A wedge of mass m_2 with an angle \alpha is inserted between two identical smooth bars of mass m_1. Determine the acceleration of bodies.

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    <h3> Statement </h3>
    <p>
        $2.1.48.$ A wedge of mass $m_2$ with an angle $\alpha$ is inserted between two identical smooth bars of mass $m_1$. Determine the acceleration of bodies.

    <h3>Solution</h3>
    <p>

Let us consider a small displacement over time $dt$ from the point of view of kinematics:

From geometric considerations:
$$dy = dx \cdot ctg\,\beta$$
Let us differentiate both parts of the expression twice:
$$\frac{dy}{d^2t} = \frac{dx}{d^2t} \cdot ctg\,\beta$$

$$a_1 = a_2 \cdot tg\,\beta\;(1)$$
Since the Author did not say anything about friction, example $F_{tr}=0$
Next, we write Newton's 2nd law for the bar $m_1$ on the $OY$ axis:
$$m_1a_1=N_1\,cos\,\beta$$

$$N_1=\frac{m_1a_1}{cos\,\beta}\;(a)$$
Similarly for $m_2$:
$$m_2a_2=m_2g-2N_1\,sin\,\beta\;(b)$$
Substitute $(a)$ into $(b)$:
$$m_2a_2=m_2g-2m_1a_1\cdot tg\,\beta$$
We substitute $(1)$ into $(c)$:
$$m_2a_2=m_2g-2m_1a_2\cdot tg^2\,\beta$$
We express $a_2$:
$$\boxed{a_2=\frac{m_2g}{m_2+2m_1\,tg^2\,\beta}}\;(d)$$
Given $(1)$, we multiply $(d)$ by $tg\,\beta$:
$$\boxed{a_1=\frac{m_2g\,tg\,\beta}{m_2+2m_1\,tg^2\,\beta}}$$

    <h4>Answer</h4>
    <p>
                    $$a_1=\frac{m_2g\\,tg\\,\beta}{m_2+2m_1\\,tg^2\\,\beta}$$
        $$a_2=\frac{m_2g}{m_2+2m_1\\,tg^2\\,\beta}$$
    </p>


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