Inclined scenario

Let the mass of the man be $\text{M kg}$ , that of the block be $\text{m kg}$ , the co-efficient of friction be $\mu$ , the angle of inclination of the inclined plane be $\theta$ , the accelerarion due to gravity he $\text{g}$ , $\text{R}$ be the reaction force and the force applied be $\text{T}$ . Note that the force applied by the man is the same as the tension in the wire.

The y-component of the forces acting on the inclined plane due to the weights of the man and the block is $\text{(M+m)g}\cos\theta$ .
So, the reaction force and the tension in the string acts opposite to the force above.
So, $\begin{aligned} \text{(M+m)g}\cos\theta & = \text{R + T}\\ \\ \text{R} & = \text{(M+m)g}\cos\theta - \text{T} \longrightarrow \boxed{1} \end{aligned}$

The frictional force acting is $\mu\text{R} = \mu\left(\text{(M+m)g}\cos\theta - \text{T}\right)$ (From $\boxed{1}$ ).
The net force acting in forward direction = $\text{T + (M + m)g}\sin\theta$ .

Since we need the block is at rest,
$\begin{aligned} \text{T + (M + m)g}\sin\theta - \left(\mu\left(\text{(M+m)g}\cos\theta - \text{T}\right)\right) & = 0\\ \\ \text{T} - \mu\text{(M+m)g}\cos\theta + \mu\text{T} & = - \text{(M+m)g}\sin\theta\\ \\ \text{T}\left(\mu + 1\right) & = \mu\text{(M+m)g}\cos\theta - \text{(M+m)g}\sin\theta\\ \\ \text{T} & = \color{#20A900}{\boxed{\dfrac{\text{g(M + m)}\left(\mu\cos\theta - \sin\theta\right)}{\left(\mu + 1\right)}}} \end{aligned}$

So, plugging in the values $\text{M} = 50$ , $\text{m} = 5$ , $\mu = 10.1$ , $\theta = {30}^°$ and $\text{g} = 10$ , we get:-

$\begin{aligned} \text{T} & = \dfrac{10(50 + 5)\left(10.1×\cos{30}^° - \sin{30}^°\right)}{(10.1 + 1)}\\ \\ & = 408.628\\ \\ & \approx \color{#3D99F6}{\boxed{409}} \end{aligned}$