Q.A mass of $4\ \text{kg}$ rests on a horizontal plane. The plane is gradually inclined until at an angle $\theta = 15^{\circ}$ with the horizontal, the mass just begins to slide. What is the coefficient of static friction between the block and the surface?

Concept: At the threshold of sliding, static friction reaches its maximum value $f_s^{\text{max}} = \mu_s N$, which exactly balances the component of weight along the incline.

When the plane is inclined at angle $\theta = 15°$, resolve the weight $mg$ into components:

• Perpendicular to plane: $N = mg \cos \theta$
• Parallel to plane (down the slope): $mg \sin \theta$

At the instant the mass just begins to slide, the static friction force equals the parallel component:

Substituting $N = mg \cos \theta$: …

Why the block stays put — until it doesn't

A block on an incline experiences two components of its weight: one pressing it into the surface (the normal component) and one trying to pull it down the slope (the tangential component). Static friction opposes the tangential pull, but it has a limit. The maximum static friction force is $f_{\text{max}} = \mu_s N$, where $\mu_s$ is the coefficient of static friction and $N$ is the normal force.

The phrase "just begins to slide" is the key: at exactly $\theta = 15°$, static friction has reached its maximum possible value and can no longer hold the block. Any steeper and the block accelerates down; any shallower and it stays put. At this critical angle, the forces are perfectly balanced.

Step-by-step force analysis

1. Resolve the weight into components.

   The block has weight $W = mg = 4 \times 9.8 = 39.2\ \text{N}$ (though we'll see the mass cancels). On an incline at angle $\theta$:

   • Component perpendicular to the plane: $W_{\perp} = mg \cos \theta$
   • Component parallel to the plane (down the slope): $W_{\parallel} = mg \sin \theta$

2. Find the normal force.

   Since there's no motion perpendicular to the surface, the normal force balances the perpendicular component:

3. Write the condition for impending motion. At the instant the block just begins to slide, static friction is at its maximum: $f_s = \mu_s N$ …

Concept: Critical Angle for Sliding on an Incline

Step 1: Resolve the weight into components on the tilted plane

Perpendicular to the plane: $mg\cos\theta$; along the plane (down-slope): $mg\sin\theta$.

Step 2: Write the normal force

No motion perpendicular to the surface, so

Step 3: Condition at the instant the block "just begins to slide"

Static friction is at its maximum and exactly balances the down-slope pull:

$\mu_s N = mg\sin\theta$ …