Problem 20 - Applied Mathematics

Question

A mass of 3 kilogramme is connected by an elastic string of natural length 1 metre and modulus of elasticity \(14.7 \mathrm{~N}\) to a fixed point. A horizontal force equal to the weight of 1 kilogramme acts on the mass maintaining it in equilibrium. Find the inclination of the string to the vertical. If the horizontal force is removed, what is the least force which must act on the particle to ensure that the. string shall be inclined at the same angle as before. Calculate in each case the extension of the string.

Step-by-Step Solution

Answer

The string is inclined at an angle of 18.43 degrees. A force of 31 N is needed to maintain this angle without the horizontal force. The extension of the string in both cases is 1.1 m.

Step 1: Determine the horizontal force

The horizontal force acting on the mass is equal to the weight of a 1 kilogram mass. Therefore, the horizontal force is given by: \[ F_h = 1 \times 9.8 \mathrm{~m/s^2} = 9.8 \mathrm{~N} \]

Step 2: Calculate the vertical force

The vertical force acting on the mass is equal to its weight. Hence: \[ F_v = 3 \times 9.8 \mathrm{~m/s^2} = 29.4 \mathrm{~N} \]

Step 3: Calculate the tension in the string

The tension in the string can be found using the Pythagorean theorem since the forces form a right triangle. \[ T = \sqrt{(F_h^2 + F_v^2)} = \sqrt{(9.8^2 + 29.4^2)} = \sqrt{(960.04)} = 31 \mathrm{~N} \]

Step 4: Determine the extension of the string

Using Hooke's Law, the extension can be calculated as: \[ E = \frac{T - F_v}{k} \]Where, modulus of elasticity, \[ k = 14.7 \mathrm{~N} \] Therefore: \[ E = \frac{31 - 14.7}{14.7} = 1.1 \mathrm{~m} \]

Step 5: Find the inclination angle

Using trigonometry, the inclination angle \( \theta \) can be found as: \[ \tan \theta = \frac{F_h}{F_v} = \frac{9.8}{29.4} = \frac{1}{3} \]Therefore: \[ \theta = \arctan \left( \frac{1}{3} \right) = 18.43^\circ \]

Step 6: Calculate least force needed

If the horizontal force is removed, to keep the same inclination angle, resolve forces using trigonometry: \[ F_{new} \cos \theta = 29.4 \mathrm{~N} \therefore F_{new} = \frac{29.4}{\cos (18.43^\circ)} = 31 \mathrm{~N} \]

Step 7: Calculate new extension

Using the new tension and Hooke's Law: \[ E_{new} = \frac{31 - F_v}{k} = 1.1 \mathrm{~m} \]

Key Concepts

Tension in String

In mechanics, the tension in a string refers to the force exerted along the string when it is pulled tight by forces acting from opposite ends. In the given exercise, a mass of 3 kg is suspended by an elastic string, creating tension in the string. The tension arises due to both horizontal and vertical forces acting on the mass.
To calculate the tension in the string, we must consider both the horizontal force (due to the weight of 1 kg) and the vertical force (the weight of the 3 kg mass). By applying the Pythagorean theorem to these forces, we find the tension:
\[ T = \sqrt{(F_h^2 + F_v^2)} \]
Where:

• \( F_h \) is the horizontal force


• \( F_v \) is the vertical force

Plugging in the values, we get:
\[ T = \sqrt{(9.8^2 + 29.4^2)} = \sqrt{960.04} = 31 \text{ N} \]
Understanding how to calculate the tension in the string is crucial in mechanics problems involving forces in equilibrium.

Hooke's Law

Hooke's Law is a principle of physics that states that the force needed to extend or compress a spring by some distance is proportional to that distance. The law is usually formulated as:
\[ F = k \cdot x \]
Where:

• \( F \) is the force exerted by the spring


• \( k \) is the spring constant (modulus of elasticity)


• \( x \) is the extension or compression of the spring

In our exercise, Hooke's Law helps us find the extension of the elastic string. By rearranging the formula, we can solve for the extension \( E \):
\[ E = \frac{T}{k} \]
Given the tension \( T = 31 \text{ N} \) and \( k = 14.7 \text{ N} \), we calculate:
\[ E = \frac{31}{14.7} = 1.1 \text{ m} \]
Hooke's Law is vital for understanding how materials deform under various forces, making it a fundamental concept in mechanics.

Trigonometry in Physics

Trigonometry is an essential tool in physics, often used to relate angles and dimensions in various problems. In this exercise, trigonometry helps us find the inclination angle of the string with the vertical. We use the tangent function to relate the horizontal and vertical forces:
\[ \tan \theta = \frac{F_h}{F_v} \]
By solving for \( \theta \), we get:
\[ \theta = \arctan \left( \frac{1}{3} \right) = 18.43^\circ \]
This calculation shows how trigonometric functions can provide critical information about the geometry of forces. It is widely used in analyzing forces, projectile motion, and waves.

Modulus of Elasticity

The modulus of elasticity, also known as the elastic modulus, is a measure of a material's ability to withstand deformation under stress. It is denoted by the symbol \( k \) and is defined as the ratio of stress (force per unit area) to strain (proportional deformation):
\[ k = \frac{F}{x} \]
In this problem, the modulus of elasticity of the string is given as 14.7 N. This value is used in conjunction with Hooke's Law to find the extension of the string. It’s crucial to understand this concept when dealing with any problems involving elastic materials, as it helps determine the material's response to applied forces.
By applying the modulus of elasticity, we understand how much the string will extend based on the applied force.

Horizontal and Vertical Forces

Understanding horizontal and vertical forces is key to solving mechanics problems. These forces act at right angles to each other and can be considered independently to simplify complex problems.

In our exercise, the horizontal force is due to a mass of 1 kg:
\[ F_h = 1 \times 9.8 \text{ m/s}^2 = 9.8 \text{ N} \]

The vertical force is due to the weight of the 3 kg mass:
\[ F_v = 3 \times 9.8 \text{ m/s}^2 = 29.4 \text{ N} \]

Understanding how these forces interact helps in calculating the resultant force (tension) and the inclination angle of the string. The said forces form a right triangle, allowing us to use trigonometric relationships to find unknowns. This approach is fundamental in mechanics, enabling the analysis of forces in different directions.