Answer
We can rank the positions according to the force on post A:
$1 \gt 2 \gt 3 \gt 4 \gt 5 \gt 6$
At position 3, the force on post A is zero.
Work Step by Step
In position 1, post A is fully supporting the weight of the safe, so the compression is the greatest.
In position 2, post A and post B share the weight of the safe, so the compression on post A in this position is half the compression of position 1.
In position 3, post B is fully supporting the weight of the safe, so there is no compression or tension in post A. In fact, post A could be removed.
In positions 4 to 6, we can consider the torque about a rotation axis at post B.
In position 4, the there is a small clockwise torque from the weight of the safe. There must be a small counterclockwise torque from the tension in post A to maintain equilibrium. Thus there is a small tension in post A.
In position 5, the there is a medium clockwise torque from the weight of the safe. There must be a medium counterclockwise torque from the tension in post A to maintain equilibrium. Thus there is a medium tension in post A.
In position 6, the there is a large clockwise torque from the weight of the safe. There must be a large counterclockwise torque from the tension in post A to maintain equilibrium. Thus there is a large tension in post A.
We can rank the positions according to the force on post A:
$1 \gt 2 \gt 3 \gt 4 \gt 5 \gt 6$
At position 3, the force on post A is zero.
