Why Pulley Problems Are Related-Rates Problems
A rope running over a pulley connects the motion of different objects. If the rope is taut and cannot stretch, its total length stays fixed. When one part of the rope lengthens, another part must shorten to compensate. That restriction links the speeds of the loads.
The main task is to count the rope segments whose lengths change, write their total as a constant, and differentiate that equation with respect to time. The signs tell you whether connected parts move in the same direction or opposite directions; the coefficients tell you how their speeds compare.
Set Up the Rope-Length Constraint
Assume the rope is taut, inextensible, and has fixed total length. Assume also that the pulleys do not change position or orientation unless a pulley is attached to a moving load. In a basic model, the lengths around the pulleys are constant, so they can be included in a constant term. Only the straight segments whose lengths change need to be written explicitly.
Choose coordinates before writing the equation. In the examples below, distances are measured from fixed reference points, and downward is the positive direction for vertical motion. Thus a positive derivative means downward motion, while a negative derivative means upward motion. State the convention so each rate has an unambiguous meaning.
A segment contributes its length once for each portion of rope that changes length. For a fixed pulley with one load on each end, there are two changing segments, one for each load. For a movable pulley supported by two rope segments, the moving load changes the length of both supporting segments. Its coordinate therefore appears twice in the rope equation.
Differentiate the Constraint
If \(x\) and \(y\) are position coordinates that depend on time \(t\), then differentiating a constant-length equation relates their instantaneous rates. This is the same procedure used for other related-rates constraints: differentiate with respect to time, and interpret each derivative as a signed velocity.
This result requires that the rope’s total length remain fixed and that the listed lengths account for all changing portions. If a segment is diagonal, its length may first need to be expressed in terms of position coordinates; its derivative then follows from the chain rule. In the vertical pulley arrangements here, the variable segment lengths are directly represented by the load coordinates.
Worked Example: A Fixed Pulley Connects Two Loads
Worked Example: A Fixed Pulley Connects Two Loads
Two loads hang from opposite ends of one rope over a fixed pulley. Let \(x\) and \(y\) be their downward distances from fixed reference points. If load A is moving downward at \(0.80\) m/s, find the signed velocity of load B and describe its direction.
The variable rope lengths are \(x\) and \(y\). The rope length around the pulley and any other fixed portion contribute a constant \(C\). Thus
Because \(C\) and \(L\) are constant, differentiating with respect to time gives
Load A moves downward at \(0.80\) m/s, so \(\frac{dx}{dt}=0.80\) m/s. Substitution gives
Since downward is positive, the negative sign means load B moves upward at \(0.80\) m/s. Check: \(0.80+(-0.80)=0\), so the two segment lengths have zero net rate of change, as the fixed rope length requires.
The fixed pulley changes the direction of motion but not the speed magnitude in this idealized arrangement. One segment lengthens at exactly the rate the other shortens.
Worked Example: A Movable Pulley Has Two Supporting Segments
Worked Example: A Movable Pulley Has Two Supporting Segments
One end of a rope is fixed overhead. The rope runs down around a movable pulley attached to load B, then up over a fixed pulley and down to load A. Let \(x\) be the downward distance of load A and \(y\) the downward distance of the movable pulley and load B. If A is pulled upward at \(1.20\) m/s, find the signed velocity of B.
Load A determines one variable segment of length \(x\). The movable pulley is held by two segments, each of which changes by the same amount when the pulley moves. Those two segments each contribute \(y\). The rope-length equation is therefore
Differentiating with respect to time gives
A is moving upward, so its downward-positive rate is \(\frac{dx}{dt}=-1.20\) m/s. Substitute this value and solve:
The positive sign means B moves downward at \(0.60\) m/s. Check: \(-1.20+2(0.60)=0\). The free end moves twice as fast as the movable load, and the two supporting segments together account for that factor of two.
Why the Speed Ratio Appears
The coefficient of a coordinate in the rope equation counts how many variable segments change with that coordinate. In the movable-pulley example, moving the pulley down by a small distance lengthens each supporting segment by that distance. The combined increase is twice the pulley’s displacement. To keep the rope length fixed, the free-end segment must shorten by twice as much.
Thus \(\frac{dx}{dt}=-2\frac{dy}{dt}\): the velocities have opposite signs, and the free end’s speed is twice the movable load’s speed. This is a kinematic rate ratio implied by the rope geometry. It does not require a calculation of forces or energy.
More generally, if one moving load changes the lengths of \(k\) supporting segments by the same amount, its coordinate contributes \(ky\) to the total-length equation. The rate equation then contains \(k\frac{dy}{dt}\). This count is valid only when those \(k\) segments really change one-for-one with the load’s displacement and the rest of the rope path is fixed.
Worked Example: Check the Ratio Using Displacements
Worked Example: Check the Ratio Using Displacements
Use the movable-pulley arrangement from the previous example. At a particular instant, let \(x=3.00\) m and \(y=1.50\) m. The free end is moving upward at \(0.90\) m/s. Find the moving load’s velocity, then check the rope length after \(2.00\) seconds if the two velocities remain constant during that interval.
Downward is positive, so the free end’s rate is \(\frac{dx}{dt}=-0.90\) m/s. The constraint and its derivative are
Substitute \(\frac{dx}{dt}=-0.90\) m/s:
The moving load travels downward. Over \(2.00\) seconds, the free-end displacement is \((-0.90)(2.00)=-1.80\) m, and the load displacement is \((0.45)(2.00)=0.90\) m. The resulting coordinates are
The variable part of the rope length initially is \(x+2y=3.00+2(1.50)=6.00\) m. After the motion it is \(1.20+2(2.40)=1.20+4.80=6.00\) m. The length is unchanged, checking both the rate and displacement calculations.
A Common Mistake: Counting the Movable Pulley Only Once
A tempting but incorrect equation for the movable-pulley arrangement is \(x+y+C=L\). It counts the free-end segment and the load’s position once, as if moving the load changed only one supporting segment. Differentiating that incorrect equation would give \(\frac{dx}{dt}+\frac{dy}{dt}=0\), predicting equal speed magnitudes.
For example, with \(\frac{dx}{dt}=-1.20\) m/s, that mistake predicts \(\frac{dy}{dt}=1.20\) m/s. But the actual rope has two supporting segments, so the correct equation is \(x+2y+C=L\). The correct rate is \(-1.20+2\frac{dy}{dt}=0\), giving \(\frac{dy}{dt}=0.60\) m/s. Substitution verifies the correct relation: \(-1.20+2(0.60)=0\). The incorrect prediction fails it: \(-1.20+2(1.20)=1.20\ne0\).
A second source of error is ignoring the sign convention. With downward positive, upward motion has a negative rate. Use the signed rate in the differentiated equation first; describe direction in words afterward. Replacing a negative rate with its positive speed too early can reverse the conclusion.
A Reliable Setup
Define each position variable and state which direction is positive.
List every segment whose length changes, counting separate supporting segments separately.
Add the variable segment lengths and represent fixed portions by a constant.
Use the chain rule if any segment length depends indirectly on a coordinate.
Substitute signed rates, solve for the unknown, state its direction, and check the rate equation.
Why the Constraint Matters
The rope equation is more than a convenient formula: it expresses the physical restriction that the rope cannot gain or lose length. Differentiating that restriction turns a relationship among positions into a relationship among instantaneous velocities. The number of times a moving load appears in the length equation determines its coefficient in the rate equation, and therefore its mechanical speed ratio with the other end.
Always build the equation from the actual rope path rather than memorizing a ratio. A different path can have a different number of changing segments, and a segment that does not change length should not be counted. Once the geometry and signs are set correctly, differentiation is direct.
Check Your Understanding
For each question, use a stated downward-positive convention for vertical coordinates unless the question specifies otherwise.
- A fixed pulley connects two loads. If one load moves downward at \(0.35\) m/s, what is the other load’s signed velocity?
- For a movable pulley supported by two changing segments, write the rope-length equation using \(x\) for the free end and \(y\) for the moving load.
- In that arrangement, if the free end moves downward at \(1.00\) m/s, find the moving load’s signed velocity and direction.
- Why does the moving load’s coordinate have coefficient \(2\) in the rope equation for a pulley supported by two changing segments?
- A proposed rate pair for a two-segment movable pulley is \(\frac{dx}{dt}=0.80\) m/s and \(\frac{dy}{dt}=-0.30\) m/s. Does it satisfy the differentiated constraint? Show the substitution.