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 $F_y$:
$F_y = \frac{(27~kg)(9.8~m/s^2)}{2} = 130~N$
In part (a), we found that the horizontal force on the door at the top hinge is $-80~N$
Since the system is in equilibrium, the horizontal force on the door at the bottom hinge is $F_x = 80~N$
We can express the force on the door at the bottom hinge:
$F = (80~N)~\hat{i}+(130~N)~\hat{j}$
