Question

Find the acceleration of the system shown in the figure.

• A. $\frac{2g}{7}$
• B. $\frac{g}{7}$
• C. $\frac{2g}{3}$
• D. $\frac{g}{4}$

Answer 1

The correct option is A $\frac{2g}{7}$

As per the given system, the FBDs of the masses are as shown

So, from the FBD of incline mass of $2m$ we have,
$T_2-2mg\sin {30}^{\circ }-T_1=2ma$ ............(i)
From the FBD of mass $4m$ we have,
$4mg-T_2=4ma$...........(ii)
On solving (i) & (ii) we get,
$4mg-2mg\sin {30}^{\circ }-T_1=6ma$
Or, $3mg-T_1=6ma$...........(iii)
Also, from the FBD of mass $m$ we have,
$T_1-mg=ma$............(iv)
On solving eq (iii) & (iv), we get
$2mg=7ma\Rightarrow a=\frac{2mg}{7m}\Rightarrow a=\frac{2g}{7}$

Alternate:

Acceleration of the given system is given by
$a=\frac{\text{(Supporting Force-Opposing force)}}{\text{Total mass}}$
$\Rightarrow a=\frac{\left(4mg\right)-(2mg\sin {30}^{\circ }+mg)}{7m}$
$\Rightarrow a=\frac{2mg}{7m}=\frac{2g}{7}$