Irodov 1.82: A Fixed Anchor Moves the Wedge – Amit & Deepali Sinha Foundation

Problem Irodov 1.82

A bar of mass $m$ rests on the sloped face of a wedge of mass $M$ whose base angle is $\alpha$ ; the wedge itself stands on a frictionless horizontal floor, so it too is free to slide horizontally. A pulley of negligible mass is mounted at the wedge’s apex and moves together with the wedge. A thread of negligible mass runs from the bar up the slope to this pulley, passes over it, and continues horizontally to a fixed wall standing apart from the wedge, where it is tied off. Every contact is frictionless: the bar on the wedge, and the wedge on the floor. Find the acceleration of the wedge $M$ .

Bar m on wedge M's sloped face. The thread runs from the bar up the slope, over the pulley at the wedge's apex, then horizontally to the fixed wall. The wedge stands on the hatched floor. — Bar $m$ on wedge $M$ ’s sloped face. The thread runs from the bar up the slope, over the pulley at the wedge’s apex, then horizontally to the fixed wall. The wedge stands on the hatched floor.

Given

$m$
$M$
$\alpha$
floor and incline frictionless
pulley and thread massless

Find

the acceleration of the wedge

M

Solution

Step 1The thread’s fixed end links the wedge’s slide to the bar’s slide

Let $X$ be the wedge’s horizontal coordinate, measured positive toward the fixed wall, and let $s$ be the bar’s distance from the pulley measured down the slope. The thread has two straight runs, of length $\ell_1=s$ from the bar to the pulley and $\ell_2$ from the pulley to the wall, and their sum is fixed. Since the pulley rides with the wedge while the wall does not, $\ell_2$ shortens exactly as fast as $X$ grows.

\begin{align*}\ell_1+\ell_2 &= \text{const} \\ \dot\ell_2 &= -\dot X \\ \dot s = \dot\ell_1 &= -\dot\ell_2 = \dot X\end{align*}

Differentiating once more, the bar slides down the slope, relative to the wedge, with exactly the wedge’s own acceleration: $\ddot s=\ddot X$ .

Step 2The bar’s acceleration in the ground frame

Call $w=\ddot X$ the wedge’s acceleration, taken positive toward the wall. In the ground frame the bar’s acceleration is the wedge’s translation plus its acceleration relative to the wedge, the latter of magnitude $w$ (Step 1) directed down the slope, i.e. along $\hat u=(-\cos\alpha,-\sin\alpha)$ .

\begin{align*}\vect a_{\mathrm{bar}} &= w\,\hat x + w\,\hat u \\ &= w(1,0)+w(-\cos\alpha,-\sin\alpha)\end{align*}

a_x = w(1-\cos\alpha), \quad a_y = -w\sin\alpha

(1)

Step 3Newton’s second law for the bar

The three forces on the bar, drawn on it as a point particle: the tension T up the slope, the normal N, and the weight m g. — The three forces on the bar, drawn on it as a point particle: the tension $\vect T$ up the slope, the normal $\vect N$ , and the weight $m\vect g$ .

With the tension $\vect T$ directed up the slope, $(\cos\alpha,\sin\alpha)$ , and the normal $\vect N$ along $(-\sin\alpha,\cos\alpha)$ (figure above), Newton’s second law for the bar in the $x$ and $y$ directions, using the components (1), reads

\begin{align*} mw(1-\cos\alpha) &= T\cos\alpha-N\sin\alpha \tag{2} \\ -mw\sin\alpha &= T\sin\alpha+N\cos\alpha-mg \tag{3}\end{align*}

Step 4Newton’s second law for the wedge

The wedge feels the bar’s reaction to $\vect N$ , $N(\sin\alpha,-\cos\alpha)$ , and the pull of the thread on its own pulley, $T$ along each of the two departing strands, $(-\cos\alpha,-\sin\alpha)$ toward the bar and $(1,0)$ toward the wall. Only the horizontal components move the wedge.

Mw = N\sin\alpha+T(1-\cos\alpha) \tag{4}

Step 5Eliminate the normal force to isolate the tension

Solving the wedge’s equation (4) for $N\sin\alpha$ and substituting into the bar’s $x$ equation (2) removes $N$ , leaving one relation between $T$ and $w$ .

\begin{align*}N\sin\alpha &= Mw-T(1-\cos\alpha) \\ mw(1-\cos\alpha) &= T\cos\alpha-\bigl[Mw-T(1-\cos\alpha)\bigr] \\ w\bigl[M+m(1-\cos\alpha)\bigr] &= T\bigl[\cos\alpha+(1-\cos\alpha)\bigr]\end{align*}

T = w\bigl[M+m(1-\cos\alpha)\bigr]

(5)

Step 6Solve for the wedge’s acceleration

Substituting the tension (5) back into the wedge’s equation (4) gives $N\sin\alpha$ in terms of $w$ alone.

\begin{align*}N\sin\alpha &= Mw-w\bigl[M+m(1-\cos\alpha)\bigr](1-\cos\alpha) \\ &= w\bigl[M-M(1-\cos\alpha)-m(1-\cos\alpha)^2\bigr] \\ &= w\bigl[M\cos\alpha-m(1-\cos\alpha)^2\bigr]\end{align*}

The bar’s $y$ equation (3), multiplied through by $\sin\alpha$ , is then ready to take both that expression and the tension (5).

\begin{align*}mg &= mw\sin\alpha+T\sin\alpha+N\cos\alpha \\ mg\sin\alpha &= mw\sin^2\alpha+T\sin^2\alpha+N\sin\alpha\cos\alpha \\ mg\sin\alpha &= mw\sin^2\alpha + w\bigl[M+m(1-\cos\alpha)\bigr]\sin^2\alpha + w\cos\alpha\bigl[M\cos\alpha-m(1-\cos\alpha)^2\bigr]\end{align*}

Every term on the right carries a factor $w$ . Write $S$ for the bracket it multiplies,

S = m\sin^2\alpha+\bigl[M+m(1-\cos\alpha)\bigr]\sin^2\alpha+\cos\alpha\bigl[M\cos\alpha-m(1-\cos\alpha)^2\bigr]

Grouping the $M$ terms first, then the $m$ terms, and substituting $\sin^2\alpha=(1-\cos\alpha)(1+\cos\alpha)$ collects $S$ one algebraic move at a time.

\begin{align*}S &= M(\sin^2\alpha+\cos^2\alpha) + m\sin^2\alpha(2-\cos\alpha) - m\cos\alpha(1-\cos\alpha)^2 \\ &= M + m(1-\cos\alpha)(1+\cos\alpha)(2-\cos\alpha) - m\cos\alpha(1-\cos\alpha)^2 \\ &= M + m(1-\cos\alpha)\Bigl[(1+\cos\alpha)(2-\cos\alpha)-\cos\alpha(1-\cos\alpha)\Bigr] \\ &= M + m(1-\cos\alpha)\bigl[2+\cos\alpha-\cos^2\alpha-\cos\alpha+\cos^2\alpha\bigr] \\ &= M + 2m(1-\cos\alpha)\end{align*}

With $S$ collected, the relation above is a single linear equation in $w$ ,

mg\sin\alpha = w\bigl[M+2m(1-\cos\alpha)\bigr]

and dividing through by that bracket isolates the wedge’s acceleration.

w = \dfrac{mg\sin\alpha}{M+2m(1-\cos\alpha)}

(6)

Check

As $M\to\infty$ , $w\to0$ : an infinitely massive wedge cannot be budged by the bar’s thread, as it must. As $\alpha\to0$ the slope flattens, the driving term $mg\sin\alpha\to0$ , and $w\to0$ as well, consistent with a bar that no longer tends to slide at all.

Remark

The denominator $M+2m(1-\cos\alpha)$ is an effective inertia larger than the wedge’s own mass $M$ : the bar contributes twice, once through the normal reaction it exerts on the slope and once through the thread pull it transmits through the pulley mounted on the wedge. Had the thread’s far end been tied to the wedge itself instead of to the fixed wall, the wedge and bar would form one rigid unit with no relative sliding at all, and this whole mechanism, and the factor of two, would disappear.