The starting point is to draw Free Body Diagrams (FBDs) for the board and the block.
Assuming there is no friction between block and walls, the horizontal forces on the block do not affect the vertical forces. So we can ignore the horizontal forces.
The only vertical forces on the block are its weight $W$ and the vertical component $N_1 \sin\theta$ of the normal reaction $N_1$ between the board and block, where $\theta$ is the angle which the board makes with the vertical. If the block is lifted with constant speed so that it does not accelerate, these 2 forces are equal at all times.
The board AB is a lever. I will assume that its base remains fixed to the ground at A, and it makes contact with the block at P. We do not need to consider reaction forces at A, because this is where the board pivots, and the reaction forces exert no moment about this point.
Just as the board does not accelerate upwards, so also it does not accelerate as it rotates. So the clockwise and anticlockwise moments of forces acting on it about any point are equal : $$N_1 p=FL\sin\theta$$ where $L=AB$ is the length of the board and $p=AP$.
The distance of P from the wall is always a fixed distance $X=a+p\sin\theta$ equal to the horizontal width of the block, where $a \le X$ is the fixed distance of A from the wall. The vertical distance of the block above the ground is $y=p\cos\theta$.
This should give you enough equations to calculate $F$ as a function of $\theta$ or as a function of $y$.
If the board is not fixed at A and is able to slip away from the wall as the angle $\theta$ increases, this is a more difficult problem.
Now $a$ depends on $\theta$ and the coefficient of friction $\mu$ between board and ground. The normal reaction at the ground is $N_2=W-F$ and the maximum static friction force is $R=\mu N_2$. The rod will not slip away from the wall until $N_1 \cos\theta = R = \mu N_2$. Thereafter it will slip as $\theta$ increases, and this condition will continue to be met as it does so.
At each value of $\theta$ the board is in quasi-static equilibrium, so we can continue to use the fact that the resultant moment of forces about A is zero.