Friction Force on an Inclined Plane — Full Solution

Inclined plane with friction is a must-master topic. It appears in NEET, JEE Main, and JEE Advanced in various forms. The procedure never changes: find the normal force, check if the block slides by comparing the gravitational component to maximum static friction, then find the actual friction force and acceleration accordingly.

Question

A block of mass 10 kg is placed on a rough inclined plane making an angle of 30° with the horizontal. The coefficient of friction between the block and the incline is $\mu = 0.3$. Determine:

1. Whether the block slides or remains stationary
2. The friction force acting on the block
3. If it slides, the acceleration down the incline

(Take $g = 10$ m/s²)

Solution — Step by Step

Step 1: Identify all forces and find the normal force.

Tilt the coordinate axes: x-axis along the incline (positive down the slope), y-axis perpendicular to the incline.

Forces acting on the block:

• Weight $mg = 100$ N (vertically downward)
• Normal force $N$ (perpendicular to incline, away from surface)
• Friction force $f$ (along incline — direction to be determined)

Along y-axis (perpendicular to incline), acceleration $= 0$:

$N = mg\cos 30° = 100 \times \frac{\sqrt{3}}{2} = 86.6 \text{ N}$

Step 2: Find the gravitational component along the incline.

$F_{\parallel} = mg\sin 30° = 100 \times 0.5 = 50 \text{ N (down the incline)}$

Step 3: Find maximum static friction.

$f_{s,\max} = \mu_s N = 0.3 \times 86.6 = 26 \text{ N}$

Step 4: Compare and decide.

Gravitational pull along incline $= 50$ N

Maximum friction available $= 26$ N

Since 50 N $>$ 26 N, the block slides.

Step 5: Find the kinetic friction force.

Once sliding, friction becomes kinetic:

$f_k = \mu_k N = 0.3 \times 86.6 = 26 \text{ N (up the incline, opposing downward motion)}$

Step 6: Find the net force and acceleration.

Net force along incline (taking down as positive):

$F_{net} = mg\sin 30° - f_k = 50 - 26 = 24 \text{ N}$

$a = \frac{F_{net}}{m} = \frac{24}{10} = 2.4 \text{ m/s}^2 \text{ (down the incline)}$

Answers:

• The block slides down the incline
• Friction force $=$ 26 N (kinetic, directed up the incline)
• Acceleration $=$ 2.4 m/s² (down the incline)

Why This Works

We resolved forces along two perpendicular axes aligned with the geometry of the problem. This is the standard "tilted axes" trick — it avoids messy components of the normal force. Along the perpendicular direction, the block has zero acceleration, which gives us $N$ directly. Along the incline, we apply $F = ma$ with the net force.

The friction force direction flipped from "adjustable static friction" to "fixed kinetic friction" once we confirmed sliding occurs. That's the critical decision point in every inclined plane problem.

Alternative Method — Using Angle of Repose

The angle of repose is the maximum angle at which a block stays stationary:

$\theta_{repose} = \tan^{-1}(\mu_s) = \tan^{-1}(0.3) \approx 16.7°$

Since $30° > 16.7°$, the incline exceeds the angle of repose — the block slides. This gives the same conclusion faster when you just need to check for sliding.

For the acceleration:

$a = g(\sin\theta - \mu_k\cos\theta) = 10(\sin 30° - 0.3\cos 30°)$

$a = 10(0.5 - 0.3 \times 0.866) = 10(0.5 - 0.26) = 10 \times 0.24 = 2.4 \text{ m/s}^2$

Common Mistake