A small block of mass 0.1 kg lies on a fixed inclined plane PQ which makes an angle θ with the horizontal. A horizontal force of 1 N acts on the block through its center of mass as shown in the figure. The block remains stationary if (take g=10 m/s²)

The net force along the plane PQ acting on the mass is given by:
F = mg sinθ - f - μcosθ
where f = frictional force towards Q
m = 0.1 kg
g = 10 m/s²
For the mass to remain stationary, the net force along the plane should be zero.
Thus, mg sinθ - f - μcosθ = 0
In the absence of friction, mg sinθ = μ cosθ
i.e., tanθ = μ
In the presence of friction,
mg sinθ = f + μcosθ
When θ > 45°, sinθ > cosθ and both are positive. Hence, f > 0.
Thus, a positive frictional force acts on the mass towards Q.