So in this problem you're asked to find the magnitude as well as direction (upwards is positive) of the friction force on the block.
The block has a mass $m$ and is not moving in any direction.
If you consider the case where $\theta = 0$ then the force of friction obviously has to point upwards to keep $mg$ from moving the block downwards.
But for $ 0 < \theta \leq \pi/2$ the direction of the friction force depends on the relationship between $mg$ and $F$ which we know nothing about.
The magnitude will be either $mg – F\sin\theta $ or $F\sin\theta – mg$
depending on the direction but how can i decide the latter?
(the $\mu_k$ in the figure is for another problem)
Best Answer
The equation of equilibrium in the vertical direction is $F\sin \theta +F_f-mg=0$, where $F_f$ is the friction force (if its sign is positive, it is directed upwards), so $F_f=mg-F\sin\theta$. However, if $|mg-F\sin\theta|>\mu_s F \cos\theta$, the problem does not have a solution (the block cannot be in equilibrium).