALL 2Does equal force imply equal acceleration?
Only if the masses are equal.
RECALL 3What changes for a massive rope?
Tension can vary along it because the rope has inertia and weight.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Interaction pairs and ideal tension
- F_A on B = −F_B on A.
- Internal pairs cancel in the combined-system force sum.
- Equal ideal tension requires the stated string and pulley assumptions.
Remember: Weight and normal force on a resting object are not a third-law pair; both act on the same object.
Conditions: Left skater mass is 40 kg. Horizontal external forces are neglected. Interaction arrows act on different objects; vertical weight and support cancel for each skater.
Refresh Kid · AP Physics C: Mechanics Unit 2 (official Unit 2) · Objectives 2.3.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 2.3, objectives 2.3.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026 alongside the Fall 2026 clarifications. This is Mechanics Unit 2: Force and Translational Dynamics. The unit covers Topics 2.1–2.10. Calculus is introduced where it is needed for continuous mass and velocity-dependent forces. Shell theorem is applied without requiring a proof; spring combinations are purely series or purely parallel. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
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