# Question #8c8b6

Nov 27, 2016

Let acceleration of the system be $a m {s}^{-} 2$ and Tension be T N

also let $g \to \text{acceleration due to gravity} = 9.8 m {s}^{-} 2$

Considering the net force acting upward on the combination of 6kg and 4kg combined block system we can write

$T - 10 g \sin {30}^{\circ} = 10 a \ldots \ldots . \left(1\right)$

Again considering the net force acting downward on the 9kg block we can write

$9 g - T = 10 a \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left(2\right)$

Adding (1) and (2) we get

$9 g - 10 g \sin {30}^{\circ} = 20 a$

So $a = \frac{9 g - 10 g \sin {30}^{\circ}}{20} = \frac{9 g - 10 g \times \frac{1}{2}}{20}$

$\implies a = \frac{g}{5} = \frac{9.8}{5} m {s}^{-} 2 = 1.96 m {s}^{-} 2$