Module 3: Kinematics · Worked Examples · Prerequisites: K14, Foundation 5
Five problems, none of which can be solved by recalling a pulley ratio.
Mahavakya
Never ask what the pulley formula is. Ask which portions of the string change length when this body moves.
Every example below is worked the same way: write the geometry as an equation, then differentiate it. Once for velocities, twice for accelerations. Nothing is memorised, and each problem is chosen because a remembered rule would get it wrong.
The five
1. A ratio that is not 1 : 2
Problem
Two blocks A and B hang in an arrangement whose string lengths satisfy
2yA + 3yB = constant
with both coordinates measured downward from a fixed point. At one instant A is moving downward at 4.0 m/s and has an acceleration of 2.0 m/s² upward.
Find the velocity and acceleration of B.
Velocity. Differentiate the constraint once:
2vA + 3vB = 0
vB = −⅔vA = −⅔(+4.0) = −2.67 m/s
Negative in a downward-positive convention, so B rises at 2.67 m/s.
Acceleration. Differentiate again. The relation has the same shape, because the coefficients are constants:
2aA + 3aB = 0
A’s acceleration is 2.0 m/s² upward, so in this convention aA = −2.0 m/s².
aB = −⅔(−2.0) = +1.33 m/s², downward
Notice what could not have been recalled. There is no standard arrangement that gives 2 : 3. The ratio came out of the geometry, and the only way to get it was to differentiate the constraint.
Notice also that B rises while its acceleration points downward. That is not a contradiction — it is decelerating, and K4’s va rule says so: opposite signs mean slowing.
2. Where the factor of two comes from — and where it does not
Problem
A load hangs from a pulley block supported by three string segments. The free end of the string is pulled downward at speed v. How fast does the load rise?
Three segments support the load, so the free end moves three times as far.
Let yL be the load’s depth and yfree the length of string paid out at the free end, both measured downward. If the load rises by h, each of the three supporting segments shortens by h, releasing 3h of string. So:
3yL + yfree = constant
3vL + vfree = 0
vL = −v/3 — the load rises at v/3
“A movable pulley gives a factor of two” is a fact about one arrangement, not a rule. Two is what you get when two segments support the load. Here three do, and the factor is three.
The reliable question is the one in the Mahavakya: which portions of the string change length when this body moves? Count those, and the factor follows for any arrangement you are given.
3. A string whose angle keeps changing
Problem
A cart runs on a straight track. A rope from the cart passes over a pulley fixed at height h = 4 m above the track, and is drawn in at a constant u = 5 m/s.
At the instant the cart is 3 m from the foot of the pulley, find (a) its speed and (b) its acceleration.
Only the component of the cart’s velocity along the string shortens it.
(a) K14’s constraint says the component of velocity along the string equals the rate the string shortens:
v cos θ = u
With x = 3 and h = 4, the string length is 5 m, so cos θ = 3/5.
v = 5 ÷ (3/5) = 8.33 m/s
The cart moves faster than the rope is drawn in, and always will, since cos θ < 1. Only part of its motion is directed along the string; the rest swings the string round rather than shortening it.
(b) Here is where the problem earns its level. The angle is not fixed — it changes as the cart moves — so differentiating v = u sec θ needs the product rule on θ(t). Carrying it through:
a = u² tan³ θ / h
= 25 × (4/3)³ ÷ 4 = 14.8 m/s²
Check it against the geometry directly. Differentiating x² + h² = L² twice gives a = u²h²/x³ = 25 × 16 ÷ 27 = 14.8 m/s². Same number by a different route, which is worth doing whenever an answer arrives through a chain of trigonometric substitutions.
The trap: treating θ as constant. It gives the right velocity and a badly wrong acceleration, and nothing in the working looks unusual.
4. How many things can move independently?
Problem
Block A hangs from a string that passes over two fixed pulleys and down to a movable pulley P. A second string passes over P, carrying blocks B and C.
(a) How many degrees of freedom does the system have? (b) Find the relation between the three block velocities.
Count the moving bodies, then the constraints. The movable pulley is a moving body too.
(a) Four things move: A, B, C and the pulley P. Two inextensible strings give two constraints.
degrees of freedom = 4 − 2 = 2
The standard error is to count three blocks and two strings, and answer 1. The movable pulley is not part of the scenery — its position changes, so it needs a coordinate of its own.
(b) Taking all coordinates downward from the ceiling:
string 1: yA + yP = L1
string 2: yB + yC − 2yP = L2
Eliminating yP:
2yA + yB + yC = constant
2vA + vB + vC = 0
One equation relating three velocities — so two of them can be chosen freely, which is the two degrees of freedom appearing a second time. The two answers agree, and each checks the other.
5. A block on a moving wedge
Problem
A wedge of face angle 37° slides along the ground at 10 m/s. A block on its inclined face slides down the face at 5 m/s relative to the wedge.
Find the block’s velocity as seen from the ground, and verify that it remains in contact with the face.
The contact condition from K14 §5 is that the two bodies have equal velocity components along the common normal:
(vblock − vwedge) · n̂ = 0
with n̂ = −sin 37° i + cos 37° j = −0.6 i + 0.8 j
The relative velocity is 5 m/s directed down the face, which is (−cos 37°, −sin 37°) in components:
vrel = −4 i − 3 j m/s
check: vrel · n̂ = (−4)(−0.6) + (−3)(0.8) = 2.4 − 2.4 = 0 ✓
Zero, as it must be — the relative velocity lies along the face, which is what “slides down the face” means. Adding the wedge’s own motion:
vblock = (10 i) + (−4 i − 3 j) = 6 i − 3 j m/s
|vblock| = √45 = 3√5 ≈ 6.71 m/s, at 26.6° below the horizontal
The block moves forward while sliding down. Its ground-frame velocity has a positive horizontal component, even though it is descending the face — because the wedge is carrying it along faster than it slides back. Nothing about that is visible from the relative velocity alone, which is why the frame has to be stated.
What the five have in common
Not one of them was solved by recalling a result. In each case the geometry was written as an equation and then differentiated — once for velocities, twice for accelerations — and the answer fell out.
That is deliberate. A student who has memorised “movable pulley means two” gets example 2 wrong, “three blocks minus two strings” gets example 4 wrong, and “the ratio is fixed” gets example 3’s acceleration wrong. The rules fail precisely where the arrangement is unfamiliar, which is where examinations put you.
The constraint equation never fails, because it is not a rule about pulleys. It is a statement that a length does not change, differentiated.
Method, in four lines
- Choose coordinates and a positive direction, and mark them on a drawing.
- Write the total length as an equation. Count segments, not pulleys.
- Differentiate once for the velocity relation, twice for the acceleration relation.
- Check the signs against the drawing before reporting anything.
Where these sit
Examples 1 and 2 are JEE Main standard. Examples 3, 4 and 5 are Advanced — 3 for the moving angle, 4 for the degrees-of-freedom count, 5 for holding two frames at once. All five assume K14, and example 5 also assumes K13.
Prerequisites: K14, Foundation 5 · Working time: 45 min · Level: JEE Main to Advanced