The problem is as follows:
The figure from below shows a block of mass $m_2=2\,kg$ and $m_1=1\,kg$. Assuming the surfaces are frictionless find the value of the acceleration of each block.
The alternatives are as follows:
$\begin{array}{ll} 1.&\frac{g}{7}\\ 2.&\frac{g}{5}\\ 3.&\frac{g}{4}\\ 4.&\frac{g}{6}\\ 5.&\frac{g}{8}\\ \end{array}$
How exactly should I make a FBD (free body diagram) in this system?. I'm not sure exactly how to assess the acceleration for both objects.
Since the object which is on top weighs more than the one which is on bottom then it makes sense that the motion for the object on top will more to the left and the one which is on bottom to move to the right.
$T-(m_1+m_2)g\sin 30^{\circ}=(m_1+m_2)(a)$
while for the object on top it would be:
$m_2g\sin 30^{\circ}-T=m_2(-a)$
But I'm not sure if this is the right interpretation for this system:
If I go further on this it would become into:
$(m_1+m_2)g\sin 30^{\circ}+(m_1+m_2)(a)=T$
$m_2g\sin 30^{\circ}+m_2(a)=T$
$m_2g\sin 30^{\circ}+m_2(a)=(m_1+m_2)g\sin 30^{\circ}+(m_1+m_2)(a)$
$m_2a=m_1g\sin30^\circ+m_1a$
$a=\frac{m_1g\sin30^\circ}{m_2-m_1}=\frac{0.5g}{1}=\frac{1}{2}g$
But it doesn't check with any of the alternatives. Where exactly did I missunderstood?. Can someone help me here?.
You would have $m_2g\sin 30^{\circ}-m_1g\sin 30^{\circ}=(m_1+m_2)a\implies a=g\sin 30^{\circ}(m_2-m_1)/(m_2+m_1)=(g/2)(2-1)/(2+1)=g/6,$ which is answer 4.