In the second case, decompose the force $\vec{F}$ into horizontal ($x$) component and a vertical ($y$) one:
$$F_x=F\cos\theta$$ And:
$$F_y=F\sin\theta$$
The $x$ component is what might overcome friction and thus cause motion acc. Newton's second.
The vertical one reduces the Normal force, caused by the weight $mg$:
$$F_N=mg-F_x=mg-F\cos\theta$$
This then gives rise to the friction force $F_f$, opposing $F_x$:
$$F_f=\mu F_N$$
If:
$$F\cos\theta > \mu(mg-F\cos\theta)$$
there will be acceleration to the right.
But there is no vertical friction at all.
Should it be the case that:
$$F\sin\theta > mg$$
there will be acceleration in the $y$ direction (upward).