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Laws of Motion — Physics STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 SciencePhysicsLaws of Motion4 Marks
Question
Figure shows a fixed pulley. A massless inextensible string with masses $m_1$ and $m_2 > m_1$ attached to its two ends is passing over the pulley. Such an arrangement is called an Atwood machine. Calculate accelerations of the masses and force due to the tension along the string assuming axle of the pulley to be frictionless.
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Answer

Method $I:$ As $m_2 > m_1,$ mass $m_2$ is moving downwards and mass $m_1$ is moving upwards.
Net downward force $= F = (m_2) g – (m_1) g$
$= (m_2 – m_1)g$
the string being inextensible, both the masses travel the same distance in the same time. Thus, their accelerations are equal in magnitude $($one upward, other downward$).$ Let it be a.
Total mass in motion, $M = m_2 + m_1$
$\therefore a =\frac{F}{M}=\left(\frac{m_2-m_1}{m_2+m_1}\right) g$ ......(i)
For mass $m_1,$ the upward force is the force due to tension $T$ and downward force is $mg.$ It has upward acceleration
$a.$ Thus, $T – m_1g = m_1a$
$\therefore T = m_1(g + a)$
Using equation $(i),$ we get,
$T=m_1\left[g+\left(\frac{m_2-m_1}{m_2+m_1}\right) g\right]=\left(\frac{2 m_1 m_2}{m_1+m_2}\right) g$
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From the free body equation for the first body,
$T – m_1g = m_1a .. (i)$
From the free body equation for the second body,
$m_2g – T = m_2a … (ii)$
Adding $(i)$ and $(ii),$ we get,
$a =\left(\frac{ m _2- m _1}{m_2+ m _1}\right) g ....(iii)$
Solving equations. $(ii)$ and $(iii)$ for $T,$ we get,
$T = m _2(g- a )=\left(\frac{2 m_1 m_2}{m_1+m_2}\right) g$

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