We establish a coordinate system with the x-axis parallel to the incline (pointing upwards) and the y-axis perpendicular to the incline. We apply Newton's first law for static equilibrium, requiring the net force in both x and y directions to be zero.

The forces perpendicular to the incline (y-direction) are the normal force $N$, the perpendicular component of gravity $mg\cos\theta$, and the perpendicular component of the horizontal force $F\sin\theta$.

$$ \sum F_y = N - mg \cos\theta - F \sin\theta = 0 $$ $$ N = mg \cos\theta + F \sin\theta $$

The forces parallel to the incline (x-direction) are the friction force $f$, the parallel component of the horizontal force $F\cos\theta$, and the parallel component of gravity $mg\sin\theta$.

$$ \sum F_x = f + F \cos\theta - mg \sin\theta = 0 $$ $$ f = mg \sin\theta - F \cos\theta $$

The sign of $f$ determines its direction. A positive $f$ means the friction force is directed up the incline, while a negative $f$ means it is directed down the incline.

For the object to remain stationary, the required static friction force $f$ must be less than or equal to the maximum possible static friction, $f_{max} = \mu N$.

$$ |f| \le \mu N $$

Substituting the expressions for $f$ and $N$:

$$ |mg \sin\theta - F \cos\theta| \le \mu (mg \cos\theta + F \sin\theta) $$

This inequality defines the valid range for $F$. We solve for the two limiting cases:

1. To prevent sliding down (friction acts up, $f \le \mu N$):

$$ mg \sin\theta - F \cos\theta \le \mu (mg \cos\theta + F \sin\theta) \implies F \ge mg \frac{\sin\theta - \mu \cos\theta}{\cos\theta + \mu \sin\theta} $$

2. To prevent sliding up (friction acts down, $-f \le \mu N$):

$$ -(mg \sin\theta - F \cos\theta) \le \mu (mg \cos\theta + F \sin\theta) \implies F \le mg \frac{\sin\theta + \mu \cos\theta}{\cos\theta - \mu \sin\theta} $$

Combining these gives the full range for $F$.