$\begin{array}{1 1}556N,F_N=1.13\times 10^3N\\278N,F_N=1.13\times 10^4N\\446N,F_N=1.13\times 10^3N\\556N,F_N=1.13\times 10^4N\end{array} $

We resolve this horizontal force into appropriate components

Newton's second law applied to the x-axis produces

$F\cos\theta-mg\sin\theta=ma$

For $a=0$,this yields $F=566$N

Applying Newton's second law to the y axis (where there is no acceleration),we have

$F_N-F\sin \theta-mg\cos \theta = 0$

Which yields the normal force $F_N=1.13\times 10^3N$

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