Problem 3 - Applied Mathematics

Question

A body of mass \(2.5\) kilogramme is attached to the end \(B\) of a light elastic string AB of natural length 2 metre and modulus \(5 g\) newton. The mass is suspended vertically in equitibrium by the string whose other end \(A\) is attached to a fixed point. (i) Find the depth below A of B when the body is in equilibrium. (ii) Find the distance through which the body must be pulled down vertically from its equilibrium position so that it will just reach A after release.

Step-by-Step Solution

Answer

i) The depth is 4.9 meters. ii) The distance is 0.9 meters.

Step 1: Understanding the Problem

A body of mass 2.5 kg is suspended by a light elastic string. The string's natural length is 2 meters and its modulus is 5g Newtons. We need to find the depth below point A when in equilibrium and the distance through which the body should be pulled down to make it reach point A after release.

Step 1: Identify forces in equilibrium

In equilibrium, the forces acting on the body are the tension in the string and the gravitational force. Tension in the string must balance the weight of the body.

Step 2: Use Hooke’s Law

Hooke's Law states that the tension in the spring is given by where k is the modulus and x is the extension beyond the natural length.

Step 3: Set up equilibrium equation

The tension in the string (due to extension by x) equals the weight of the mass: T = mg Using Hooke’s Law: T = (5g/2) x / 2 Equating these, we can solve for x.

Step 4: Solve for extension x

Using Hooke's Law and the equilibrium equation we get: mg = (5g * x) / 2 Simplifying: 2.5 * 9.8 = 5 * x / 2 Solving for x: x = (2.5 * 9.8 * 2) / 5 x = 9.8 m.

Step 5: Calculate the depth below A in equilibrium

The total length of the string = natural length + extension = 2 m + x = 2 m + 9.8 m = 4.9 m.

Step 7: Part (ii): Calculate the distance to pull down

To make the body reach point A, it has to travel a further distance equal to the total length of the string minus the equilibrium position: distance = total length - 4.9 m.

Step 6: Answer part (ii)

Total length of the string = 4.9 m, Distance = 4 - 4.9 m = 0.9 m.

Key Concepts

Hooke's Law

Hooke's Law is a foundational principle in understanding elastic strings in equilibrium. It states that the force needed to extend or compress a spring is directly proportional to the distance it is stretched or compressed.
In mathematical terms, Hooke's Law is expressed as:
\[ F = k \times x \]where:

• F is the force applied
• k is the spring constant (modulus)
• x is the extension or compression of the spring

In the given problem, the modulus of the string is given by 5g Newton, and we use this constant to find out how much the string stretches when a mass is hung from it.

Gravitational Force

Gravitational Force is the force exerted by the Earth that pulls objects towards its center.
This force is calculated using the formula:
\[ F_g = m \times g \]where:

• m is the mass of the object
• g is the acceleration due to gravity (approximately 9.8 m/s²)

In our exercise, the body has a mass of 2.5 kg, hence the gravitational force acting on it would be:
\[ F_g = 2.5 \times 9.8 = 24.5 \text{ N} \]This force must be balanced by the tension in the string for the body to remain in equilibrium.

Equilibrium Equation

The equilibrium equation is critical in solving problems involving elastic strings in equilibrium.
In equilibrium, the upward tension in the string is equal to the downward gravitational force.
This is given by:
\[ T = F_g \]Using Hooke's Law within this equilibrium state, we can set:
\[ k \times x = m \times g \]In our problem, substituting the known values gives:
\[ \frac{5g \times x}{2} = 2.5 \times 9.8 \]Solving for x (extension), we find:
\[ x = \frac{2.5 \times 9.8 \times 2}{5} = 9.8 \text{ m} \]

Tension in Strings

The tension in an elastic string is the force it exerts to return to its natural length.
When a mass is suspended at the end of the string, this tension balances the gravitational pull.
Using Hooke's Law, the tension is given by:
\[ T = k \times x \]where k is the modulus and x is the extension.
For equilibrium, we equate the tension to the gravitational force:
\[ k \times x = m \times g \]For our problem, with the modulus as 5g Newtons and the extension found previously, we solve:
\[ T = \frac{5g \times 9.8m}{2} \=24.5 N \]Thus, the tension in the string at equilibrium is 24.5 Newtons.

Extension of Elastic Strings

The extension of an elastic string happens when a force stretches it beyond its natural length.
This extension is a key factor in determining the tension in the string and is calculated using Hooke's Law.
In the given problem, we already derived that the extension x is 9.8 meters.
The total stretched length of the string is then the natural length plus the extension:
\[ L = L_0 + x \]where:

• L_0 = natural length of the string (2 meters)
• x = extension (9.8 meters)

So, the total length is:
\[ L = 2 + 9.8 = 4.9 \text{ meters} \]Finally, to make the body just reach point A after being pulled down from its equilibrium position, you'd have to pull it through the total length minus the equilibrium position: