Question

A block of base \(10 \mathrm{~cm} \times 10 \mathrm{~cm}\) and height \(15 \mathrm{~cm}\) is kept on an inclined plane. The coefficient of friction between them is \(\sqrt{3}\). The inclination \(\theta\) of this inclined plane from the horizontal plane is gradually increased from \(0^{\circ}\). Then (A) at \(\theta=30^{\circ}\), the block will start sliding down the plane (B) the block will remain at rest on the plane up to certain \(\theta\) and then it will topple (C) at \(\theta=60^{\circ}\), the block will start sliding down the plane and continue to do so at higher angles (D) at \(\theta=60^{\circ}\), the block will start sliding down the plane and on further increasing \(\theta\), it will topple at certain \(\theta\)

Short answer

The block will start sliding down the plane at \(\theta=60^\circ\) and will not topple before sliding since the angle for toppling (approximately 56.31 degrees) is less than 60 degrees. Thus, option (C) is correct.

Step-by-step solution

Identify forces at play

When the block is on the inclined plane, there are three main forces acting on it: gravitational force (weight), normal reaction from the plane, and frictional force. Gravity acts downward, normal reaction perpendicular to the surface of the incline, and frictional force opposes the motion parallel to the surface of the incline.

Calculate normal force and maximum frictional force

The normal force is given by \( N = mg\cos(\theta) \) where \( m \) is the mass of the block, \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of inclination. The maximum static frictional force before the block starts sliding is \( f_{max} = \mu N = \mu mg\cos(\theta) \), where \( \mu \) is the coefficient of friction.

Determine the angle of sliding

The block will start sliding once the component of gravitational force along the incline exceeds the maximum static frictional force. This occurs when \( mg\sin(\theta) > \mu mg\cos(\theta) \), simplifying to \( \tan(\theta) > \mu \). Plug in the given coefficient of friction, \( \mu = \sqrt{3} \), to find the critical angle \( \theta \).

Check for toppling

A block topples when the line of action of the resultant force (gravity in this case) falls outside its base. For a uniform block, this condition is met first along the edge farthest from the incline. The critical angle for toppling \( \theta_{topple} \) can be found by setting the height-to-base ratio equal to the tangent of the inclination angle: \( \tan(\theta_{topple}) = \frac{15 cm}{10 cm} = 1.5 \).

Compare sliding and toppling angles

Compare the critical angle for sliding with the critical angle for toppling to determine which will occur first and at what angles the block will slide or topple. Since \( \mu = \sqrt{3} \), the angle for sliding is \( \theta = 60^\circ \). For toppling, \( \theta_{topple} = \arctan(1.5) \), which is approximately 56.31 degrees. Therefore, the block will reach the sliding condition before the toppling condition since 60 degrees is greater than 56.31 degrees.

Key concepts

Physics JEE Advanced

Physics problems in JEE Advanced often involve a deep understanding of fundamental concepts, and they challenge students with practical applications. In the scenario of a block on an inclined plane, JEE Advanced aspirants must employ principles from mechanics, such as the resolution of forces, friction, and equilibrium conditions. Understanding such concepts not only helps in solving problems efficiently but also lays a strong foundation for future studies or careers in physics and engineering.

For this particular problem, aspirants would need to analyze forces that act on a block resting on an inclined plane and determine the conditions under which the block would start sliding down or topple. Comprehensive knowledge of static and kinetic friction, normal force, gravitational forces and how they relate to the geometry of the block and the inclined plane are crucial to cracking such questions in the examination.

Inclined Plane Problem

Inclined plane problems are crucial in the study of physics, serving as a classic example of how an object's motion can be influenced by multiple forces. When an object is on an inclined plane, several forces act upon it including the gravitational force, normal reaction, and frictional force. The gravitational force attempts to pull the object directly downwards, while the normal force acts perpendicular to the surface, and the frictional force opposes the object's motion along the plane.

To analyze an inclined plane problem, one needs to decompose the gravitational force into two components: one parallel to the plane's surface and the other perpendicular to it. This breakdown allows us to then understand how these forces balance out or lead to motion. In a JEE Advanced problem, determining when a block will begin sliding down the plane or topple becomes an exercise in balancing these forces.

Coefficient of Friction

The coefficient of friction (denoted as \( \mu \) ) is a critical factor in problems involving objects on surfaces. It's a dimensionless scalar value that represents the ratio of the frictional force resisting motion between two bodies and the normal force pressing them together. The coefficient of friction depends on the materials of the surfaces in contact and is crucial in determining whether an object will slide or remain stationary. For example, in the provided inclined plane problem, the coefficient of friction will help students calculate the maximum static friction and thus predict the angle at which the block will start sliding.

In the context of the JEE Advanced syllabus, questions could challenge students to understand the nuance of static versus kinetic friction, and knowing the coefficient of friction for various materials might be essential. The higher this coefficient, the greater the frictional force – a fundamental understanding that underpins many physics problems.

Static Friction

Static friction is the force that keeps an object at rest when it is subjected to other forces and prevents it from starting to move. It's a self-adjusting force that increases to counteract an applied force up until it reaches its maximum value, proportional to the normal force and dependent on the coefficient of friction between the surfaces in contact. Unlike kinetic friction, which comes into play once an object is in motion, static friction varies from zero up to a maximum value that, once exceeded, results in motion.

Understanding the maximum static friction is essential in this type of problem, as it defines the conditions required for an object to start slipping on an inclined plane. For JEE Advanced students, mastering the concept of static friction is pivotal, as it frequently appears in questions that explore the initiation of motion in mechanics.

Toppling Condition

The toppling condition of an object is reached when the line of action of the gravitational force (or resultant force) no longer falls within its base of support, leading to a moment that causes the object to rotate about a pivot point. In the context of a block on an inclined plane, the critical angle at which toppling occurs depends on the dimensions of the block and the angle of inclination of the plane. By applying concepts like center of gravity and moments, students can predict the point of instability.

For JEE aspirants, understanding toppling is not merely about applying formulas. It's about visualizing how the forces would affect an object's stability and applying rotational dynamics principles. In the given problem, for instance, one must compare the angle at which sliding starts with the angle at which toppling occurs to predict the block's behavior accurately. Therefore, a clear conceptual understanding of the toppling condition is essential for solving such physics problems.