Massless string → tension is the same everywhere along it. (A massive string would need net force to accelerate itself, so T would differ between ends.)
Inextensible string → both masses have the same acceleration magnitude. If m1 drops by x, m2 rises by x.
Massless, frictionless pulley → it just redirects the rope; it does not store/rob any force, so tension is unchanged across the pulley.
We use only Newton's second law, Fnet=ma, on each mass separately. This is Derivation-from-scratch.
Step 1 — Choose which way it accelerates.
Assume m1>m2, so m1 goes down and m2 goes up.
Why this step? We must commit to a sign convention. We'll define "the direction of motion" as positive for each mass — if we guessed wrong, a just comes out negative, which self-corrects.
Step 2 — Forces on each mass.
Each mass feels two forces: gravity mg (down) and tension T (up, the rope always pulls toward the pulley).
For m1 (accelerating down, take down as +):
m1g−T=m1a(1)Why this step? Down is positive here, so the downward weight is +m1g and the upward tension is −T. Net = mass × acceleration.
For m2 (accelerating up, take up as +):
T−m2g=m2a(2)Why this step? Up is positive here, so tension is +T, weight is −m2g. Same a because the rope is inextensible.
Step 3 — Add the two equations to eliminate T.(m1g−T)+(T−m2g)=m1a+m2am1g−m2g=(m1+m2)aWhy this step? The unknown T appears as −T and +T; adding cancels it, leaving only a.
Step 4 — Find tension. Substitute a back into (2): T=m2(g+a).
T=m2(g+m1+m2(m1−m2)g)=m2⋅m1+m2(m1+m2)g+(m1−m2)g
What does an inextensible string guarantee about the two masses?
They have equal acceleration magnitude (and equal-but-opposite displacement/velocity).
What does a massless string guarantee?
Tension is the same at every point of the rope.
Acceleration of an Atwood machine?
a=m1+m2(m1−m2)g
Tension in an Atwood machine?
T=m1+m22m1m2g
Why add the two Newton equations during the derivation?
To eliminate the unknown tension T (it appears as +T and −T).
Check: if m1=m2, what are a and T?
a=0 and T=mg.
Check: if m2=0, what are a and T?
a=g (free fall) and T=0.
Why is tension less than the heavy mass's weight?
Because m1 is accelerating downward, so net force on it is downward → T<m1g.
What is the driving force of the system?
The weight difference (m1−m2)g.
What inertia resists that force?
The total mass (m1+m2).
Recall Feynman: explain to a 12-year-old
Imagine a rope over a wheel with a bucket on each end. If both buckets weigh the same, nothing happens. If you put a rock in one bucket, that side sinks and the other rises — but they're tied together, so they move at the same speed. The rope pulls up on both buckets with the same strength (because the rope is light and the wheel is smooth). How fast it speeds up depends on how much extra weight you added compared to how much stuff you're trying to move in total. More extra weight → faster; more total stuff → slower (heavier to get going).
Atwood machine ka idea bilkul simple hai: do masses ek halki (massless) aur na khinchne wali (inextensible) rope se judi hain, jo ek smooth pulley ke upar se guzarti hai. Kyunki rope ek hi hai, dono masses ki acceleration ka magnitude same hota hai — bas direction opposite. Aur kyunki rope ka apna koi weight nahi, rope mein tension T har jagah equal rehta hai. Yeh do baatein puri derivation ki jaan hain.
Derivation mein hum har mass par alag se Newton ka second law Fnet=ma lagate hain. Heavy mass m1 neeche jaa rahi hai, toh us par: m1g−T=m1a. Light mass m2 upar jaa rahi hai, toh us par: T−m2g=m2a. Dono equations ko add karte hain taaki T cancel ho jaye, aur seedha mil jaata hai a=m1+m2(m1−m2)g. Yeh wapas daalo toh tension T=m1+m22m1m2g.
Intuition yeh hai: system ko aage badhane wala "driving force" sirf weight ka difference(m1−m2)g hai, aur jo cheez resist karti hai woh hai total mass(m1+m2). Isiliye agar dono masses barabar ho jaayein toh a=0 — koi motion nahi. Yaad rakho: tension kabhi m1g ya m2g ke equal nahi hota jab tak acceleration ho — woh dono ke beech mein hoti hai.
Exam tip: pehle hamesha free body diagram banao, sign convention fix karo (motion direction ko positive lo), phir dono equations add karke T uda do. Yeh same trick inclined plane aur multi-pulley problems mein bhi kaam aati hai.
