Here, we’ll build the first stage of a house of cards. The two cards rest on a table with coefficient of static friction, ${\mu }_s$, as in the figure below.

The cards don’t fall, provided the total force and total torque on each card vanishes. By symmetry, we can examine the forces and torque on just one of the cards.

Let’s take the point of contact with the floor as the axis about-which to evaluate the torque. Then only two of the four forces (gravity and the force due to the other card, $N_c$) contribute to the torque. If we call the length of the card, $L$, the displacement at which gravity acts is $L/2$, and the perpendicular component of the gravitational force is $-mg\sin \theta$. The force from the other card acts at a displacement $L$, and the perpendicular component is $+N_c\cos \theta$. So the total torque is

The $x,y$ components of the total force are

We also must have

Combining (3) with (2), we have $N_c\le {\mu }_{s}mg$. Making the torque vanish in (1) requires

So we need

or

If we try an angle larger than that, the cards will slide.