Answer
$F = (-80~N)~\hat{i}+(130~N)~\hat{j}$
Work Step by Step
Each hinge supports half the door's weight. We can find $F_y$:
$F_y = \frac{(27~kg)(9.8~m/s^2)}{2} = 130~N$
We can consider the torque about the rotation axis at the bottom hinge. The horizontal force on the door from the top hinge $F_x$ produces a torque that opposes the torque from the door's weight.
We can find the magnitude of $F_x$:
$\sum \tau_i = 0$
$(1.5~m)~F_x-(\frac{0.91~m}{2})(27~kg)(9.8~m/s^2) = 0$
$(1.5~m)~F_x = (\frac{0.91~m}{2})(27~kg)(9.8~m/s^2)$
$F_x = \frac{(\frac{0.91~m}{2})(27~kg)(9.8~m/s^2)}{1.5~m}$
$F_x = 80~N$
Since the horizontal force on the door from the top force produces a counter-clockwise torque, $F_x = -80~N$
We can express the force on the door at the top hinge:
$F = (-80~N)~\hat{i}+(130~N)~\hat{j}$
