Connected bodies: one a, two free-body diagrams
"Connected bodies: one a, two free-body diagrams" is a free, interactive Class 11 Physics simulation for NEET and JEE, built on the NCERT section "Solving problems in mechanics" in Laws of Motion. Atwood machine and blocks-pushed-together: system acceleration, then a separate FBD for each body gives tension or contact force. NEET has asked about this idea 3 times in 10 years of papers. JEE Main asked 39 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).
From the lesson Solving problems in mechanics in Laws of Motion. Change the values and watch what happens.
What it shows
NCERT § Solving Problems
Atwood machine and blocks-pushed-together: system acceleration, then a separate FBD for each body gives tension or contact force.
Controls
4 to try
- ToggleAtwood / two blocks pushed / block on table with hanging mass
- Sliderm₁, m₂ 1–10 kg
- Sliderpush force 0–50 N
- Slidertable μ 0–0.5
Try this
On the Atwood machine, choose masses so the acceleration is g/5 and read the tension; then push two blocks so the contact force is 12 N.
Common mistake
Taking tension as m₁g, or forgetting each body needs its own FBD.
In NEET
10 years of papers
Asked 3 times in 10 years of NEET.
See the past questions: Newton's Laws of Motion, Laws of Motion (chapter)
In JEE Main
42 shifts analysed
39 questions from Laws of Motion across the 42 JEE Main shifts Lumi has analysed (2017–2026).
The idea behind it
NCERT § "Solving Problems in Mechanics"
- Pick the body (or system) and draw a free-body diagram showing every force acting on it, and only those forces.
- Choose axes, usually along and perpendicular to the direction of acceleration, and write the second law separately for each axis.
- For connected bodies, the string condition gives the same magnitude of acceleration for both bodies and, for a light string, the same tension at both ends.
- Treating connected bodies as one system gives the common acceleration quickly, because internal tensions cancel; a separate free-body diagram is then needed to find the tension.
- For two masses m₁ > m₂ hanging over a light frictionless pulley, a = (m₁ − m₂)g/(m₁ + m₂) and T = 2m₁m₂g/(m₁ + m₂).
- With friction on connected bodies, first check whether the driving force exceeds limiting static friction; if not, the system stays at rest and friction equals the driving force.
- When a force is applied at an angle to a floor, its vertical component changes the normal reaction and therefore the friction.
Questions
5 answers
- What does the "Connected bodies: one a, two free-body diagrams" simulation show?
- Atwood machine and blocks-pushed-together: system acceleration, then a separate FBD for each body gives tension or contact force.
- What mistake does the "Connected bodies: one a, two free-body diagrams" simulation fix?
- A common mistake it clears up: taking tension as m₁g, or forgetting each body needs its own FBD.
- Has NEET asked about this idea?
- Yes. NEET has asked about this idea 3 times in 10 years of papers.
- Does JEE Main ask questions from Laws of Motion?
- Yes. JEE Main asked 39 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).
- Is this simulation free?
- Yes. It is free and runs in the browser on a phone or a laptop. It belongs to the lesson "Solving problems in mechanics" in Laws of Motion (Class 11 Physics).
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