Classical-Mechanics · Unit 9 · Video 3 · Interactive Practice

The Length That Can't Change: Constraints in Pulley Systems

IKey Formulas

Relation	Where it comes from
$a_{y,P} = -\,a_{y,1}$	String A over the fixed pulley: $l_A = y_1 + y_P + \pi R = \text{const}$
$a_{y,2} + a_{y,3} + 2\,a_{y,1} = 0$	String B on the moving pulley, with $a_{y,P} = -a_{y,1}$ substituted
$T_A = 2\,T_B$	Massless movable pulley: $2T_B - T_A = m_P a = 0$
$T_B = \dfrac{4\,g\,m_1 m_2 m_3}{m_1 m_3 + m_1 m_2 + 4\,m_2 m_3}$	Five equations, five unknowns, solved together

Key Insight: A length that can't change has a second derivative of zero — differentiate a rope's length twice and out drops a relation among accelerations.

IIString A — One Length, Locked Motion

String A can't stretch, so block 1 and pulley P are locked into equal and opposite motion.

IIIString B — Why the Moving Pulley Counts Twice

The pulley's coordinate enters string B's length as $-2y_P$ , so moving it shifts both blocks together.

IVThe Masses Decide Everything

The three masses alone fix every acceleration — even the sign of block 1's.

m_1

m_2

m_3

💡 Challenge: Block 1 accelerates upward exactly when $m_1(m_2 + m_3) < 4\,m_2 m_3$ — find masses that flip its direction.

VQuiz Questions

Problem 1 · Differentiate the Length (Basic)

Given: String A has constant length $l_A = y_1 + y_P + \pi R$ . Differentiating twice — a constant gives zero — find the relation between the accelerations.

✅ Correct! The

\pi R

arc term is constant and differentiates away, leaving

a_{y,1} + a_{y,P} = 0

❌ Close, but… That factor of 2 belongs to string B's moving pulley. String A has each coordinate appear once.

❌ Not quite.

l_A = y_1 + y_P + \pi R

has

y_1

and

y_P

each to the first power — no factor, and the sum is zero.

Show solution

Differentiate the constant length twice:

\frac{d^2 l_A}{dt^2} = \frac{d^2 y_1}{dt^2} + \frac{d^2 y_P}{dt^2} + \underbrace{\frac{d^2(\pi R)}{dt^2}}_{=\,0} = 0

So $a_{y,1} + a_{y,P} = 0$ , i.e. $a_{y,P} = -a_{y,1}$ : when block 1 drops, pulley P rises by exactly as much.

Problem 2 · The Moving-Pulley Factor of 2

Given: String B has $l_B = y_2 + y_3 - 2y_P + \pi R = \text{const}$ . Using $a_{y,P} = -a_{y,1}$ , find the three-block constraint. ⚠️ Mind the factor of 2 and the sign.

✅ Correct!

a_{y,2}+a_{y,3}-2a_{y,P}=0

, and

a_{y,P}=-a_{y,1}

turns

-2a_{y,P}

into

+2a_{y,1}

❌ Close, but check the sign. Substituting

a_{y,P} = -a_{y,1}

into

-2a_{y,P}

gives

+2a_{y,1}

, not

-2a_{y,1}

❌ You dropped the 2. The pulley coordinate enters

l_B

-2y_P

, so its acceleration carries a coefficient of 2.

❌ Not quite. Differentiate

l_B

twice to get

a_{y,2}+a_{y,3}-2a_{y,P}=0

, then substitute

a_{y,P}=-a_{y,1}

Show solution

Differentiate $l_B = y_2 + y_3 - 2y_P + \pi R$ twice:

a_{y,2} + a_{y,3} - 2\,a_{y,P} = 0

Now bring in string A's result $a_{y,P} = -a_{y,1}$ :

a_{y,2} + a_{y,3} - 2(-a_{y,1}) = a_{y,2} + a_{y,3} + 2\,a_{y,1} = 0

The moving pulley shortens both B-strands at once, which is where the factor of 2 comes from.

Problem 3 · The Massless Pulley's Bonus Equation

Given: The movable pulley is massless. String A pulls it up with $T_A$ ; string B pulls down on both strands with $T_B$ each. Find the tension relation from Newton's second law with $m_P = 0$ .

✅ Correct!

2T_B - T_A = m_P a_{y,P} = 0

, so

T_A = 2T_B

regardless of how the pulley accelerates.

❌ Close — count the strands. String B pulls down on two strands, so the downward force is

2T_B

, not

T_B

❌ Not quite. Up force

T_A

must balance the two down forces

T_B + T_B

, since

m_P a = 0

Show solution

Newton's second law on the pulley, downward positive:

2T_B - T_A = m_P\,a_{y,P}

The pulley is massless, so $m_P = 0$ and the right side vanishes no matter how it accelerates:

2T_B - T_A = 0 \quad\Longrightarrow\quad T_A = 2\,T_B

A massless pulley forces a fixed ratio between the two string tensions.

Problem 4 · Equal Masses (Video Example, Transfer)

Given: $m_1 = m_2 = m_3 = m$ . Using $a_{y,1} = \dfrac{g(m_1 m_3 + m_1 m_2 - 4 m_2 m_3)}{m_1 m_3 + m_1 m_2 + 4 m_2 m_3}$ — find both the direction and the magnitude of block 1's acceleration.

Which way does block 1 accelerate?

What is the magnitude?

✅ Excellent! With equal masses the numerator is

m^2 + m^2 - 4m^2 = -2m^2 < 0

, so block 1 rises at

\tfrac{g}{3}

even though nothing is lighter.

❌ Check the sign. The numerator

m^2 + m^2 - 4m^2 = -2m^2

is negative; down is positive, so

a_{y,1} < 0

means upward.

❌ Check the arithmetic.

a_{y,1} = \dfrac{-2m^2}{6m^2}\,g = -\dfrac{g}{3}

Show solution

Substitute $m_1 = m_2 = m_3 = m$ into the denominator:

D = m^2 + m^2 + 4m^2 = 6m^2

And the numerator of $a_{y,1}$ :

m_1 m_3 + m_1 m_2 - 4 m_2 m_3 = m^2 + m^2 - 4m^2 = -2m^2

Therefore

a_{y,1} = \frac{-2m^2}{6m^2}\,g = -\frac{g}{3}

The negative sign (down is positive) means block 1 accelerates upward at $\tfrac{g}{3}$ ; by symmetry $a_{y,2} = a_{y,3} = +\tfrac{g}{3}$ , and $2(-\tfrac{g}{3}) + \tfrac{g}{3} + \tfrac{g}{3} = 0$ satisfies the constraint.

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