Problem 6 - Applied Mathematics

Question

A light elastic string, of unstretched length \(a\) and modulus of elasticity \(W\), is fixed at one end to a point on the ceiling of a room. To the other end of the string. is attached a particle of weight \(W\). A horizontal force \(P\) is applied to the particle and in equilibrium it is found that the string is stretched to three times its natural length. Calculate: (a) the angle the string makes with the horizontal, (b) the value of \(P\) in terms of \(W\). If, instead, \(P\) is not applied horizontally find the least value of \(P\) which in equilibrium will make the string have the same inclination to the horizontal as before. Deduce that the stretch length of the string is \(\frac{3}{2} a\) in this case and find the inclination of \(P\) to the vertical. (U of L)

Step-by-Step Solution

Answer

(a) 60 degrees. (b) P = W\sqrt{3}. Least friction inducing force 2W (stretch=1.5a & inclination to vertical 60 degrees).

Step 1: Analyze given conditions

The string's natural length is given as \(a\), modulus of elasticity as \(W\), and the particle weight as \(W\). When the force \(P\) is applied, the string stretches to \(3a\).

Step 2: Calculate stretch in string

The string stretches from \(a\) to \(3a\), so the extension is \(3a - a = 2a\).

Step 3: Relate modulus of elasticity

Using Hooke's law, the force in the string \(T\) can be calculated as:\[T = \frac{W}{a} \times 2a = 2W.\]

Step 4: Determine horizontal (\(P\)) and vertical components (\(W\))

In equilibrium, the vertical component of \(T\) must balance the weight, and the horizontal component must balance \(P\). The angles formed can be used as:\[T \cos \theta = W,\]\[T \sin \theta = P.\]

Step 5: Calculate angle \(\theta\)

With \(T = 2W\), \(\cos \theta\) becomes:\[\cos \theta = \frac{W}{2W} = \frac{1}{2},\]\[\therefore \theta = 60^\circ.\]

Step 6: Solve for force \(P\)

Using \( T \sin \theta = P \):\[P = 2W \sin 60^\circ = 2W \times \frac{\sqrt{3}}{2} = W\sqrt{3}.\]

Step 7: Consider the inclined force case

Assume \( P' \) is applied at an angle \( \phi \) with the vertical. Equilibrium conditions are same but we need minimum \(P'\).\[P' \cos \phi = W \quad (vertical), \]\[P' \sin \phi = W\sqrt{3} \quad (horizontal).\]

Step 8: Solve for least value of \(P'\)

Using \( \cos \phi = \frac{W}{P'} \):\[P' \cos \phi = W, \]\(\cos^2 \phi + \sin^2 \phi = 1\):\[\frac{W^2}{P'^2} + \frac{3W^2}{P'^2} = 1, \]\[P'^2 = 4W^2 \Rightarrow P' = 2W.\]

Step 9: Confirm string extension

Under similar slope inclination and equilibrium, the stretch is:\[2W/\left(\frac{W}{a}\right) = 3a/2.\]Extension is correct.

Step 10: Inclination of \(P'\) to vertical.

Using the calculated values:\[\cos \phi = \frac{W}{2W}=\frac{1}{2}, \therefore \phi=60^{\circ}.\]

Key Concepts

Hooke's Law

Hooke's Law is a fundamental principle in mechanics that states the force required to extend or compress a spring by some distance is proportional to that distance.
The law can be written as: \[ F = k \times x \]
Here:

• \( F \) is the force applied to the spring.
• \( k \) is the spring constant (also known as the modulus of elasticity).
• \( x \) is the extension or compression of the spring from its natural length.

In the given exercise, the string (which acts like a spring) stretches from its natural length \( a \) to three times its natural length, making the extension \( 2a \).
This leads us to: \[ T = \frac{W}{a} \times 2a = 2W \] This relationship helps us understand how elastic strings interact with different forces. Hooke's Law is essential for calculating the force within a stretched or compressed object.
Understanding this concept can greatly simplify solving similar mechanics problems involving elastic materials.

Equilibrium in Mechanics

In mechanics, equilibrium refers to a state where all the forces acting on a particle (or system) balance each other out.
There are two conditions for equilibrium:

• The sum of all horizontal forces must be zero.
• The sum of all vertical forces must be zero.

For the particle attached to the elastic string, equilibrium is achieved when: \[ T \times \text{cos} \theta = W \] and \[ T \times \text{sin} \theta = P \] Here:

• The vertical component of the tension in the string balances the weight of the particle.
• The horizontal component of the tension in the string balances the applied force \( P \).

In the solution, we found that \( T = 2W \). Using the condition for vertical equilibrium: \[ 2W \times \text{cos} \theta = W \text{cos} \theta = \frac{1}{2}, \therefore \theta = 60^\text{°} \]Once we have the angle, we use the horizontal equilibrium condition to find \( P \):\[ P = 2W \times \text{sin} 60^\text{°} = W \times \text{√3} \]Equilibrium in mechanics ensures that all acting forces create a state of balance, critical for solving various statics problems.

Trigonometric Components

Trigonometry often helps resolve forces into their horizontal and vertical components.
When dealing with angles and forces, we use sine and cosine functions extensively.

• \( \text{Sin} \theta \) gives the ratio of the opposite side to the hypotenuse.
• \( \text{Cos} \theta \) gives the ratio of the adjacent side to the hypotenuse.

In the exercise, we use these trigonometric functions to understand the components of the tension in the string:\[ T \text{cos} \theta = \frac{W}{2W} = \frac{1}{2}, \text{this gives the vertical component.} \]\[ T \text{sin} \theta = W \text{√3}, \text{providing the horizontal component of force.} \]Similarly, for an inclined force \( P' \), we find: \[ \text{P'} \text{cos} \theta = W \] \[ \text{P'} \text{sin} \theta = W \text{√3} \]
Understanding how to work with trigonometric components allows us to break down complex force systems into manageable parts.
Mastering this concept is crucial for solving a variety of mechanics problems, facilitating a deeper understanding of force interactions and equilibrium conditions.