Connected objects and tension
You will be able to: Solve a two-mass pulley problem using system and individual force equations.
How can one system equation simplify two moving objects?
Two hanging masses share a taut cord over a light, freely turning pulley. The heavier side falls and pulls the lighter side up. Both travel the same distance in the same time.
A useful starting point: Net force and acceleration →
Words and symbols before equations
- Ideal cord
- Massless and inextensible; it does not stretch.
- Ideal pulley
- Negligible rotational inertia and friction in this model.
- Constraint
- A connection that relates the objects’ motions.
What this picture assumes
Left mass 2 kg. Ideal massless inextensible cord and frictionless pulley with negligible inertia; g=10 m/s². Signed a is downward for the right mass.
Connect the picture to the physics
An ideal continuous cord over an ideal pulley has the same tension magnitude on both sides. The connected masses have equal acceleration magnitudes but opposite vertical directions. A massive hanging cord is different: the upper portion supports more cord below it, so its tension can be greater. We treat that variation qualitatively here.
Choose downward positive for the heavier mass and upward positive for the lighter mass. Write one equation for each. Adding them eliminates the unknown tension; this is algebraic cancellation using the connected-motion constraint.
For a horizontal pair of carts, selecting both carts makes their mutual contact internal. For this pulley problem, do not confuse adding equations along opposite positive axes with a vector sum in one common vertical direction.
A worked example, step by step
A 3 kg mass hangs opposite a 2 kg mass. Use g=10 m/s². Find acceleration and cord tension.
- For 3 kg, downward positive: 30−T = 3a.
- For 2 kg, upward positive: T−20 = 2a.
- Add: 10 = 5a, so a = 2 m/s². The 3 kg side descends.
- Substitute into the light-side equation: T=20+2(2)=24 N. Check heavy side: 30−24=3(2).
Tension is not automatically the weight of either hanging mass when the system accelerates.
Why do equal masses give zero acceleration?
Compare with an explanation
Their equal gravitational driving terms cancel; with the ideal constraints, each tension equals its weight.
Predict. Change one thing. Explain.
Keep the light mass at 2 kg. Change the other mass. Predict the zero-acceleration case and compare tension with each weight when the masses differ.
Right-side signed acceleration 2 m/s² downward; left-side acceleration has equal magnitude and opposite direction. Tension 24 N. Force scale 2 drawing units/N; cord geometry schematic.
Left mass 2 kg. Ideal massless inextensible cord and frictionless pulley with negligible inertia; g=10 m/s². Signed a is downward for the right mass.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use a force or motion 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 questionAn ideal pulley connects 4 kg and 1 kg masses. (a) Write the two equations, (b) find acceleration, (c) find tension, and (d) check both equations.
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Compare with the answer and four-point rubric
- 1 point: 40−T=4a and T−10=a.
- 1 point: a=30/5=6 m/s².
- 1 point: T=10+6=16 N.
- 1 point: 40−16=24=4×6 and 16−10=6=1×6.
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 equal acceleration magnitudes?
The taut inextensible cord fixes the combined length.
RECALL 2When is tension uniform?
For the stated massless cord and ideal pulley.
RECALL 3How do you check a solution?
Substitute into each original force equation.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Connected objects and tension
- a = (m_heavy−m_light)g/(m_heavy+m_light).
- T = m_light(g+a) = m_heavy(g−a), for this ideal setup.
Remember: Tension is not automatically the weight of either hanging mass when the system accelerates.
Conditions: Left mass 2 kg. Ideal massless inextensible cord and frictionless pulley with negligible inertia; g=10 m/s². Signed a is downward for the right mass.
Refresh Kid · Unit 2 · Objectives 2.3.A, 2.5.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 2.5, objectives 2.3.A, 2.5.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. Fall-2026 corrections also checked. 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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