Module 3: Kinematics · Worked Examples · Advanced · Prerequisites: K13, K14, Foundation 5
Six mechanisms, and one question asked of each: which direction do I resolve along?
Mahavakya
A string fixes the component along it. A surface fixes the component across it. Choosing which, and in which direction, is the whole of the problem.
This is the third constraints set. The first was strings and pulleys, the second was geometric linkages. These six are chosen for a different reason: in each, the difficulty is not the differentiation but deciding what to project onto before you start.
The six, and what to resolve along
- A ring, a pulley and a block the string
- Two carts, one cable each segment separately
- A rigid rod over a peak the rod
- A four-bar linkage each link in turn
- A telescoping boom nothing — add the frames
- A closing V the geometry, not the velocities
1. A ring, a pulley and a block
Problem
A ring slides along a horizontal rod at a steady 5.0 m/s. A string from the ring runs over a pulley fixed 1.5 m above the rod and down to a hanging block.
When the ring is 2.0 m horizontally from the pulley, find the block’s speed and acceleration.
s = √(x² + h²) = √(4 + 2.25) = 2.5 m
vB = (x/s)vR = (2.0/2.5)(5.0) = 4.0 m/s
aB = vR²h²/s³ = 25(2.25)/15.625 = 3.6 m/s²
Only the component of the ring’s velocity along the string lengthens it. The factor x/s is exactly the cosine of the angle between the rod and the string — so the block always moves slower than the ring, and the two speeds converge only as the ring runs far from the pulley.
The ring is unaccelerated and the block is not. Nothing changes about the ring’s motion; the angle changes, and that alone produces 3.6 m/s². Note that h² survives in the numerator — if the pulley sat on the rod, the angle would never change and the acceleration would vanish.
2. Two carts, one cable
Problem
A cable of fixed length runs from cart A on an upper rail, horizontally to a pulley at P, then down to cart B on a lower rail 6 m below. Cart A is 6 m from P and moving away at 3 m/s. Cart B is 8 m from P horizontally.
Find cart B’s velocity, and state which way it moves.
L = xA + √(xB² + h²) with √(64 + 36) = 10 m
3 + (8/10)vB = 0
vB = −30/8 = −3.75 m/s — B moves toward the pulley
The sign is the answer here, not a decoration on it. Cart A is taking cable, so cart B must give it back — and the only way B can shorten its segment is to move toward P. Nothing in the problem said which way B travels; the constraint decided.
And B moves faster than A, at 3.75 against 3. Its segment is slanted, so only 8/10 of its motion shortens the cable — it has to move further to give back what A took.
3. A rigid rod over a peak
Problem
Blocks A and B sit on opposite faces of a ridge and are joined by a rigid rod passing over the peak. A slides at 4.0 m/s. At this instant the rod makes 30° with A’s face and 40° with B’s.
Find B’s speed.
vA cos θA = vB cos θB
4.0 cos 30° = vB cos 40° → vB = 4.52 m/s
This is the string condition with a rod in place of the string, and the derivation is identical — square the rod’s fixed length and differentiate. The components along the connector must match, or the connector would stretch.
The face angles never appear. Only the angles between each velocity and the rod matter. A student who resolves along the slopes instead of along the rod gets a plausible number and a wrong one.
4. A four-bar linkage
Problem
A planar four-bar mechanism has fixed frame AD = 0.9 m, crank AB = 0.3 m, coupler BC = 0.5 m and follower CD = 0.5 m. At the instant the crank stands vertical and turns at 10 rad/s, the coupler lies horizontal.
Find the follower’s angular velocity.
Resolve along each link in turn. The crank pin moves perpendicular to the crank, so horizontally:
vB = ωAB × AB = 10(0.3) = 3.0 m/s, horizontal
The coupler is horizontal, so the whole of vB lies along it — and C’s component along it must match:
vC,x = 3.0 m/s
C circles D, so vC ⊥ CD, and sin φ = 0.3/0.5 = 0.6:
3.0 = ωCD(0.5)(0.6) → ωCD = 10 rad/s
The follower turns at exactly the crank’s rate, and that is no coincidence. Since sin φ = AB/CD, the follower’s contribution is ωCD × CD × (AB/CD) = ωCD × AB — and CD cancels entirely.
So in this configuration ωCD = ωAB whatever the follower’s length. Turn the crank a few degrees and it stops being true — which is what makes a four-bar linkage useful rather than a rigid frame.
Check that the linkage closes before trusting any answer. With the crank vertical and the coupler horizontal, the joints sit at A(0, 0), B(0, 0.3), C(0.5, 0.3) and D(0.9, 0) — giving CD = √(0.4² + 0.3²) = 0.5 m, exactly as stated. A four-bar whose link lengths do not close cannot be built, and a velocity computed for it means nothing.
5. A telescoping boom
Problem
A two-stage boom extends along a slope at 30°. Stage 1 extends from the base at 1.2 m/s; stage 2 extends from stage 1 at 0.8 m/s. A cable anchored at the base runs over a pulley at the boom’s tip and hangs vertically to a crate.
Find the crate’s velocity.
Nothing here needs resolving — the frames simply add. Stage 2’s tip moves relative to stage 1, which moves relative to the base:
vtip = 1.2 + 0.8 = 2.0 m/s along the slope
Every metre the tip advances pays out a metre of cable, so the crate hangs 2.0 m/s further below the tip each second — while being carried along with it:
vx = 2.0 cos 30° = 1.73 m/s vy = −2.0 sin 30° + 2.0 = 1.0 m/s downward
|v| = √(3 + 1) = 2.0 m/s, at 30° below the horizontal
The crate’s speed equals the tip’s speed exactly, and its direction is the mirror of the slope. The upward part of the tip’s motion is precisely cancelled by the cable paying out, leaving a descent equal to the rise. Worth checking rather than assuming — it holds because the payout rate happens to equal the full tip speed, not merely its vertical part.
6. A closing V
Problem
Two wedges approach each other along the ground — A moving right at 3.0 m/s with a face at 45°, B moving left at 1.0 m/s with a face at 30°. A small pin rests in the V where the two faces meet.
Find how fast the pin descends.
Do not resolve the velocities — locate the vertex. The pin sits where the two sloping lines cross, and that crossing point has a height set purely by how far apart the wedges are:
y = (xB − xA) · tan θA tan θB / (tan θA + tan θB)
The gap closes at VA + VB = 4.0 m/s, so:
vy = 4.0 × (1)(0.577)/(1 + 0.577) = 1.46 m/s downward
Only the closing rate matters, not the individual speeds. Swap the wedges to 2.0 and 2.0 m/s and the pin descends at exactly the same rate. The constraint knows about the gap, and nothing else.
The angle factor is the harmonic-mean form — ab/(a + b) in the tangents. It appeared in K3 for averaging speeds over equal distances, and it appears here for an entirely unrelated reason. The mathematics does not know which problem it is in.
What the six have in common
In four of them the answer came from projecting onto something: a string, a rod, a link. In the fifth nothing needed projecting — the frames simply added. In the sixth, projecting the velocities would have been a mistake; the geometry had to be located first.
So the skill is not how to resolve. It is knowing what to resolve onto, and when not to — and that decision is made from the drawing, before any component is written down.
Three sets of constraint problems now, and no formula has appeared in any of them that was worth memorising. Every answer came from a fixed length, a fixed angle or a fixed contact, differentiated once.
Carried forward
- A rigid rod obeys the same condition as a string — components along it are equal.
- Resolve onto the connector, not onto the surfaces the bodies happen to sit on.
- A sign that emerges from a constraint is an answer, not a formality: it tells you which way a body must move.
- Check that a mechanism closes geometrically before computing anything about it.
- Sometimes nothing needs resolving. Nested frames just add.
Prerequisites: K13, K14, Foundation 5 · Working time: 45 min · Level: JEE Advanced