Problem 68 - A Complete Resource Book in Physics for JEE Main

Question

A man of mass \(70 \mathrm{~kg}\) stands on a weighing scale in lift which is moving. Choose the correct statement. (A) Reading of weighing scale is \(700 \mathrm{~N}\) upwards with a uniform speed of \(10 \mathrm{~ms}^{-1}\). (B) Reading of weighing scale is \(700 \mathrm{~N}\) downwards with a uniform acceleration of \(5 \mathrm{~ms}^{-2}\) (C) Reading of weighing scale is \(700 \mathrm{~N}\) upwards with a uniform acceleration of \(5 \mathrm{~ms}^{-2}\) (D) Reading of weighing scale is \(700 \mathrm{~N}\) if the lift mechanism failed and it fall down freely under gravity.

Step-by-Step Solution

Answer

The correct statement is (D): Reading of the weighing scale is \(700 \mathrm{~N}\) if the lift mechanism failed and it falls down freely under gravity.

Step 1: Determine the Applied Force in Each Scenario

We will analyze the given statements based on the motion of the lift and the man inside it. (A) In this case, the lift is moving upwards with a uniform speed of \(10 \mathrm{~ms}^{-1}\). This means there is no acceleration as the velocity is constant. According to Newton's second law, the net force acting on the man is \(F = m \times a\) Since "a" is zero (velocity being constant), the force F is also zero, and the force of gravity acting on the man is equal to his weight, \(mg = 70 \times 9.8 = 686 \mathrm{~N}\). This statement is incorrect as the reading on the weighing scale would be 686 N (approx.). (B) In this case, the lift is moving downwards with a uniform acceleration of \(5\mathrm{~ms}^{-2}\). Using Newton's second law of motion, the force acting on the man is \(F = m \times a\) \(F = 70 \times 5 = 350 \mathrm{~N}\) Since the man is moving downwards, the net force acting on him is the difference between the gravitational force (his weight) and the force due to the lift's acceleration. As the lift is accelerating downwards, the weighing scale would show less force. So, \(F_{net} = mg - F = 686 - 350 = 336 \mathrm{~N}\) This statement is also incorrect. (C) In this case, the lift is moving upwards with a uniform acceleration of \(5\mathrm{~ms}^{-2}\). As the lift is accelerating upwards, the weighing scale would show increased force. So, the net force acting on the man would be the sum of the gravitational force and the force due to the lift's acceleration. \(F_{net} = mg + F = 686 + 350 = 1036 \mathrm{~N}\) This statement is also incorrect. (D) In this case, the lift's mechanism fails, and it falls freely under gravity. In this situation, the man inside will also fall with the same acceleration due to gravity. Therefore, the man will feel weightless, and the weighing scale will effectively show zero force acting on him. This statement is correct. So, the correct statement is: (D) Reading of the weighing scale is \(700 \mathrm{~N}\) if the lift mechanism failed and it falls down freely under gravity.

Key Concepts

Newton's Second Law of Motion

Understanding Newton's Second Law of Motion is crucial when analyzing lift physics problems. This law states that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass. The mathematical representation is given by the formula:

\[ F = m \times a \]
\( F \) is the net force, \( m \) is the object's mass, and \( a \) is the acceleration.

When a person stands on a scale in an elevator, the reading on the scale is a measure of the net force that the scale exerts on them. If the elevator is accelerating, the net force is the sum of gravity and the force due to acceleration. During uniform motion or at rest (where acceleration is zero), the force is just the person's weight. It's important to grasp that movement doesn't necessarily mean acceleration; moving with constant velocity also has no net force other than weight, due to no change in speed or direction.

Free Body Diagram

A Free Body Diagram (FBD) is a simple yet powerful tool used to understand the forces acting upon an object. Drawing an FBD helps isolate and visualize different forces in a physics problem, such as lift problems.

When creating an FBD, first draw a simple outline of the object (in this case, a person standing on a scale in a lift). Then, draw vectors representing all the forces: gravity pulling downwards and any reaction force exerted by the scale upwards. If the lift is accelerating, we must also include that force vector. Direction and magnitude are key; the length of the vector arrow should correlate with the force's strength, and the direction reflects whether it's pushing or pulling on the object. These diagrams do not represent movement but let us focus on the physical forces and their vectors, granting us a clearer path to apply Newton's laws and find solutions to problems.

Weightlessness in Free Fall

The sensation of Weightlessness in Free Fall is an intriguing aspect of lift physics. Weightlessness occurs when there are no contact forces supporting an object, and it falls due to gravity alone -- known as free fall.

When an elevator falls freely as its mechanism fails, the only force acting on people inside is gravity. However, since they are accelerating at the same rate, there is no relative motion between the person and the lift. Therefore, the scale reads zero, as there is no force from the person pressing down on it. This feeling of weightlessness is the same that astronauts experience in orbit. In orbit, astronauts are in continuous free fall towards Earth, but they never reach the ground because their forward motion keeps them curving around the planet. Thus, understanding free fall is not just about elevators and scales, but it's also central to comprehending how astronauts float in space stations!