Irodov 1.77: One Contact Point Links Two Accelerations

Problem Irodov 1.77

A rod $A$ is held in a fixed vertical guide, so it can only translate up or down. Its lower tip rests on the sloped upper surface of a wedge $B$ , a right triangle of base angle $\alpha$ standing on a horizontal floor. Wedge $B$ is free to slide horizontally on the floor; as it does, the changing height of its slope beneath the fixed rod pushes rod $A$ up or lets it move down. All contact surfaces (rod tip on wedge, wedge on floor) are frictionless, and the mass of wedge $B$ is $\eta$ times the mass of rod $A$ . Find the magnitudes of the accelerations of rod $A$ and of wedge $B$ .

Rod A in a fixed guide that lets it slide only vertically, its lower end cut at the base angle α so that face lies flat on wedge B's slope. Wedge B slides freely on the hatched floor. — Rod $A$ in a fixed guide that lets it slide only vertically, its lower end cut at the base angle $\alpha$ so that face lies flat on wedge $B$ ’s slope. Wedge $B$ slides freely on the hatched floor.

Given

$\eta = M_B/m_A$
$\alpha$

Find

the accelerations

w_A

(rod

A

, vertical) and

w_B

(wedge

B

, horizontal)

Solution

Step 1One contact point ties the two accelerations together

Let $s$ be the horizontal distance, measured along the floor, from the wedge’s own foot to the point of its slope currently under the rod, $y_A$ the rod’s height and $X_B$ the wedge’s horizontal position. The slope rises $s\tan\alpha$ above the foot, and the rod’s own horizontal line never moves, so $s$ changes one for one with the wedge,

\begin{align*}y_A &= s\tan\alpha\\ \frac{ds}{dt} &= -\frac{dX_B}{dt}\end{align*}

Differentiating the height twice with respect to time, with $\alpha$ fixed,

\begin{align*}\frac{d^2y_A}{dt^2} &= \frac{d^2s}{dt^2}\tan\alpha\\ &= -\frac{d^2X_B}{dt^2}\tan\alpha\end{align*}

The answer wants magnitudes, $w_A=\left|d^2y_A/dt^2\right|$ and $w_B=\left|d^2X_B/dt^2\right|$ , and $\tan\alpha>0$ over the whole range $0<\alpha<90^\circ$ , so the minus sign (which only records that the rod descends while the wedge advances) drops out on taking absolute values,

w_A = w_B\tan\alpha \tag{1}

Step 2Newton’s second law for the rod

Taking $w_A$ positive downward, the vertical component of the wedge’s normal force $N$ (which acts along the slope’s own normal) opposes gravity.

m_A w_A = m_A g - N\cos\alpha

Solving for the normal force,

N = \dfrac{m_A(g-w_A)}{\cos\alpha}

(2)

Step 3Newton’s second law for the wedge

The only horizontal force on the frictionless wedge is the horizontal component of the rod’s reaction, equal and opposite to $\vect N$ above.

M_B w_B = N\sin\alpha

Solving for $N$ , then substituting $w_B = w_A\cot\alpha$ from (1) and $M_B = \eta m_A$ ,

\begin{align*}N &= \dfrac{M_B w_B}{\sin\alpha}\\ &= \dfrac{\eta m_A w_A\cot\alpha}{\sin\alpha}\end{align*}

and writing $\cot\alpha=\cos\alpha/\sin\alpha$ ,

N = \eta m_A w_A\dfrac{\cos\alpha}{\sin^2\alpha}

(3)

Step 4Eliminate

N

and solve

Equating the two expressions for $N$ , (2) and (3), and clearing $\cos\alpha$ ,

\begin{align*}\dfrac{m_A(g-w_A)}{\cos\alpha} &= \eta m_A w_A\dfrac{\cos\alpha}{\sin^2\alpha} \\ g - w_A &= \eta w_A\cot^2\alpha \\ g &= w_A\bigl(1+\eta\cot^2\alpha\bigr)\end{align*}

w_A = \dfrac{g}{1+\eta\cot^2\alpha}

(4)

Multiplying by $\cot\alpha$ per (1) gives the wedge’s own acceleration.

w_B = w_A\cot\alpha = \dfrac{g\cot\alpha}{1+\eta\cot^2\alpha}

w_B = \dfrac{g}{\tan\alpha+\eta\cot\alpha}

(5)

Check

For any fixed $\alpha<90^\circ$ (so $\cot\alpha\neq0$ ), as $\eta\to\infty$ (an immovably heavy wedge) both (4) and (5) $\to0$ : the wedge cannot be budged, so the rod simply rests in static equilibrium on a fixed incline, exactly as it should. Dimensionally, both results reduce to $g$ times a pure number, as an acceleration must.

Remark

At $\alpha=90^\circ$ , $\cot\alpha=0$ and $w_A=g$ for every $\eta$ , however heavy the wedge: a vertical face has no horizontal reach into the rod’s own vertical line, so its normal force is purely vertical and can never budge the wedge sideways, leaving the rod in free fall regardless of what sits below it. The mass ratio $\eta$ only ever matters through $\eta\cot^2\alpha$ , so a steep wedge (small $\cot\alpha$ ) needs a far larger $\eta$ than a shallow one to slow the rod’s fall by the same factor, consistent with the $\alpha=90^\circ$ case above needing an infinite one.