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Read each concept carefully, then check the formula, common mistake, and exam tip before moving to the next. This is Part 2 of Laws of Motion, focusing entirely on Friction and rough surfaces for CBSE Class 11 Physics.
Key Concepts
Friction
The force that opposes relative motion
1
Static Friction & Limiting Friction
Formula
Static friction ($f_s$) is a self-adjusting force that exactly balances the applied force up to a maximum limit. This maximum limit is called Limiting Friction ($f_{\text{max}}$). It depends on the nature of surfaces and the Normal reaction ($N$).
$$f_s \le \mu_s N \quad | \quad f_{\text{max}} = \mu_s N$$
2
Kinetic (Dynamic) Friction
Formula
Once the applied force overcomes limiting friction, the object starts moving. The friction drops slightly to a constant value called kinetic friction ($f_k$). It is independent of the velocity of the object.
$$f_k = \mu_k N \quad (\text{Note: } \mu_k < \mu_s)$$
3
Angle of Friction ($\theta$)
Formula
The angle made by the resultant of the Normal reaction ($N$) and limiting friction ($f_{\text{max}}$) with the Normal reaction. The tangent of this angle equals the coefficient of static friction.
$$\tan \theta = \frac{f_{\text{max}}}{N} = \mu_s$$
4
Angle of Repose ($\alpha$)
Formula
The minimum angle of an inclined plane at which a body placed on it just begins to slide down. Mathematically, it is exactly equal to the angle of friction.
$$\tan \alpha = \mu_s \quad \implies \quad \alpha = \theta$$
5
Acceleration on a Rough Horizontal Plane
Formula
When a force $P$ pulls a mass $m$ horizontally, the net force is $P – f_k$. Since $N=mg$, the kinetic friction is $\mu_k mg$.
$$a = \frac{P – \mu_k mg}{m}$$
6
Motion on a Rough Inclined Plane
Case Formulas
For a block of mass $m$ on an incline of angle $\theta$. The normal reaction is $N = mg \cos\theta$. Friction is $\mu_k mg \cos\theta$.
$$\text{Sliding DOWN: } a = g(\sin\theta – \mu_k \cos\theta)$$
$$\text{Pushed UP: } a = g(\sin\theta + \mu_k \cos\theta)$$
7
Work Done Against Friction
Formula
Friction is a non-conservative force. Work done against friction dissipates as heat energy. It depends on the path taken.
$$W = f_k \times s = (\mu_k N) s$$
Concept Deep Dive
01
Why is Pulling Easier Than Pushing?
It’s all about the Normal forceImagine a lawnmower.
When you PUSH it at an angle $\theta$ downwards, your force resolves into a horizontal component ($F \cos\theta$) moving it forward, and a vertical component ($F \sin\theta$) pushing it into the ground. This increases the Normal reaction ($N = mg + F \sin\theta$). Since friction depends on $N$ ($f = \mu N$), the friction increases, making it harder to move.
When you PULL it at an angle $\theta$ upwards, the vertical component ($F \sin\theta$) lifts it slightly off the ground. This decreases the Normal reaction ($N = mg – F \sin\theta$). Friction decreases, making it much easier to move!
When you PUSH it at an angle $\theta$ downwards, your force resolves into a horizontal component ($F \cos\theta$) moving it forward, and a vertical component ($F \sin\theta$) pushing it into the ground. This increases the Normal reaction ($N = mg + F \sin\theta$). Since friction depends on $N$ ($f = \mu N$), the friction increases, making it harder to move.
When you PULL it at an angle $\theta$ upwards, the vertical component ($F \sin\theta$) lifts it slightly off the ground. This decreases the Normal reaction ($N = mg – F \sin\theta$). Friction decreases, making it much easier to move!
02
The Microscopic Cause of Friction
Cold WeldingWhy does friction exist even on polished surfaces? At a microscopic level, no surface is perfectly smooth; they have hills and valleys (asperities). When two surfaces touch, they only make contact at the tips of these asperities. The pressure at these tiny points is so immense that the molecules actually bond together, forming temporary microscopic welds (Cold Welding). To move the object, you have to apply enough force to sheer (break) these millions of tiny welds!
Compare & Contrast
✗ Static Friction ($f_s$)
- Acts when there is no relative motion (body is at rest).
- It is a self-adjusting force (it changes magnitude to match the applied force).
- Does not have a fixed formula until the verge of motion ($f_s \le \mu_s N$).
- Has a higher maximum value ($\mu_s > \mu_k$).
✓ Kinetic Friction ($f_k$)
- Acts when the body is in actual relative motion (sliding).
- It is a constant force, independent of the applied force.
- Has a strict, fixed formula ($f_k = \mu_k N$).
- Has a lower value, because moving asperities have less time to form strong cold welds.
Remember
Friction does NOT always oppose motion. It opposes relative motion. When you walk, friction pushes your foot forward! Without friction acting in the direction of your motion, you would slip backward.
Common Mistakes to Avoid
Mistake 1
Using $f = \mu N$ when the object isn’t moving: If you push a 100 kg block with 10 N of force, and the limiting friction is 400 N, the actual static friction acting on the block is only 10 N, not 400 N! Static friction only grows as large as it needs to be to prevent motion.
Mistake 2
Assuming Normal Force $N = mg$ on an Incline: On an inclined plane of angle $\theta$, gravity splits into two components. The component pulling it down the slope is $mg \sin\theta$. The component pressing it into the surface is $mg \cos\theta$. Therefore, the Normal reaction is $N = mg \cos\theta$, which makes the friction $f = \mu mg \cos\theta$.
Mistake 3
Block on Block Problems: When a block rests on another block and you pull the bottom one, students often assume they slip immediately. Always calculate the maximum static friction between them first. If the required acceleration demands a force less than this maximum static friction, the blocks move together as a single system!
Exam Tips
Tip 1
If a question asks for the minimum force required to move a block on a horizontal surface, it’s a trap to just say $P = \mu_s mg$. The true minimum force is achieved by pulling at an angle equal to the angle of friction ($\theta$). The formula for this absolute minimum force is $P_{\text{min}} = \frac{\mu_s mg}{\sqrt{1 + \mu_s^2}}$.
Tip 2
If a block is sliding down a rough inclined plane at a constant velocity, it means acceleration is zero. This tells you immediately that the downward force equals the frictional force: $mg \sin\theta = \mu_k mg \cos\theta$, which means $\mu_k = \tan\theta$.
Expected Exam Questions
SQ
Board Pattern Questions
Class 11 · Friction · CBSE Exam1
A block of mass $2 \text{ kg}$ rests on a rough inclined plane making an angle of $30^\circ$ with the horizontal. The coefficient of static friction between the block and the plane is $0.7$. Will the block slide down? ($g = 9.8 \text{ m/s}^2$) [2 marks]
Answer
No, it will not slide.
📝
Explanation
The downward force along the incline is $F_{\text{down}} = mg \sin(30^\circ) = 2 \times 9.8 \times 0.5 = 9.8 \text{ N}$.
The maximum limiting static friction is $f_{\text{max}} = \mu_s N = \mu_s mg \cos(30^\circ) = 0.7 \times 2 \times 9.8 \times 0.866 \approx 11.88 \text{ N}$.
Since $F_{\text{down}} < f_{\text{max}}$ ($9.8 \text{ N} < 11.88 \text{ N}$), the block will not slide. The actual static friction acting on it will simply be $9.8 \text{ N}$ to keep it at rest.
2
Prove that the angle of friction is equal to the angle of repose. [2 marks]
Answer
Derivation
📝
Explanation
1. By definition, Angle of Friction ($\theta$) is given by $\tan \theta = \frac{f_{\text{max}}}{N} = \mu_s$.
2. Let $\alpha$ be the angle of repose. At this exact angle, the block is on the verge of sliding. Therefore, the forces are balanced:
Downward force along incline = Limiting Friction $\implies mg \sin\alpha = f_{\text{max}}$
Normal reaction $\implies N = mg \cos\alpha$
3. Dividing the two equations: $\frac{mg \sin\alpha}{mg \cos\alpha} = \frac{f_{\text{max}}}{N} \implies \tan \alpha = \mu_s$.
4. Since $\tan \theta = \mu_s$ and $\tan \alpha = \mu_s$, we conclude $\theta = \alpha$.
3
A car is moving on a straight horizontal road with a speed $v$. If the coefficient of kinetic friction between the tires and the road is $\mu_k$, find the minimum distance in which the car can be stopped. [3 marks]
Answer
$s = \frac{v^2}{2\mu_k g}$
📝
Explanation
When the brakes are fully applied, the only horizontal force acting on the car is kinetic friction opposing motion: $F = -f_k = -\mu_k N$.
Since $N = mg$, the retarding force is $F = -\mu_k mg$.
By Newton’s second law, $ma = -\mu_k mg \implies a = -\mu_k g$.
Using the 3rd equation of motion: $v_f^2 = u^2 + 2as$.
Final velocity $v_f = 0$, and initial velocity $u = v$.
$0 = v^2 + 2(-\mu_k g)s \implies 2\mu_k gs = v^2 \implies s = \frac{v^2}{2\mu_k g}$.
Concept Map
Friction connects to →
Dynamics
Newton’s Laws
Work & Heat Dissipation
Circular Motion (Banking of roads)
Rolling Friction (Rigid Bodies)
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