Comment
Share
Q)

# 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 $\;m_1 = 2 m_2$?

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}$