Pulley and Atwood-type systems

Every pulley problem starts from the same three idealisations. The rope is massless, so no net force is needed along it and its tension is the same from end to end. The pulley is massless and turns without friction, so it needs no unbalanced pull to spin, and the tension is the same on both sides of it. The rope is inextensible and taut, so its length is a fixed number. Together they leave one unknown T per rope, which a pulley can turn through any angle but never enlarge. Then count before solving. Two blocks over one fixed pulley bring three unknowns, a₁, a₂ and T; two second-law equations and one length equation make three. Each extra rope adds one tension and one length equation, and each massless movable pulley adds a force balance tying its ropes' tensions together. If the count will not close, a body is missing its free-body diagram.

Hang m₁ = 2.0 kg and m₂ = 3.0 kg over one ideal pulley, and take each body's positive direction along its expected motion: m₂ down, m₁ up. Then m₂g − T = m₂a and T − m₁g = m₁a, with one a because one rope links them. Adding the equations eliminates T: a = (m₂ − m₁)g/(m₁ + m₂) = 1.96 m s⁻². Substituting back gives T = 2m₁m₂g/(m₁ + m₂) = 23.5 N. Now read the result instead of boxing it. The driving force is only the weight difference, 9.81 N, yet it must accelerate all 5.0 kg, which is why George Atwood described the machine in 1784 as a way to slow a fall enough to time it. Equal masses give a = 0 and T = mg. Let m₁ → 0 and a → g while T → 0, the free fall of m₂. A formula that fails either limit has an error in it.

Two conventions work; mixing them does not. The quick one gives each body its own positive direction along the rope's expected motion, so one a serves every body on that rope and a negative answer only means the rig runs the other way. The strict one measures every coordinate from a fixed point: with y₁ and y₂ both measured down from the axle, y₁ + y₂ = constant forces a₁ = −a₂, and each equation carries its own sign. The quick convention licenses a shortcut: treat the rope as one bent axis, add the forces along it and divide by the total mass. For the 2.0 kg and 3.0 kg pair, (29.43 − 19.62) N ÷ 5.0 kg = 1.96 m s⁻² in one line. The shortcut needs every body to share one speed, so it dies at the first movable pulley. Friction sets a second trap. Kinetic friction opposes the relative sliding at the contact, so if a guessed direction returns a < 0, redo the case with friction reversed; if that also fails, the rig stays at rest.

Hang a 12 kg load from a movable pulley on two segments of one rope, and run the free end over a fixed pulley to a 5.0 kg counterweight. The rope pulls the load up with 2T, and its length, 2y_load + y_cw = L, makes the counterweight's acceleration −2 times the load's. Those are the same 2 for one reason: a massless rope gives out exactly the energy it takes in, so T times the free end's speed equals 2T times the load's. With down positive, Mg − 2T = Ma for the load and mg − T = −2ma for the counterweight. Eliminating T gives a = (M − 2m)g/(M + 4m) = 0.613 m s⁻² down, a counterweight rising at 1.23 m s⁻², and T = 55.2 N. Notice the 4. Statics sees only the factor 2 and correctly predicts balance at M = 2m. It cannot see that the counterweight, moving twice as fast and geared by 2, loads the dynamics with 2 × 2 = 4 times its mass. A shortcut that keeps one factor of 2 and drops the other gets the motion wrong.

An internal force deserves an external check. A massless fixed pulley with both segments vertical is held up by 2T, so for the 2.0 kg and 3.0 kg Atwood machine the hook carries 2 × 23.544 N = 47.09 N, not the 49.05 N the two weights add up to. Now take pulley, rope and blocks as one system. Its only external vertical forces are the weights and the hook, so (m₁ + m₂)g − F = (m₁ + m₂)a_cm, with a_cm positive down. The heavy block sinks at 1.962 m s⁻² while the light one rises at the same rate, so a_cm = (3.0 − 2.0)(1.962)/5.0 = 0.392 m s⁻² downward, and 5.0 kg × 0.392 m s⁻² = 1.96 N, exactly the shortfall. The two routes agree, so the tension is right. It also explains a hook scale under a running Atwood machine: it reads less than the weight it holds, and reads the full weight only when the masses are equal and nothing accelerates.

Every real rig breaks the model somewhere; the useful question is by how much. A pulley with rotational inertia needs unequal tensions to spin up. For a uniform disc of mass mₚ that adds mₚ/2 to the moving mass, so a 1.0 kg disc under the 2.0 kg and 3.0 kg pair lowers a from 1.96 to 1.78 m s⁻², 9% less, with T₂ − T₁ = 0.89 N. Axle friction removes a roughly constant force, and it matters against the net driving force, not the weights: 0.010 N is lost against 9.81 N but is 2.5% of the 0.392 N driving a 0.480 kg and 0.520 kg pair. A rope with mass shifts rope from the rising side to the falling side, so the driving force grows as the rig runs. And a rope can only pull: if solving returns T < 0, the rope is slack, each body moves as if the rope were gone, and the length constraint is void. The near-balanced rigs that make g easy to time are the ones most exposed to all three.