Consider the situation shown in figure. Both the pulleys and the string are light and all the surfaces are frictionless. Find the acceleration of the mass M.

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Solution

$M a - 2 T = 0$ $T + M a - M g = 0$
$\Rightarrow M a = 2 T \Rightarrow T = M a / 2$ $\Rightarrow M a / 2 + m a = M g$ (because $T = M a / 2$ )
$\Rightarrow 3 M a = 2 M g \Rightarrow a = 2 g / 3$
acceleration of mass $M$ is $2 g / 3$

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Consider the situation shown in figure. Both the pulleys and the string are light and all the surfaces are frictionless. Find the acceleration of the mass M.

Ma−2T=0 T+Ma−Mg=0⇒Ma=2T⇒T=Ma/2 ⇒Ma/2+ma=Mg (because T=Ma/2) ⇒3 Ma=2 Mg⇒a=2g/3 acceleration of mass M is 2g/3

$M a - 2 T = 0$ $T + M a - M g = 0$
$\Rightarrow M a = 2 T \Rightarrow T = M a / 2$ $\Rightarrow M a / 2 + m a = M g$ (because $T = M a / 2$ )
$\Rightarrow 3 M a = 2 M g \Rightarrow a = 2 g / 3$
acceleration of mass $M$ is $2 g / 3$