core31 min studyLesson 112 in Physics

Constraints determine motion; reactions enforce them

Translate rope lengths and contact geometry into acceleration relations, then solve for tensions without assuming every pulley is ideal.

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What you’ll understand

  • Differentiate a geometric constraint to connect accelerations.
  • State the assumptions under which one rope has uniform tension.
  • Check whether a proposed solution requires a slack rope or lost contact.

A rigid track or taut rope restricts which motions are possible. The restriction is geometric, while the normal reaction or tension is the force required to enforce it. Keeping these roles separate lets you solve problems with several connected bodies systematically. Guessing equal accelerations or equal tensions from a picture can work accidentally, but a written constraint explains when that guess is justified and when a pulley changes the answer.

Differentiate the geometry, not the sketch

Consider two masses hanging from a rope over a fixed pulley. Define both vertical coordinates downward from fixed reference marks. The variable portions of the rope have total length equal to the sum of these coordinates; the wrapped segment and other fixed portions add a constant. An inextensible taut rope therefore makes the coordinate sum constant. Differentiating once gives opposite velocities, and differentiating again gives opposite accelerations. Their magnitudes match because this particular geometry has one variable segment attached to each mass, not because ropes universally produce equal accelerations.

A movable pulley changes the length relation. If two supporting rope segments both lengthen when a movable pulley descends, its displacement contributes twice to the total variable length. A free end must then move twice as far in the opposite direction. Draw each segment and count its change before differentiating. Use a single time variable and fixed coordinate directions throughout. A geometrical equation is especially valuable when the pulley itself moves: it prevents a factor-of-two error from being hidden inside a plausible-looking force diagram or an intuitive claim about mechanical advantage.

y1+y2=ℓ0⟹a1+a2=0y_1+y_2=\ell_0\quad\Longrightarrow\quad a_1+a_2=0
For two hanging ends over a fixed pulley, downward coordinates y1 and y2 have a constant sum ℓ0 when the rope is taut and inextensible. Acceleration magnitudes are equal, but their signed components are opposite.↔ Scroll wide equations horizontally.

Uniform tension requires a mechanical model

Isolate a short piece of straight rope. If its mass is negligible, a finite imbalance of the tensions at its ends would contradict the zero-mass idealization. The tension is consequently uniform along that ideal segment. To carry the same conclusion around a pulley, also assume a freely turning pulley with negligible rotational inertia and no relevant bearing torque, or an ideal frictionless guide. A massive rotating pulley generally needs different tensions on its two sides to supply angular acceleration. A rough fixed surface can also support a tension difference through distributed friction.

Tension acts along the local rope direction and pulls away from the object it acts upon. It is an unknown reaction to be solved from dynamics together with the length constraint. Do not set tension equal to the weight of a hanging mass unless that mass has zero vertical acceleration. A rope can bend around a support, so its direction can change even when its tension magnitude stays constant. Conversely, a real elastic rope changes length under load and can transmit waves; replacing it by an instantaneous inextensible constraint discards those dynamics and is appropriate only on suitable time and length scales.

m1g−T=m1a,T−m2g=m2a(m1>m2)m_1g-T=m_1a,\qquad T-m_2g=m_2a\qquad(m_1>m_2)
For an ideal Atwood apparatus, a is the positive downward acceleration of the heavier mass and the positive upward acceleration of the lighter mass. The opposite coordinate choices explain the different weight signs.↔ Scroll wide equations horizontally.

Solve the forces and audit the constraint

The unknown reactions do not replace Newton's laws. Write one force equation for each body, then add the independent constraint relations until the unknown accelerations and reactions are determined. Adding two suitable equations often eliminates a common tension, revealing the acceleration first. Substitution then gives the internal load. This order is efficient, but solving for a negative tension does not mean a rope pushes. It means the assumed taut-rope branch cannot describe the motion. Switch to the appropriate slack-rope motion, and impose the constraint again only when the geometry permits tension to develop.

Energy offers a separate audit when the ideal constraint does no net work on the connected system. Equal and opposite tensions can transfer energy between bodies while their total power cancels because the constrained velocities are related. This cancellation is not a claim that each tension does zero work individually. With a massive pulley, include rotational kinetic energy; with friction, include dissipation. Agreement between force and energy calculations checks algebra, but a measurement of pulley acceleration or rope stretch is still needed to test the idealization itself. A correct solution always belongs to a specified apparatus model.

An ideal two-mass pulley

An ideal massless rope over a massless frictionless pulley connects 3 kg3\,\mathrm{kg} and 2 kg2\,\mathrm{kg} masses. Use g=10 m s−2g=10\,\mathrm{m\,s^{-2}}. Find acceleration and tension after release.

  1. Choose downward positive for the heavier mass and upward positive for the lighter one. The taut fixed-length rope gives the same positive acceleration magnitude a for both.

  2. Add 30−T=3a30-T=3a and T−20=2aT-20=2a. This gives 10=5a10=5a, so a=2 m s−2a=2\,\mathrm{m\,s^{-2}}.

  3. Substitute into the lighter mass equation: T=20+2(2)=24 NT=20+2(2)=24\,\mathrm N. Check the heavier mass: 30−24=3(2)30-24=3(2). The positive tension is compatible with a taut rope.

The result

The common acceleration is 2 m/s² and the tension is 24 N, between the two weights. Unequal side tensions would be needed if a non-negligible pulley inertia were included.

Explore the model

Interactive lab

Interactive lab · vectors

Inspect a rope force's components

Set angle to 30° and magnitude to 2 for one normalized tension vector. Its vertical projection is 1; at 90° it becomes 2. A symmetric pair would contribute twice each projection, but the lab draws only one vector and does not solve a pulley constraint.

Inspect a rope force's componentsPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forGeometry fixes directions; dynamics fixes the reaction magnitude.

Check your understanding

A movable pulley descends by d while two supporting rope segments each lengthen by d. How far must the free end move if the rope stays taut and inextensible?

Try it yourself

In the same ideal fixed-pulley arrangement, replace the masses by 4 kg4\,\mathrm{kg} and 1 kg1\,\mathrm{kg}. Find a and T using g=10 m s−2g=10\,\mathrm{m\,s^{-2}}, and check the limit as the lighter mass approaches zero.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Add the two force equations to eliminate tension; then solve either one for it.

Compare with the solution

The acceleration is a=(4−1)10/(4+1)=6 m s−2a=(4-1)10/(4+1)=6\,\mathrm{m\,s^{-2}}. The lighter mass gives T=1(10+6)=16 NT=1(10+6)=16\,\mathrm N. In the ideal limit m2→0m_2\to0 at fixed m1m_1, a approaches g and tension approaches zero. That limit approaches free fall rather than allowing a finite tension on a massless free end.

Take it with you

  • A rope-length relation must be derived before relating accelerations.
  • Uniform tension relies on negligible rope mass and an appropriate pulley or guide model.
  • Negative tension or impossible contact reactions signal a change of physical branch.

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