Coefficient of Friction on Inclined Plane

 

❓ Question

On an inclined plane of angle $\theta$, a block slides down with given acceleration.

How do we find the coefficient of kinetic friction $\mu_k$?

---

🖼 Concept Image

---

✍️ Short Solution

This is a force resolution + Newton’s 2nd Law question.

Golden idea:

👉 Resolve gravity
👉 Apply $F = ma$ along the plane
👉 Extract $\mu_k$

---

🔷 Step 1 — Forces on an Inclined Plane 💯

Inclined angle = $\theta$

Gravity splits into:

Along plane → $mg \sin\theta$
Perpendicular → $mg \cos\theta$

Normal reaction:

$$N = mg \cos\theta$$

---

🔷 Step 2 — Direction of Kinetic Friction

If block slides downwards:

Friction always acts upwards (opposes motion).

Friction force:

$$f_k = \mu_k N$$
$$f_k = \mu_k mg \cos\theta$$

⚠️ Direction mistake = direct negative marking in JEE.

---

🔷 Step 3 — Apply Newton’s 2nd Law Along Plane

Net force along plane:

$$mg \sin\theta - \mu_k mg \cos\theta$$

And that equals:

$$ma$$

So,

$$mg \sin\theta - \mu_k mg \cos\theta = ma$$

👉 Yahin se $\mu_k$ niklega.

---

🔷 Step 4 — Acceleration Given ⇒ Friction Fixed

If acceleration already given (example $a = g/2$):

$$mg \sin\theta - \mu_k mg \cos\theta = m \left(\frac{g}{2}\right)$$

Cancel m and g:

$$\sin\theta - \mu_k \cos\theta = \frac{1}{2}$$

👉 Difference between gravity component and actual acceleration
= friction ka effect.

---

🔷 Step 5 — Special Angles = Easy Math

JEE loves:

$$\theta = 30^\circ, 37^\circ, 45^\circ, 60^\circ$$

Example:

$$\sin 60^\circ = \frac{\sqrt{3}}{2}$$ $$\cos 60^\circ = \frac{1}{2}$$

👉 Calculation clean ho jata hai
👉 $\mu_k$ nice fraction mein aata hai

---

✅ Final Takeaway

$$\boxed{\mu_k = \frac{\sin\theta - \frac{a}{g}}{\cos\theta}}$$

Bas ye formula yaad nahi — derive in 20 seconds using Newton.

---

⭐ Golden JEE Insight

Inclined plane problems are always:

1️⃣ Resolve forces
2️⃣ Choose direction
3️⃣ Apply $F = ma$

Direction clarity = full marks.