Question

consider the system of 2 pullies as shown in fig. both the pully are smooth and strings are light if the acceleration of m1 was 5m/s^2 downward then find the value of m1/ m2 .
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Answer 1

Since masses of the blocks B and C are same, there will be no relative motion between them, and so the acceleration of B and C each is zero.

Free Body Diagram of B or C:-

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\put(0,0){\framebox(5,5){ \sf{m_2} }}\put(2.5,0){\vector(0,-1){10}}\put(2.5,5){\vector(0,1){10}}\put(-1,-13){ \sf{m_2g} }\put(1.3,16){\sf{T}}\end{picture}$

Since acceleration is zero, so is the net force.

$\sf{\longrightarrow m_2g-T=0}$

$\sf{\longrightarrow T=m_2g}$

Free Body Diagram of the pulley connecting B and C:-

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\put(0,0){\circle{6}}\multiput(-3,0)(6,0){2}{\vector(0,-1){10}}\put(0,0){\vector(0,1){13}}\multiput(-7,-12)(11.5,0){2}{\sf{T}}\put(-1,14){\sf{T'}}\end{picture}$

Assume the pulley to be massless so that the net force acting on it is zero.

$\sf{\longrightarrow T'-2T=0}$

$\sf{\longrightarrow T'=2T}$

$\sf{\longrightarrow T'=2m_2g}$

Free Body Diagram of A:-

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\put(0,0){\framebox(5,5){ \sf{m_1} }}\put(2.5,0){\vector(0,-1){10}}\put(2.5,5){\vector(0,1){10}}\put(-1,-13){ \sf{m_1g} }\put(1.3,16){\sf{T'}}\put(10,7.5){\vector(0,-1){10}}\put(11.5,1.5){ \sf{5\ m\,s^{-2}} }\end{picture}$

The net force acting on A,

$\sf{\longrightarrow m_1g-T'=5m_1}$

$\sf{\longrightarrow m_1g-2m_2g=5m_1}$

$\sf{\longrightarrow (m_1-2m_2)g=5m_1}$

Taking $\sf{g=10\ m\,s^{-2},}$

$\sf{\longrightarrow (m_1-2m_2)10=5m_1}$

$\sf{\longrightarrow 2m_1-4m_2=m_1}$

$\sf{\longrightarrow m_1=4m_2}$

$\sf{\longrightarrow\underline{\underline{\dfrac{m_1}{m_2}=4}}}$