Irodov 1.72: One Movable Pulley, Two Tensions – Amit & Deepali Sinha Foundation

Problem Irodov 1.72

A bar $1$ of mass $m_1$ sits on a frictionless incline that makes an angle $\alpha$ with the horizontal. A thread runs from the bar, up over a fixed pulley mounted at the top of the incline, and down to the axle of a second, movable pulley. A separate, independent thread is anchored to a fixed wall, wraps around the movable pulley, and carries a hanging body $2$ of mass $m_2 = \eta m_1$ on its other end. All pulleys and threads are massless and frictionless. Find the acceleration $w$ of body $2$ , covering every possible case.

Bar 1 on the slope of a fixed wedge, joined by a thread over the apex pulley to the axle of a movable pulley; a second thread, anchored to the wedge's back face, wraps that pulley and carries body 2. — Bar $1$ on the slope of a fixed wedge, joined by a thread over the apex pulley to the axle of a movable pulley; a second thread, anchored to the wedge’s back face, wraps that pulley and carries body $2$ .

Given

$m_2 = \eta m_1$
$\alpha$ = incline angle
no friction, massless pulleys and threads

Find

The acceleration

w

of body

2

Solution

Step 1Two independent tensions

The thread over the fixed pulley (tension $T_1$ ) runs from bar $1$ to the axle of the movable pulley; a second, separate thread (tension $T_2$ ) is anchored to the wall, wraps the movable pulley, and carries body $2$ on its other end.

Step 2Force balance on the massless pulley

The movable pulley is massless, so the net force on it is zero: the single strand $T_1$ holding its axle balances the two strands of $T_2$ pulling down on its rim.

Force balance on the massless movable pulley: one strand at T_1 against two strands at T_2. — Force balance on the massless movable pulley: one strand at $T_1$ against two strands at $T_2$ .

T_1 = 2T_2

(1)

Step 3The 2:1 kinematic constraint

Both threads are inextensible, so each contributes a fixed-length equation. Measure $s_1$ along the incline from the fixed pulley to bar $1$ , and let $y_p$ , $y_2$ and $y_a$ be the depths below that pulley of the movable pulley, of body $2$ and of thread $2$ ’s wall anchor. Thread $1$ runs from the bar over the fixed pulley to the movable pulley’s axle,

\begin{align*}s_1 + y_p &= \text{const}\end{align*}

Thread $2$ climbs from the anchor to the movable pulley and comes back down to body $2$ , so its two strands are $y_a-y_p$ and $y_2-y_p$ long,

\begin{align*}(y_a - y_p) + (y_2 - y_p) &= \text{const}\end{align*}

Differentiating each twice with respect to time, with the fixed anchor giving $\ddot y_a = 0$ ,

\begin{align*}\ddot s_1 + \ddot y_p &= 0\\ \ddot y_2 - 2\,\ddot y_p &= 0\end{align*}

The first line makes the movable pulley descend exactly as fast as bar $1$ climbs toward the fixed pulley; the second doubles that descent at body $2$ ’s end. Writing $w_1$ , $w_p$ and $w$ for the magnitudes of the three accelerations,

w = 2w_p = 2w_1

(2)

Step 4Newton’s second law on each body

Free body of bar 1: normal N, weight m_1 g, and thread tension T_1 up the slope. — Free body of bar $1$ : normal $\vect N$ , weight $m_1\vect g$ , and thread tension $\vect T_1$ up the slope.

Body $2$ (taking downward, its direction of motion when $w>0$ , as positive):

m_2 w = m_2 g - T_2 \tag{3}

Bar $1$ (taking up the slope, its direction of motion when $w>0$ , as positive), from the free body above, with its acceleration $w_1=w/2$ read off (2):

m_1\frac{w}{2} = T_1 - m_1 g\sin\alpha \tag{4}

Step 5Combining the three equations

Solving (3) for $T_2$ , then substituting $m_2=\eta m_1$ and applying the pulley relation (1), gives both tensions in terms of $w$ alone.

\begin{aligned} T_2 &= m_2(g-w) = \eta m_1(g-w), \\ T_1 &= 2T_2 = 2\eta m_1(g-w). \end{aligned}

Substituting $T_1$ into bar $1$ ’s equation (4),

m_1\frac{w}{2} = 2\eta m_1(g-w) - m_1 g\sin\alpha,

and dividing through by $m_1$ and collecting the $w$ terms,

\begin{align*}\frac{w}{2} + 2\eta w &= 2\eta g - g\sin\alpha\\ w\left(\frac{1+4\eta}{2}\right) &= g(2\eta - \sin\alpha)\end{align*}

Solving for $w$ ,

w = \dfrac{2g(2\eta - \sin\alpha)}{4\eta+1}

(5)

Check

The sign of $w$ tracks the sign of $2\eta - \sin\alpha$ : for $2\eta > \sin\alpha$ body $2$ descends and bar $1$ is hauled up the slope, for $2\eta < \sin\alpha$ the motion reverses (bar $1$ slides down, body $2$ rises), and $2\eta = \sin\alpha$ is the equilibrium case, $w=0$ , where the thread tensions alone balance bar $1$ ’s weight component along the slope. All three are covered by the one formula, which is the ”possible cases” the problem asks for.

Remark

The force relation $T_1=2T_2$ and the kinematic relation $w=2w_1$ are reciprocal, and that is not a coincidence: a massless, frictionless pulley can store no energy, so the power the axle strand delivers to it must equal the power the two rim strands carry away, $T_1 v_p = 2T_2\,v_p$ , which forces the tension ratio to be the inverse of the speed ratio. Either relation could have been derived from the other by that virtual-work argument alone, without ever writing Newton’s second law for the pulley itself.