RRB NTPC & Group D Physics: Master Friction – Complete Guide with Solved Problems & Tricks

Introduction to Friction for RRB Exams

Friction is one of the most fundamental and heavily tested topics in General Science for Indian Railway Recruitment Board (RRB) exams, including RRB NTPC, RRB Group D, and Technician posts. In our daily lives, friction is the invisible force that allows us to walk without slipping, enables vehicles to brake safely, and keeps nails fixed in wood. For railway aspirants, understanding friction is crucial not just for solving numerical problems in physics, but also for appreciating real-world mechanical applications such as train braking systems and track adhesion. This comprehensive guide covers all aspects of friction from basic definitions to advanced problem-solving techniques tailored specifically for RRB exams.

Topic Weightage and Importance

In the General Science section of RRB NTPC and Group D examinations, physics constitutes a major portion of the questions. Within physics, mechanics—specifically force, motion, and friction—frequently yields 2 to 3 direct questions per shift. Aspirants can expect both conceptual questions (such as factors affecting friction or methods to reduce it) and direct numerical problems involving coefficients of friction, normal reaction force, and inclined planes. Scoring full marks in this topic gives candidates a significant edge in the computer-based test (CBT).

Key Concepts and Formulas

Friction is the opposing force that comes into play when one body moves or tends to move over the surface of another body. It always acts tangentially to the contact surfaces and opposite to the direction of relative motion. Let us review the core types and formulas:

1. Types of Friction

• Static Friction ($f_s$): The frictional force that opposes impending motion. It is a self-adjusting force ranging from zero up to a maximum value known as limiting friction ($f_{s,max}$).
• Limiting Friction ($f_L$): The maximum value of static friction up to which the body does not move. Formula: $f_L = \mu_s N$, where $\mu_s$ is the coefficient of static friction and $N$ is the normal reaction.
• Kinetic or Sliding Friction ($f_k$): The frictional force that opposes relative motion when a body is actually sliding over a surface. Formula: $f_k = \mu_k N$, where $\mu_k$ is the coefficient of kinetic friction.
• Rolling Friction ($f_r$): The opposing force when one body rolls over another. Rolling friction is generally much smaller than sliding friction, which is why ball bearings are widely used in railway axles.

2. Important Formulas

• Normal Reaction on a horizontal surface: $N = mg$
• Normal Reaction on an inclined plane at angle $\theta$: $N = mg \cos\theta$
• Angle of Friction ($\lambda$): The angle between the normal reaction and the resultant of limiting friction and normal reaction. Relation: $\mu = \tan\lambda$
• Angle of Repose ($\alpha$): The minimum angle of an inclined plane at which a body placed on it just begins to slide down. Relation: $\mu = \tan\alpha$ (Note: Angle of repose equals the angle of limiting friction).

Solved Examples (Step-by-Step)

Example 1

Problem: A wooden block of mass 10 kg is placed on a horizontal rough floor. If the coefficient of static friction between the block and the floor is 0.4, find the maximum static frictional force acting on the block. (Take $g = 10 \text{ m/s}^2$).

Solution:
1. Identify the given values: Mass $m = 10 \text{ kg}$, coefficient of static friction $\mu_s = 0.4$, acceleration due to gravity $g = 10 \text{ m/s}^2$.
2. Calculate the normal reaction $N$:
$N = mg = 10 \times 10 = 100 \text{ N}$.
3. Use the limiting friction formula $f_{s,max} = \mu_s N$:
$f_{s,max} = 0.4 \times 100 = 40 \text{ N}$.
Answer: The maximum static frictional force is 40 N.

Example 2

Problem: A body of mass 5 kg is pulled along a horizontal floor with a constant acceleration of $2 \text{ m/s}^2$ by a horizontal force of 25 N. Find the coefficient of kinetic friction between the body and the floor. ($g = 10 \text{ m/s}^2$).

Solution:
1. Given: $m = 5 \text{ kg}$, applied force $F = 25 \text{ N}$, acceleration $a = 2 \text{ m/s}^2$, $g = 10 \text{ m/s}^2$.
2. Calculate the net force acting on the body using Newton's second law ($F_{net} = ma$):
$F_{net} = 5 \times 2 = 10 \text{ N}$.
3. The net force is the difference between the applied force and the kinetic friction ($f_k$):
$F - f_k = F_{net}$
$25 - f_k = 10 \implies f_k = 15 \text{ N}$.
4. Calculate the normal reaction $N = mg = 5 \times 10 = 50 \text{ N}$.
5. Find the coefficient of kinetic friction $\mu_k = \frac{f_k}{N}$:
$\mu_k = \frac{15}{50} = 0.3$.
Answer: The coefficient of kinetic friction is 0.3.

Example 3

Problem: An iron box is placed on an inclined plane. If the plane is tilted to an angle of $30^\circ$ with the horizontal, the box just begins to slide. What is the coefficient of static friction between the box and the plane?

Solution:
1. Recognize that the angle at which the body just begins to slide is the angle of repose ($\alpha = 30^\circ$).
2. Apply the relationship between the coefficient of friction and the angle of repose: $\mu = \tan\alpha$.
3. Substitute the value: $\mu = \tan(30^\circ) = \frac{1}{\sqrt{3}} \approx 0.577$.
Answer: The coefficient of static friction is approximately 0.577.

Common Mistakes to Avoid

• Confusing Static and Kinetic Coefficients: Always remember that the coefficient of static friction is generally greater than or equal to the coefficient of kinetic friction ($\mu_s \geq \mu_k$).
• Ignoring the Normal Reaction: Do not assume $N$ is always equal to $mg$. On inclined planes or when an \texternal force has a vertical component, $N$ changes.
• Misinterpreting Self-Adjusting Nature: Static friction is not constant; it matches the applied force up to its maximum limit. Do not use the formula $f = \mu N$ when the applied force is less than the limiting friction.

Practice Questions with Solutions

Question 1

A block of mass 20 kg rests on a rough horizontal surface. If $\mu_s = 0.5$ and $\mu_k = 0.4$, what is the frictional force if a horizontal force of 80 N is applied? ($g = 10 \text{ m/s}^2$)

Solution:
Normal reaction $N = 20 \times 10 = 200 \text{ N}$.
Maximum static friction $f_{s,max} = 0.5 \times 200 = 100 \text{ N}$.
Since the applied force (80 N) is less than the maximum static friction (100 N), the block will not move, and the static friction adjusts itself to equal the applied force.
Answer: 80 N.

Question 2

What is the primary reason for using ball bearings in railway wagon axles?

Solution:
Ball bearings convert sliding friction into rolling friction. Because rolling friction is significantly smaller than sliding friction, it reduces energy loss and wear and tear.
Answer: To reduce friction by converting sliding friction into rolling friction.

Question 3

A body slides down a smooth inclined plane of inclination $45^\circ$. If $g = 10 \text{ m/s}^2$, find its acceleration.

Solution:
For a smooth (frictionless) inclined plane, the acceleration down the plane is given by $a = g \sin\theta$.
$a = 10 \times \sin(45^\circ) = 10 \times \frac{1}{\sqrt{2}} = 5\sqrt{2} \approx 7.07 \text{ m/s}^2$.
Answer: $7.07 \text{ m/s}^2$.

Question 4

If the coefficient of friction between two surfaces is $\sqrt{3}$, what is the angle of friction?

Solution:
We know $\mu = \tan\lambda = \sqrt{3}$.
Therefore, $\lambda = \tan^{-1}(\sqrt{3}) = 60^\circ$.
Answer: $60^\circ$.

Question 5

Which of the following methods cannot be used to reduce friction? (A) Polishing (B) Lubrication (C) Using ball bearings (D) Increasing normal load

Solution:
Increasing the normal load increases the contact force between surfaces, thereby increasing the frictional force ($f = \mu N$).
Answer: (D) Increasing normal load.

Frequently Asked Questions (FAQs)

Q1: Is friction a conservative or non-conservative force?

Friction is a non-conservative force because the work done by friction in moving an object depends on the path taken, and mechanical energy is dissipated as heat.

Q2: Can friction ever be zero?

Practically, absolute zero friction is impossible in macroscopic systems, though it can be made \textremely small in vacuum or frictionless surfaces (like superconductors or air tracks).

Q3: Why does walking on ice become difficult?

Ice offers very low friction compared to normal ground. Without adequate static friction between our shoes and the ice, we cannot push backward against the ground to propel ourselves forward.

Conclusion and Final Tips

Mastering friction is essential for cracking the physics section in RRB NTPC, Group D, and Technician examinations. Ensure you practice both conceptual reasoning questions and numerical calculations involving normal reactions and coefficients of friction. Consistency, revision of formulas, and solving previous years' question papers will guarantee your success in the upcoming railway exams. Keep practicing and stay focused!