Connected Particles and Pulleys
For two hanging masses joined by a taut, light, inextensible string over a fixed smooth pulley, acceleration magnitudes match and tension is uniform. Write F = ma for each mass, then solve the equations for acceleration and tension.
What a learner can do afterwards
- Write the equation of motion for each of two connected masses
- Solve the pair for the acceleration magnitude and the tension
- Explain why tension has the same magnitude throughout a light string over a smooth pulley
1 · Read
As with one body, choose a positive direction before writing F = ma. Here two hanging masses use a taut, light, inextensible string over a fixed smooth pulley. Taut means no slack; inextensible means no stretch. Their acceleration magnitudes match, but their directions are opposite. Light means negligible string mass. A smooth pulley supplies no friction along the string, so tension has one magnitude throughout. Ignore air resistance. This is our ideal pulley model.
For each hanging mass, weight acts down and tension acts up. Let M be the heavier mass and m the lighter mass. Take down as positive for M and up for m. Write Mg - T = Ma and T - mg = ma, where a is the acceleration magnitude. Add the equations: -T and +T cancel, giving (M - m)g = (M + m)a. Divide by M + m to find a. Then use T = mg + ma to find T.
Hang 9 kg and 1 kg in the ideal pulley model, with g = 10 N/kg. Their weights are 90 N and 10 N. Write 90 - T = 9a and T - 10 = a. Add the left sides and the right sides: 80 = 10a, so a = 8 m/s squared. Substitute into T - 10 = a: T = 18 N. Check the heavier mass: 90 - 18 = 9 times 8.
Equal hanging masses give a = 0. They may rest or move at constant speed. Equal tension does not balance unequal weights: T lies between their weights. With both axes positive down, the heavier mass has acceleration +a and the lighter mass -a: they accelerate down and up respectively. These signs describe acceleration, regardless of velocity. Keep each axis fixed.
A mass p on a smooth horizontal table can connect to a hanging mass q over a fixed smooth pulley at the table edge. Use a taut, light, inextensible string with its table section horizontal. Ignore air resistance. Take positive toward the pulley for p and down for q. Write T = pa and qg - T = qa. Adding gives qg = (p + q)a. Find a, then use T = pa.
The same equations can find a missing mass or g. In the ideal hanging-mass model, if the lighter mass m, g and acceleration magnitude a are known, first find T = m(g + a). Then M(g - a) = T, so M = T/(g - a). If both masses are known but g is missing, start with (M - m)g = (M + m)a. Divide by M - m to find g. Check the result in both original equations.
Write F = ma for each mass with clear positive directions. Use equal acceleration magnitudes and one tension in the ideal model, then solve and check both equations.
Lesson, practice questions and flashcards written for this dataset.
Readings, questions and flashcards written for this dataset. Referenced videos: Khan Academy, "Analyzing systems" and "Intro to forces (part 1)".
More to read
What the string assumptions mean
In this model, one string joins two hanging masses over a pulley fixed in place. The string stays taut and does not stretch. When one end moves down by a distance, the other moves up by that same distance. Their speeds match, and their acceleration magnitudes match. Their directions are opposite.
The string is also light, meaning its mass is neglected. Along a straight section, the two end pulls must balance. The smooth pulley supplies no friction along the string. These assumptions let us use the same tension magnitude everywhere in it.
The string pulls upward on both hanging masses. Those two upward forces are not an equal-and-opposite pair. For either mass, list its own weight and the string's pull before writing F = ma.
Choosing directions for each mass
For the heavier hanging mass M, choose downward as positive. Its weight contributes +Mg and its tension contributes -T. Its equation is Mg - T = Ma.
For the lighter mass m, choose upward as positive. Tension contributes +T and weight contributes -mg. Its equation is T - mg = ma. With these choices, a is the common acceleration magnitude.
Another valid choice is to point both axes downward. The lighter mass then has acceleration -a, and its equation is mg - T = -ma. The physics has not changed; only the signs used to describe it have changed.
Keep each chosen axis fixed. Check the direction of every force and the sign of the acceleration against that axis.
Solving the two motion equations
The equations Mg - T = Ma and T - mg = ma contain the same two unknowns, a and T. Adding them removes T because -T + T = 0. The result is (M - m)g = (M + m)a.
Divide by M + m to find the acceleration magnitude. Both masses appear in this total because both accelerate. Then put a into T = mg + ma to calculate the tension.
The subtraction of weights comes from adding equations that use different positive directions. It does not mean the actual weights point opposite ways. Both weights point downward.
Use the final a and T in both original equations. A proposed pair must satisfy both. Checking only one equation can miss a sign or arithmetic error in the other.
What a pulley answer should satisfy
For two unequal hanging masses in the ideal model, the heavier mass accelerates downward. Its downward weight must exceed the upward tension: Mg > T. The lighter mass accelerates upward, so its tension must exceed its weight: T > mg. Thus the tension lies between the two weights.
This check belongs to the two-hanging-mass setup with no other forces. A mass on a table or an extra applied force needs a different force list.
Equal hanging masses have zero acceleration. They can remain at rest or continue at constant speed while the string stays taut. Zero acceleration does not tell you their speed.
Tension is measured in newtons. Acceleration is measured in m/s squared. Check units and substitute the results into both motion equations.
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2 · Watch
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