Irodov 1.75: A Second Pulley Halves the Pull – Amit & Deepali Sinha Foundation

Problem Irodov 1.75

A fixed pulley $P_1$ carries a thread with ball $1$ (mass $m_1$ ) hanging from one end; the other end holds the axle of a second, movable pulley $P_2$ . A separate thread wraps around $P_2$ : one end carries rod $2$ (mass $m_2$ , length $l$ ), the other end is anchored to a fixed support. All pulleys and threads are massless and frictionless, and $m_1 = \eta m_2$ with $\eta = 1.8$ , $l = 100\ \text{cm}$ . Ball $1$ starts at rest level with rod $2$ ’s lower end. Find the time $t$ until ball $1$ is level with rod $2$ ’s upper end.

Ball 1 hangs from the fixed pulley P_1's far strand; P_1's other strand carries the axle of the movable pulley P_2. A second thread wraps P_2, with rod 2 on its open (left) strand and a fixed anchor on the other. — Ball $1$ hangs from the fixed pulley $P_1$ ’s far strand; $P_1$ ’s other strand carries the axle of the movable pulley $P_2$ . A second thread wraps $P_2$ , with rod $2$ on its open (left) strand and a fixed anchor on the other.

Given

$m_1 = \eta m_2$ , $\eta = 1.8$
$l = 100$ cm
massless pulleys and threads, no friction

Find

the time

t

Solution

Step 1The movable pulley’s force balance

Force balance on the massless movable pulley P_2: one strand at T_1 (from P_1) against two strands at T_2 (wrapped, to rod 2 and the anchor). — Force balance on the massless movable pulley $P_2$ : one strand at $T_1$ (from $P_1$ ) against two strands at $T_2$ (wrapped, to rod $2$ and the anchor).

$P_2$ is massless, so the single strand $T_1$ holding its axle (continued from $P_1$ ’s thread, which also carries ball $1$ ) must balance the two strands of $T_2$ pulling down on its rim.

T_1 = 2T_2

(1)

Step 2Two kinematic constraints, chained through

P_2

Taking downward as positive throughout, and writing $w_1$ , $w_2$ , $w_Q$ for the (constant) accelerations of ball $1$ , rod $2$ , and pulley $P_2$ itself: $P_1$ ’s thread has fixed length, with ball $1$ on one end and $P_2$ ’s axle on the other, so as one descends the other rises by the same amount,

w_Q = -w_1.

The second thread wraps $P_2$ with rod $2$ ’s end at position $y_2$ on its free strand and a fixed anchor at the constant position $y_a$ on the other; the sum of the two strand lengths, measured down from $P_2$ at $y_Q$ , is fixed,

(y_a-y_Q) + (y_2-y_Q) = \text{const}.

Differentiating twice, with $y_a$ constant,

\begin{align*} -2w_Q + w_2 &= 0 \\ w_2 &= 2w_Q \tag{2}\end{align*}

Combining the two,

w_2 = -2w_1

(3)

Step 3Newton’s second law for the two loads

\begin{align*} m_1 w_1 &= m_1 g - T_1 \tag{4} \\ m_2 w_2 &= m_2 g - T_2 \tag{5}\end{align*}

Step 4Solving for the accelerations

Solving (4) and (5) for the two tensions and putting both into the pulley balance (1),

\begin{align*}T_1 &= m_1(g-w_1)\\ T_2 &= m_2(g-w_2)\\ m_1(g-w_1) &= 2m_2(g-w_2)\end{align*}

Replacing $w_2$ by $-2w_1$ with (3), then expanding and collecting the $w_1$ terms with $m_1=\eta m_2$ ,

\begin{align*}m_1(g-w_1) &= 2m_2(g+2w_1)\\ m_1 g - 2m_2 g &= m_1 w_1 + 4m_2 w_1\\ (\eta-2)m_2 g &= (\eta+4)m_2 w_1\end{align*}

so that

w_1 = \dfrac{(\eta-2)g}{\eta+4}

(6)

Rod $2$ ’s acceleration then follows from (3),

w_2 = -2w_1 = \frac{2(2-\eta)g}{\eta+4} \tag{7}

Step 5The time for ball 1 to reach the rod’s upper end

Ball $1$ starts level with rod $2$ ’s lower end; it must end up level with the upper end, a distance $l$ higher on the rod, so rod $2$ ’s own descent must exceed ball $1$ ’s descent by exactly $l$ (the same bookkeeping as Step 1 of the sliding-ball problem earlier in this section):

x_2(t)-x_1(t) = l, \qquad x_1=\dfrac{w_1t^2}{2},\ x_2=\dfrac{w_2t^2}{2}

Dividing out the common factor $t^2/2$ and solving for $t^2$ ,

t^2 = \frac{2l}{w_2-w_1}

Substituting the accelerations (6) and (7),

\begin{align*}w_2-w_1 &= \frac{2(2-\eta)g}{\eta+4} - \frac{(\eta-2)g}{\eta+4} \\ &= \frac{2(2-\eta)g+(2-\eta)g}{\eta+4}\\ &= \frac{3(2-\eta)g}{\eta+4}\end{align*}

t = \sqrt{\dfrac{2l(4+\eta)}{3g(2-\eta)}}

(8)

Check

With $\eta=1.8$ , $l=1$ m and $g=9.8\ \text{m/s}^2$ : $2l(4+\eta)=2(5.8)=11.6$ , $3g(2-\eta)=3(9.8)(0.2)=5.88$ , so $t=\sqrt{11.6/5.88}=\sqrt{1.973}=1.4$ s, matching the book’s numeric answer. Also, $\eta<2$ makes $w_1<0$ in Step 4: ball $1$ (only $1.8\times$ rod $2$ ’s mass) is not heavy enough to overcome $P_2$ ’s 2:1 mechanical disadvantage, so it actually accelerates upward ( $w_1<0$ ) while rod $2$ descends. The geometry itself only demands the relative displacement $x_2-x_1=l$ , exactly as in problem 1.74, where both bodies moved in the same direction, downward, while still satisfying an analogous relative condition; ball $1$ ’s own upward acceleration here is a consequence of $\eta<2$ ’s specific force balance, not a requirement of the setup.

Remark

The force relation $T_1=2T_2$ from (1) and the kinematic relation $w_2=2w_Q$ from (2) are reciprocal, as they must be for a massless, frictionless pulley: it stores no energy, so the power the axle strand delivers to $P_2$ must equal the power its two rim strands carry away, $T_1v_Q=2T_2v_Q$ , which forces the tension ratio to be the inverse of the speed ratio. That reciprocity is why rod $2$ ’s weight, carried back through $P_2$ ’s two rim strands to its axle, is felt at $P_1$ ’s thread doubled, $T_1=2T_2$ , the same factor of two by which rod $2$ ’s own end outpaces $P_2$ .