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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 tension in the light cord that connects the two masses?

When the bodies are released, the heavier one moves downward and the lighter one moves upward.
Net force in the direction of motion of m1m_1 is F1=m1gTF_1 = m_1g - T which is equal to m1am_1a as per Newton's second law.
m1gT=m1a(i)\Rightarrow m_1g - T = m_1 a \quad (i)
Net force in the direction of motion of m2m_2 is F2=Tm2gF_2 = T-m_2g which is equal to m2am_2a as per Newton's second law.
Tm2g=m2a(ii)\Rightarrow T-m_2g = m_2a \quad (ii)
Solving (i)(i) and (ii)(ii) for acceleration, aa=m1m2m1+m2a \rightarrow a = \large\frac{m_1-m_2}{m_1+m_2}gg
Substituting in (ii)(ii), we get T=m2(a+g)=m2(gm1m2m1+m2T = m_2 (a+g) = m_2 (g \large\frac{m_1-m_2}{m_1+m_2}+g) + g)
T=gm2m1m2+m1+m2m1+m2\Rightarrow T = gm_2 \large\frac{m_1 - m_2 + m_1 + m_2}{m_1+m_2}=2m1m2m1+m2 = \large\frac{2m_1m_2}{m_1+m_2}gg