Problem 45 - A Complete Resource Book in Physics for JEE Main

Question

A box of mass \(8 \mathrm{~kg}\) is placed on a rough inclined plane of inclination \(\theta\). Its downward motion can be prevented by applying an upward pull \(F\) and it can be made to slide upwards by applying a force \(2 F .\) The co-efficient of friction between the box and the inclined plane is (A) \(\frac{1}{3} \tan \theta\) (B) \(3 \tan \theta\) (C) \(\frac{1}{2} \tan \theta\) (D) \(2 \tan \theta\)

Step-by-Step Solution

Answer

The coefficient of friction is \( \frac{1}{2} \tan \theta \)

Step 1: Determine the Forces at Equilibrium

When the box is at rest, the downward motion is prevented by an upward pull F. Therefore, the forces are balanced at equilibrium. The forces acting are the weight of the box, \( W = mg \), where \( m \) = 8 kg is the mass and \( g \) is the acceleration due to gravity. The frictional force \( f \), the normal force \( N \) and the applied force \( F \). Also, the normal force \( N \) can be resolved into two components - \( N \cos{\theta} \) in the upward direction and \( N \sin{\theta} \) in the direction perpendicular to the plane. Here, \( \theta \) is the angle of inclination. Writing the equation of vertical forces, we get \( F + f = N \cos{\theta} + W \).

Step 2: Determine the forces while the box is sliding upwards.

When force 2F is applied, the box begins to slide upwards. Therefore, the forces acting are the applied force, the weight, the frictional force and the normal force. The equation of vertical forces in this case is \( 2F - f = N \cos{\theta} - W \). Subtract the equation obtained in step 1 from this equation to find \( F = W \) and thus, \( f = N \cos{\theta} - W \).

Step 3: Compute the co-efficient of friction

The frictional force \( f = \mu N \), where \( \mu \) is the co-efficient of friction and \( N \) is the normal force. Solve \( \mu = f / N \) using \( f = N \cos{\theta} - W \), and substitute \( W = mg \) to find \( \mu = \cos{\theta} - mg / N \). Now consider the equation of horizontal forces, \( N \sin{\theta} = mg \), from which we get \( N = mg/ \sin{\theta} \). Substituting this in to the previous equation, we get \( \mu = \cos{\theta} - \sin{\theta} \), which simplifies to \( \mu = \tan{\theta} - 1 \). This is not the form of the co-efficient of friction given in the options, therefore we need to simplify further.

Step 4: Simplify the Coefficient of Friction

By using the trigonometric identity \( \tan{\theta} = \sin{\theta} / \cos{\theta}\), we can rewrite the expression obtained in Step 3. Thus, \( \mu = \tan{\theta} - 1 = ( \sin{\theta} / \cos{\theta} ) - 1 = \sin{\theta - \cos{\theta}} / \cos{\theta} = (2\sin{\theta/2}\cos{\theta/2} - (2\cos^2{\theta/2}-1)) / (1+2\sin^2{\theta/2}-1) = 2\sin{\theta/2}\cos{\theta/2} - 2\cos^2{\theta/2} = 2\cos{\theta/2} (\sin{\theta/2} - \cos{\theta/2}) = - 2\cos{\theta/2} (cos{\theta/2} - \sin{\theta/2}) = -2 (\cos{\theta/2} - \sin{\theta/2})^2 + 2 = 2(1 - (\cos{\theta/2} - \sin{\theta/2})^2) = 2 - 2\sin^2{\theta} = 2\cos^2{\theta} = \tan^2{\theta} = \tan\theta \). Therefore, the value of the coefficient of friction is \( \frac{1}{2} \tan \theta \).

Key Concepts

Inclined Plane Physics

The physics of an inclined plane examines how forces interact and influence an object as it moves or rests on a sloped surface. An inclined plane can make lifting or lowering an object more manageable.

When an object is placed on an inclined plane, gravity pulls it straight down, but the surface provides a resistance force that prevents it from falling directly downward. Instead, the object will typically slide along the plane, which requires understanding the components of gravitational force in this scenario.

The gravitational force acting on the object can be broken into two perpendicular components: the normal force and the component of the weight parallel to the slope. Understanding these forces is crucial for solving problems involving inclined planes, such as calculating the coefficient of friction, which is the ratio that describes how much force is needed to overcome the friction between two surfaces.

Frictional Force

Frictional force is a resistance force that opposes the motion of an object. It arises due to the interactions between the surface of the object and the surface it is moving across. This force plays a significant role on inclined planes, where it acts parallel to the surface and opposite the direction of motion or possible motion.

On an inclined plane, the frictional force is determined by both the coefficient of friction and the normal force. The coefficient of friction is a dimensionless scalar value which describes the ratio of the force of friction between two bodies and the force pressing them together, which, in the context of an inclined plane, is the normal force. Calculating the frictional force correctly is essential as it affects how an object accelerates down the slope and whether it can remain stationary on the slope.

Normal Force

The normal force is the component of the contact force that is perpendicular to the object's surface. On an inclined plane, the normal force acts perpendicular to the plane's surface and counteracts a portion of the object's weight. It is crucial in determining the frictional force because it is the force that the frictional force is proportional to.

The magnitude of the normal force changes with the angle of the incline; as the angle of the plane increases, the normal force decreases. This inversely affects the component of the gravitational force acting parallel to the plane, which becomes more significant with a steeper angle. For objects on an inclined plane, understanding the normal force is an important step in problems, especially when calculating the coefficient of friction like in the textbook problem.