Module 3: Kinematics · Theory · Advanced · Prerequisites: K13, Foundation 5
When a string or a surface links two bodies, the motion of one determines the motion of the other — and the link is pure geometry.
Mahavakya
A constraint does not tell a particle how to move. It tells the particle which motions are allowed.
On this page
1. Three levels of one statement
A constraint is a geometric fact about a system: a string has a fixed length, a block stays on a surface, a rod does not bend. Write it as an equation relating the coordinates:
F(q1, q2, …, t) = 0
Differentiating that equation once, then again, produces the velocity and acceleration relations. Three levels, one statement:
| Level | Question it answers | Expression |
|---|---|---|
| Position | Which configurations are possible? | F = 0 |
| Velocity | How must the bodies move? | dF/dt = 0 |
| Acceleration | How must their accelerations relate? | d²F/dt² = 0 |
This replaces every pulley formula you have been asked to memorise. Write the geometry, differentiate. The velocity relation and the acceleration relation come out, and they come out correct for whatever arrangement you happened to draw.
2. The inextensible string, derived
Two points A and B joined by a taut string of fixed length ℓ. The geometric statement is |rAB| = ℓ, a constant.
Square it, so the square root disappears:
rAB · rAB = ℓ² = constant
Differentiate with respect to time. The right side is constant, so its derivative vanishes:
2 rAB · (vA − vB) = 0
(vA − vB) · n̂ = 0
where n̂ is the unit vector along the string. Rearranged:
vA · n̂ = vB · n̂
The component of velocity along the string is the same at both ends. This is the whole of string constraint kinematics, and it took three lines.
Different speeds, different directions — equal components along the string.
Notice what is not required. The two ends need not have equal speeds, and need not move in the same direction, or even in remotely similar directions. Only the components along the string must match.
Motion perpendicular to the string does not change its length — not at that instant — so the string does not constrain it.
The dot product is why Foundation 5 was set as this part’s prerequisite. It is not a technique invented for pulleys; it is the differential form of the distance between the ends stays constant.
3. Count segments, not pulleys
Everyone learns that a movable pulley gives a factor of two. Rather fewer can say why, and the “why” is what tells you when the factor is not two.
The factor comes from how many segments change length when the body moves.
Fixed pulley. One segment on each side, so L = x + y. Differentiating: vx + vy = 0, so the blocks move at equal speeds in opposite senses.
Movable pulley. Two segments support it, so if A rises by h, both shorten by h and 2h of string is released. B descends by 2h. Hence vB = 2vA.
The rule is not “a movable pulley gives 2”. It is: count how many string segments change length when the body moves. Arrangements exist where a movable pulley gives 3, or where two pulleys give a factor that is not a power of two. Counting segments is always right; remembering “2” is right only for the standard picture.
So the habit worth building is a question rather than a formula: which portions of the string change length when this body moves?
4. Strings at an angle
When the string is not aligned with the motion, resolve along it — which is what section 2’s result instructs you to do.
Worked example
A block slides horizontally at vA. A string from it passes over a pulley and lifts a second block vertically at vB. At the instant shown the string makes an angle θ with the horizontal. Relate the two speeds.
Component of A’s velocity along the string: vA cos θ. Component of B’s velocity along its portion: vB (the string is vertical there).
vA cos θ = vB
The horizontal block moves faster than the hanging one, since cos θ < 1. Only the part of its motion directed along the string does any lifting; the rest is perpendicular and changes the string’s direction rather than its length.
θ is not constant. As the block slides, the angle changes — so vA cos θ = vB holds instant by instant, and differentiating it for the acceleration relation requires the product rule on cos θ(t). Treating θ as fixed is the commonest error in angled-string problems.
5. Contact surfaces and wedges
A block resting on a moving wedge obeys a constraint of exactly the same kind — with the string replaced by a surface, and “along” replaced by “perpendicular to”.
The bodies remain in contact: they neither interpenetrate nor separate. So the components of their velocities along the common normal must be equal:
(vblock − vwedge) · n̂ = 0
for an incline of angle θ, n̂ = −sin θ i + cos θ j
Stating it in relative form makes the connection visible: this is the same statement as the string condition, with a different unit vector. A string fixes the component along a line; a surface fixes the component across one.
Watch the frame. “The block slides down the wedge” describes the velocity of the block relative to the wedge — not relative to the ground. Mixing vblock, ground with vblock, wedge is the standard error here, and K13’s subscript convention exists precisely to prevent it.
6. The virtual power shortcut
For an ideal string over light frictionless pulleys, the tensions do no net work on the system, so the total instantaneous power they deliver vanishes:
∑ Ti · vi = 0
Applied to the movable-pulley system: force balance on the massless pulley gives the upper tension as 2T, so with A moving down at vA and B up at vB:
−2TvA + TvB = 0
vB = 2vA
Same answer as segment-counting, obtained without tracing any string. For multi-pulley trees this is considerably faster, and it is worth having.
One thing to be careful about. It is tempting to differentiate ∑T·v = 0 and claim ∑T·a = 0. That does not follow, because the tension varies with time and the product rule gives an extra term.
The acceleration relation is correct — but for a different reason. Tension is common to every segment of one string, so it factors out of the sum entirely, leaving a purely geometric relation. To get accelerations, differentiate the length constraint twice (section 1’s third level), not the power equation.
7. Traps
1. Assuming the pulley is massless. Ttop = 2Tside holds only for a massless, frictionless pulley. If the problem gives the pulley a mass or a moment of inertia, extra terms appear. Check that the problem says so.
2. Misidentifying the angle. The angle in the dot product is between the string and the velocity — not between the velocity and the surface, and not the angle of the incline. Write the unit vector explicitly if there is any doubt.
3. Assuming a constant string angle. Relative motion changes the string’s orientation, so θ = θ(t). For acceleration relations, differentiate carefully.
4. Mixing reference frames. “Slides along the wedge” means relative to the wedge. Say which frame every velocity is measured in.
5. Differentiating before understanding the geometry. Ask what the constraint equation expresses before asking how it changes. A wrongly written F = 0 differentiates smoothly into a wrong answer.
Contemplation
An inextensible string does not push a block, and does not supply it with an acceleration. It removes possibilities. Before the string, the block could move in any direction at any speed; after it, a whole family of motions has been struck out, and what remains is what we observe.
The same is true of a contact surface. It does not prescribe a velocity; it forbids motion through itself. What is left over is the motion.
So motion is not determined only by equations of evolution — by what pushes what. It is also restricted by equations of constraint, which say what was never available. That is a different kind of physical law, and it is worth noticing that mechanics contains both.
Common misconceptions
1. Connected bodies always have equal speeds
Only their components along the string are equal. With a movable pulley or an angled string the speeds differ, sometimes by a large factor.
2. A movable pulley always gives a factor of two
It gives whatever the segment count gives. Two is the answer for the standard arrangement, not a property of movable pulleys.
3. The string angle stays fixed
It changes as the bodies move. The velocity relation holds instantaneously; the acceleration relation needs the product rule.
4. Constraints are about forces
They are pure geometry, and everything in this part was derived before any force was mentioned. The forces that enforce them come later, in Newton’s laws.
5. You can differentiate the power equation for accelerations
Tension varies with time, so the product rule adds a term. Differentiate the length constraint twice instead.
Reflection
- Two blocks joined by a string over a fixed pulley. Under what circumstances could the string go slack, and what happens to the constraint then?
- A system has three blocks and two independent inextensible strings. How many constraint equations are there, and how many coordinates can move freely?
- Section 5 said a string fixes the component along a line and a surface the component across one. Is there a physical situation that fixes both at once?
Key takeaways
- Write the geometry as F = 0, then differentiate once for velocities and twice for accelerations.
- For a taut string, vA·n̂ = vB·n̂ — derived by squaring the length and differentiating.
- The two ends may have different speeds and directions. Only the components along the string match.
- Count string segments, not pulleys. That is where the factor comes from.
- Contact surfaces: (v1 − v2)·n̂ = 0 — the same idea across a normal instead of along a line.
- ∑T·v = 0 is a fast route to velocity relations. For accelerations, go back to the length constraint.
- Constraints are geometry, not force. No force was mentioned anywhere in this part.
What comes next
One constraint has appeared repeatedly without being named: a body kept at a fixed distance from a point moves in a circle. K15 takes that case seriously — angular position, angular velocity, and the acceleration that circular motion demands.
Prerequisites: K13 and Foundation 5 · Reading: 18 min · Practice: 45 min · Difficulty: Advanced