Pulleys and Connected Bodies: Mastering Teams of Masses in Physics

JEE/NEET Physics · Laws of Motion series · Part 3 of 8 · All parts →

Hitch two masses to one rope over a pulley and they become a single team: one acceleration, one shared fate. Pulley problems look like puzzles — they’re actually bookkeeping. Once you accept two simple rules and apply the free-body-diagram method from Part 2 to each body in turn, every connected-body problem collapses into routine algebra. This is Part 3 of the Laws of Motion series, and it covers the machinery behind half the mechanics questions you’ll meet in JEE and NEET.

The Rules of Connected Teams

Two rules govern every connected system, and both come straight from what an ideal rope is. (1) One ideal rope = one tension everywhere. An ideal rope is massless and frictionless through its pulleys, so the tension it transmits is identical at every point along its length. If you ever calculate two different tensions for the same rope, you’ve made a slip — differing values belong to different ropes or to real pulleys with mass. (2) A taut, inextensible rope makes all attached bodies move with the same speed. The rope cannot stretch or slacken, so if one end moves a metre, the other end must move exactly a metre too. One speeding up exactly as much as the other slows means their acceleration magnitudes are equal.

Combine these two constraints with the standard method — a separate FBD for each body, Newton’s second law written for each, then solve the equations together — and the system effectively solves itself. The rope’s rules supply the extra equations (“same T, same |a|”) that turn two unknowns-per-body into a solvable set.

The Atwood Machine

The classic setup: two masses hanging from a single rope over a frictionless, massless pulley. The heavier one sinks, the lighter one rises, and both share the same acceleration magnitude. Draw the FBDs — for m₁ the tension pulls up and weight pulls down; for m₂ it’s the same story — and write:

m₂g − T = m₂a (the heavier mass accelerates downward) and T − m₁g = m₁a (the lighter mass accelerates upward). Add the two equations and T vanishes, leaving:

Read the tension formula with intuition: T always sits between m₁g and m₂g. The heavier mass’s weight exceeds T (so it accelerates down) and the lighter mass’s weight falls short of T (so it accelerates up). If m₁ = m₂, then a = 0 and T = mg exactly — a perfectly balanced rope, static despite two forces fighting.

A useful one-line summary: the difference in weights pushes the total mass. The net driving force is (m₂ − m₁)g and the inertia being moved is (m₁ + m₂). That’s the entire Atwood machine in a sentence.

The Inclined Plane Split

On a slope at angle θ, gravity still points straight down — but the motion (or the tendency of motion) runs along the surface. So we resolve gravity into two perpendicular components:

• mg sin θ along the slope — the pulling part, the only piece that accelerates the block downhill;
• mg cos θ into the surface — the pressing part, which the surface must balance: N = mg cos θ.

Why these components? Because the block never leaves the surface, there’s no acceleration perpendicular to it — the surface’s normal force must exactly cancel mg cos θ. Everything left over, mg sin θ, is free to accelerate along the slope. For a frictionless slide:

Notice what’s missing: mass. Slope acceleration is mass-blind, exactly like free fall — the angle decides everything. Galileo’s insight about falling bodies applies just as well to sliding ones.

Combining: Slope + Pulley

The exam favourite: a block on a frictionless incline, connected by a rope over a pulley at the top to a hanging mass. Which way does the system move? Compare the drivers — the hanger’s weight mg against the slider’s down-slope component mg sinθ.

The method is the same three moves every time:

1. FBD the slider: tension pulls up-slope; mg sinθ pulls down-slope; mg cosθ is cancelled by N.
2. FBD the hanger: tension pulls up; weight mg pulls down.
3. Add the equations to eliminate T — the rope’s tension is internal to the team and cancels in the sum.

You can even shortcut the addition: the net external driving force is the difference of the two “engines” (here mg − mg sinθ), and the total inertia is the sum of all masses. One division later, you have a.

Solved Examples

This Physics in Your Daily Life

Practice set (answers hidden — try first)