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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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The device shown here is called an Atwood's machine. m$_1$ and m$_2$ are suspended at the ends of a string of negligible mass passing over a friction-less pulley as shown here. What is the acceleration in the light cord that connects the two masses if $\;m_1 = 2 m_2$?

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Answer: $\large\frac{1}{3}$$g$
When the bodies are released, the heavier one moves downward and the lighter one moves upward.
Net force in the direction of motion of $m_1$ is $F_1 = m_1g - T$ which is equal to $m_1a$ as per Newton's second law.
Net force in the direction of motion of $m_2$ is $F_2 = T-m_2g $ which is equal to $m_2a$ as per Newton's second law.
Solvingfor acceleration, $a \rightarrow a = \large\frac{m_1-m_2}{m_1+m_2}$$g$
Substituting for $a$ and solving for $T$, we get $T = m_2 (a+g) = m_2 (g \large\frac{m_1-m_2}{m_1+m_2}$$ + g)$$= \large\frac{2m_1m_2}{m_1+m_2}$$g$
$\quad a = \large\frac{2m_2 - m_2}{2m_2+m_2}$$g = \large\frac{g}{3}$
answered Aug 20, 2014 by balaji.thirumalai
 

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