Two bodies of mass m_1 and m_2 are connected by a thread that can withstand the tension force T. Forces F_1 = \alpha t and F_2 = 2 \alpha t are applied to the bodies, where \alpha is a constant coefficient, and t is the time of action of the force. Determine at what point in time the thread will break.

Solutions of Savchenko Problems in Physics

Aliaksandr Melnichenka
October 2023

    <h3 id="back-link"><a href="/#2.1">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $2.1.6.$ Two bodies of mass $m_1$ and $m_2$ are connected by a thread that can withstand the tension force $T$. Forces $F_1 = \alpha t$ and $F_2 = 2 \alpha t$ are applied to the bodies, where $\alpha$ is a constant coefficient, and $t$ is the time of action of the force. Determine at what point in time the thread will break.

    <h3>Solution</h3>
    <p>
        <center>

Let's write Newton's second law for both bars along the horizontal axis $$\left\{\begin{matrix} m_2 a = F_2 - T\\ m_1 a = T - F_1 \end{matrix}\right.$$

Reduce for acceleration $a$ $$\frac{m_2}{m_1} = \frac{F_2 - T}{T - F_1}$$

$$\frac{m_2}{m_1} = \frac{2\alpha t - T}{T - \alpha t}$$
Express $T$
$$T = \alpha t \frac{2m_1+m_2}{m_1+m_2}$$
Where does the time until the thread breaks $T$ come from?
$$\fbox{$t = \frac{T}{\alpha}\frac{m_1 + m_2}{(2m_1 + m_2)}$}$$

    <h4>Answer</h4>
    <p>
        $$t = \frac{T}{\alpha}\frac{m_1 + m_2}{(2m_1 + m_2)}$$
    </p>


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