# In the arrangement shown in the figure M_a\ "and"\ M_b are 2\ "Kg and"\ 1\ "Kg" respectively .Find the acceleration of center of massof both the blocks Neglect friction everywhere . ?

Aug 2, 2017

See below.

#### Explanation:

Considering

${x}_{g} = \frac{{m}_{a} {x}_{a} + {m}_{b} {x}_{b}}{{m}_{a} + {m}_{b}}$ as the system center of mass

we have

${\ddot{x}}_{g} = \frac{{m}_{a} {\ddot{x}}_{a} + {m}_{b} {\ddot{x}}_{b}}{{m}_{a} + {m}_{b}}$

For each mass we have

$\left\{\begin{matrix}T - {m}_{a} g = {m}_{a} {\ddot{x}}_{a} \\ T - {m}_{b} g = {m}_{b} {\ddot{x}}_{b}\end{matrix}\right.$

but due to the fact that the rope is inextensible

${\ddot{x}}_{a} = - {\ddot{x}}_{b} = a$ and substituting and resolving for $a$ we have

$a = \left(\frac{{m}_{b} - {m}_{a}}{{m}_{a} + {m}_{b}}\right) g$ now substituting into the formula for ${\ddot{x}}_{g}$ we have

${\ddot{x}}_{g} = - {\left(\frac{{m}_{a} - {m}_{b}}{{m}_{a} + {m}_{b}}\right)}^{2} g$