A block of mass 10 kg is placed on an inclined plane. When the angle of inclination is 30° , the block just begins to slide down the plane. The force of static friction is

The correct answer is 5 kg wt.

When a block is placed on an inclined plane and is on the verge of sliding (i.e., it "just begins to slide"), the system is in a state of limiting equilibrium. At this precise moment, the force of static friction is exactly equal to the component of the block's weight acting along (down) the inclined surface.

Let us identify all the forces acting on the block on the inclined plane:

1. Weight (W) acting vertically downward = mg
2. Normal Reaction (N) acting perpendicular to the inclined surface
3. Force of Static Friction (f_s) acting up the incline (opposing the tendency to slide down)

We resolve the weight mg into two components:

- Component perpendicular to the incline = mgcos\theta (balanced by Normal Reaction N)
- Component parallel to the incline (down the slope) = mgsin\theta (balanced by Static Friction f_s)

Applying equilibrium along the incline:

Since the block is just about to slide but has not yet moved, the net force along the plane is zero:

f_s=mgsin\theta

Substituting the given values:

- Mass, m = 10 kg
- Angle of inclination, θ = 30°
- sin 30° = 0.5

f_s=10\times g\times sin30°

f_s=10\times 0.5

f_s=5\text{kg wt}

Note: Here we express the answer in kg wt (kilogram-weight), which is a unit of force equal to the weight of 1 kg mass. Since mass = 10 kg and sin 30° = 1/2, the friction force comes out cleanly as 5 kg wt.

Therefore, the force of static friction acting on the block is 5 kg wt.