Solution

For the forces acting the person, with mass $m_1$, we have:
\begin{align*} \sum F_x = f_{1,s} - T \cos \theta = m_1 \, a_{1,x} \end{align*}
\begin{align*} \sum F_y = n_1 - m_1 g - T \sin \theta = m_1 \, a_{1,y} \end{align*}

and for the block, $m_2$, we have:
\begin{align*} \sum F_x = T \cos \theta - f_{2,s} = m_2 \, a_{2,x} \end{align*}
\begin{align*} \sum F_y = n_2 - m_2 g + T \sin \theta = m_2 \, a_{2,y} \end{align*}

We'll need to place some constraints on the expected motion to simplify the problem. First, neither the person nor the block is pulled so hard by the rope that they lift off the ground; i.e., the acceleration in the y-direction for each object is zero. Second, we need to ensure that the person remains in place and \emph{does not slip}, requiring that the friction force holding the person in place is \emph{greater} than the x-component of the tension force pulling on the person. Third, we must consider the x-component of the tension in the rope sufficiently large to overcome the static friction force holding the block in place. Noting that $f_{i,s} \leq \mu n_i$ We can then write:
\begin{align*} \sum F_x &= \mu n_1 - T \cos \theta > 0 \end{align*}
\begin{align*} \sum F_y &= n_1 - m_1 g - T \sin \theta = 0 \end{align*}