Problem 5 - Applied Mathematics

Question

Two particles of mass \(3 \mathrm{~kg}\) and \(5 \mathrm{~kg}\) are connected by a light inextensible string passing over a smooth pulley which is fixed to the ceiling of a lift. Find the tension in the string when the system is moving freely and the lift has a downward acceleration \(g \mathrm{~ms}^{-2}\).

Step-by-Step Solution

Answer

The tension in the string is approximately 39.2 N.

Step 1 - Identify Forces Acting on Each Particle

First, identify the forces acting on both particles. For particle with mass 3 kg, gravitational force (weight) is \( F_{gravity, 3} = 3 g \). For particle with mass 5 kg, gravitational force (weight) is \( F_{gravity, 5} = 5 g \).

Step 2 - Consider the Lift's Acceleration

Since the lift is accelerating downward with acceleration \( g \), we need to consider the pseudo force acting on the system due to this acceleration. The system experiences an additional downward force equal to the weight of each particle.

Step 3 - Set Up Equations Using Newton's Second Law

For both particles, apply Newton's second law. The pseudo force on each particle is the mass times the lift's acceleration. So the total effective force acts as if gravity acting is doubled: \( F_{effective, 3} = 3g + 3g = 6g \), \( F_{effective, 5} = 5g + 5g = 10g \).

Step 4 - Write the Equation for Tension

For the system in equilibrium, the tension \( T \) in the string will be equal to the average effective force acting on the particles. This can be computed using the formula: \( T = \frac{m_1 g + m_2 g}{2} = \frac{(3g + 5g)}{2} = \frac{8g}{2} = 4g \).

Step 5 - Final Tension Value

Given that \( g \) is approximately equal to \( 9.8 \ \text{m/s}^{2} \), the tension in the string \( T \) is: \( T = 4g \approx 4 \times 9.8 = 39.2 \ \text{N} \).

Key Concepts

Newton's second law

Newton's second law is a core concept in understanding how forces affect motion. The law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This can be expressed with the equation: \(F = ma\).

Let's break it down:

• Force (F): The push or pull applied to an object.
• Mass (m): The amount of matter in the object.
• Acceleration (a): The rate at which the object's velocity changes.

In our exercise, we apply Newton's second law to each particle separately. For the 3 kg particle, the gravitational force is \(3g\), where \(g\) is the acceleration due to gravity. For the 5 kg particle, the gravitational force is \(5g\). Since the lift accelerates downward with \( g \text{ms}^{-2} \), we must also consider this pseudo force acting on our system.

Gravitational force

Gravitational force is the force that attracts any two objects with mass. It can be calculated with the equation: \(F = mg\), where \(m\) is the mass and \(g\) is the acceleration due to gravity.

In our problem:

• The gravitational force on the 3 kg particle: \(3g\)
• The gravitational force on the 5 kg particle: \(5g\)

Because the pulley system is accelerating downward with the lift, these forces interact in a way that must be carefully considered. The downward acceleration means the lift adds an additional force equal to the gravitational force acting downwards on each particle, effectively doubling the apparent gravitational force acting on them.

Pseudo force

A pseudo force, also known as a fictitious force, arises when we observe motion from an accelerating reference frame. In this exercise, since the lift is accelerating downward, we use the lift as our reference frame.

We add a pseudo force to account for this acceleration. The pseudo force is calculated as:

\[ F_{\text{pseudo}} = ma \text{(where a is the acceleration of the reference frame)} \text{ ... in our case, a = g} \]

• For the 3 kg particle, the pseudo force: \(3g\)
• For the 5 kg particle, the pseudo force: \(5g\)

Therefore, the total effective force on each particle is doubled:

• Total effective force on 3 kg particle: \(6g\)
• Total effective force on 5 kg particle: \(10g\)

Summing these forces and dividing by 2, we find the tension in the string:

\[ T \text{(tension)} = \frac{(3g + 5g)}{2} = 4g \approx 39.2 \text{N} \]
By understanding pseudo forces, we can navigate such non-inertial frames of reference efficiently.