Irodov 1.81: Gravity Slides Both Bodies at Once – Amit & Deepali Sinha Foundation

Problem Irodov 1.81

Prism $1$ , a right triangular wedge of mass $m_1$ and base angle $\alpha$ , rests on a horizontal floor and is free to slide on it. Bar $2$ , of mass $m_2$ , sits on the prism’s sloped surface. All contact surfaces, floor to prism and prism to bar, are frictionless. Find the magnitude of prism $1$ ’s horizontal acceleration.

Bar 2 on the frictionless slope of prism 1, base angle α; prism 1 is itself free to slide along the horizontal floor beneath it, which is frictionless too. — Bar $2$ on the frictionless slope of prism $1$ , base angle $\alpha$ ; prism $1$ is itself free to slide along the horizontal floor beneath it, which is frictionless too.

Given

$m_1$
$m_2$
$\alpha$

Find

the acceleration

w

of prism

1

Solution

Step 1Decompose bar 2’s acceleration

Let $w$ be the prism’s horizontal acceleration and $a_r$ bar $2$ ’s acceleration relative to the prism, up-slope positive. Bar $2$ ’s acceleration in the lab frame is then

\vect w_2 = (w+a_r\cos\alpha,\ a_r\sin\alpha)

Step 2Newton’s second law for bar 2

The same arrangement with the two forces acting on bar 2: the slope's normal reaction N, perpendicular to the sloped face because that contact is frictionless, and the weight m_2 g. — The same arrangement with the two forces acting on bar $2$ : the slope’s normal reaction $\vect N$ , perpendicular to the sloped face because that contact is frictionless, and the weight $m_2\vect g$ .

The only forces on bar $2$ are its weight $m_2\vect g$ and the slope’s normal reaction $N$ , along the outward normal $(-\sin\alpha,\cos\alpha)$ .

\begin{align*} m_2(w+a_r\cos\alpha) &= -N\sin\alpha \tag{1} \\ m_2 a_r\sin\alpha &= N\cos\alpha-m_2 g \tag{2}\end{align*}

Step 3Newton’s second law for the prism

The only horizontal force on the frictionless prism is the horizontal component of bar $2$ ’s reaction on it, equal and opposite to $N$ above.

m_1 w = N\sin\alpha

(3)

Step 4Eliminate

N

and

a_r

Substituting $N=m_1w/\sin\alpha$ from (3) into the bar’s horizontal equation (1) and solving for $a_r$ ,

\begin{align*}m_2 w+m_2 a_r\cos\alpha &= -m_1 w \\ a_r &= -\dfrac{w(m_1+m_2)}{m_2\cos\alpha}\end{align*}

Substituting this $a_r$ and $N=m_1w/\sin\alpha$ into the bar’s vertical equation (2),

\begin{align*}-w(m_1+m_2)\tan\alpha &= m_1 w\cot\alpha - m_2 g \\ m_2 g &= w\bigl[m_1\cot\alpha+(m_1+m_2)\tan\alpha\bigr]\end{align*}

w = \dfrac{m_2 g}{m_1\cot\alpha+(m_1+m_2)\tan\alpha}

(4)

Multiplying the numerator and the denominator of (4) by $\sin\alpha\cos\alpha$ clears the cotangent and the tangent; regrouping the denominator then lets the Pythagorean identity collect the two $m_1$ terms.

\begin{align*}w &= \frac{m_2 g\sin\alpha\cos\alpha}{m_1\cos^2\alpha+(m_1+m_2)\sin^2\alpha} \\ &= \frac{m_2 g\sin\alpha\cos\alpha}{m_1(\cos^2\alpha+\sin^2\alpha)+m_2\sin^2\alpha} \\ &= \frac{m_2 g\sin\alpha\cos\alpha}{m_1+m_2\sin^2\alpha}\end{align*}

Dividing the numerator and the denominator by $m_2$ then puts the result in the book’s own form.

w = \dfrac{g\sin\alpha\cos\alpha}{\sin^2\alpha+m_1/m_2}

(5)

Check

As $m_2/m_1\to0$ (a negligible bar), $w\to0$ : nothing pushes the prism if there is nothing on it. As $\alpha\to90^\circ$ (a vertical face bar $2$ cannot actually rest on), $\sin\alpha\cos\alpha\to0$ and $w\to0$ too, consistent with $N\sin\alpha\to0$ carrying no horizontal push at all.

Remark

Bar $2$ ’s vertical acceleration being nonzero is not, by itself, what sets this problem apart: even a block sliding down a fixed, stationary incline has a nonzero vertical acceleration ( $g\sin\alpha$ along the slope has a vertical component $g\sin^2\alpha$ ). What is different here is that bar $2$ ’s lab-frame acceleration is no longer directed along the slope at all, since the slope itself is accelerating out from under it; conflating “the acceleration along the slope” with “the acceleration relative to the ground” is the standard error this problem is built to catch.