CM.3 · Coupled acceleration and relative motion

For a mass m1 on a frictionless horizontal table connected over an ideal massless pulley to a hanging mass m2, choose table motion and downward hanging motion as positive. A taut inextensible string gives one acceleration magnitude. Write T=m1a and m2g−T=m2a. Adding eliminates the internal tension, so a=m2g/(m1+m2) and T=m1m2g/(m1+m2). Thus T<m2g when both masses are positive. Setting tension equal to hanging weight silently assumes zero acceleration.

The common tension assumption requires an ideal massless string and pulley with negligible friction and inertia. A pulley with rotational inertia can have different tensions, with (T2−T1)R=Iα and a=Rα if the string does not slip. Draw each body separately: a force internal to the combined system remains an external force on one chosen body. Constraint forces can cancel from a system equation while still being needed for individual motion.

Check limiting cases after solving. As m1 approaches zero, the ideal coupled acceleration approaches g and tension approaches zero; as m1 grows very large, acceleration approaches zero and tension approaches m2g from below. An energy derivation gives m2g s=½(m1+m2)v² for release from rest and an ideal pulley, agreeing with v²=2as. This agreement checks the equations; it does not permit energy conservation if friction or other unaccounted work is present.

A table mass m1 and hanging mass m2 share an acceleration through a taut ideal string. Choose rightward table motion and downward hanging motion positive.

T is less than the hanging weight 20 N. A nonzero net downward force is needed for acceleration.

A massless ideal pulley equates tension, not tension and weight. Subtract frame velocities component by component before taking a speed magnitude.

constraint: A condition linking allowed positions or motions of connected bodies.

relative velocity: Velocity of one object measured in another moving frame.