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 . ?

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Answer 1 · 2017-08-02T09:05:13

See below.

Considering

$x_g = (m_ax_a+m_b x_b)/(m_a+m_b)$ as the system center of mass

we have

$ddot x_g = (m_a ddot x_a+m_b ddot x_b)/(m_a+m_b)$

For each mass we have

${(T-m_a g = m_a ddot x_a),(T-m_b g = m_b ddot x_b):}$

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

$a = ((m_b-m_a)/(m_a+m_b)) g$ now substituting into the formula for $ddot x_g$ we have

$ddot x_g = -((m_a - m_b)/(m_a + m_b))^2g$

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