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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 tension in the light cord that connects the two masses?

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Answer: 2m1m2m1+m2 \large\frac{2m_1m_2}{m_1+m_2}gg
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
answered Aug 19, 2014 by balaji.thirumalai
 

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