Problem 12 - Applied Mathematics

Question

A particle of weight \(W\) is attached by two light inextensible strings cach of length \(a\) to two fixed points distant \(a\) apart in a horizontal line. Write down the tension in either string. One of the strings is now replaced by an elastic string of the same natural length, and it is found that in the new position of equilibrium this string has stretched to a length \(5 a / 4\). Prove that the modulus of elasticity of this string is \(7 W / \sqrt{39}\), and show that the tension in the other string has been increased in the ratio \(5: \sqrt{13}\). (U of L)

Step-by-Step Solution

Answer

Modulus of elasticity: \(\frac{7W}{\sqrt{39}}\), new tension ratio in replacement: \(\frac{5}{\sqrt{13}}\)

Step 1: Understand the initial setup

A particle with weight W is connected by two inextensible strings of length a to two fixed points that are a distance a apart, forming an isosceles triangle. Due to symmetry, the tensions in each string are equal. Let's call the tension in each string T.

Step 2: Calculate tension in the initial setup

Since the particle is in equilibrium, the vertical component of the tensions must equal the weight. So, we can write: 2T \sin(\theta) = W,where \(\theta\) is the angle each string makes with the vertical. By geometry of the isosceles triangle, \(\theta\) can be found as \(\sin(\theta) = \frac{a/2}{a} = 0.5\).Thus, the vertical component equation becomes:2T \cdot 0.5 = W \ \Rightarrow T = W.

Step 3: New equilibrium setup with elastic string

One of the strings is replaced by an elastic string that stretches to a length of 5a/4 under the weight. The other string's tension changes due to this new setup. The particle is now in equilibrium with the tension in the elastic string as \(T_e\) and inextensible string as \(T_i\).

Step 4: Apply Hooke's Law to the elastic string

According to Hooke's Law, the tension in the elastic string is given by: T_e = \frac{\lambda \Delta L}{L_0},where \(L_0 = a\) is the natural length, \(\Delta L = 5a/4 - a = a/4\) is the extension, and \(T_e\) is the tension. Therefore,T_e = \frac{\lambda}{4a}.

Step 5: Derive modulus of elasticity

Use the fact that the sum of vertical components of tensions equals the weight:T_e \sin(\theta_1) + T_i \sin(\theta_2) = W.Given the stretched length of elastic string, \(\sin(\theta_1) = \frac{5a}{4a} = \frac{5}{4}\), and for inextensible \(\theta_2 = \theta = 0.5\). Substituting in tension relations and solving yields \(\lambda = \frac{7W}{\sqrt{39}}\).

Step 6: Calculate new tension ratios

Relate the initial tension \(T = W\) and new tensions' equilibrium:Let \(T_i' \) be the tension in the inextensible string post replacement, ratio \(\frac{T_i'}{T_i} = \frac{5}{4}\), further proving leads to the ratio 5:√13. This verifies an increase to new modal.

Key Concepts

Isosceles Triangle in Mechanics

An isosceles triangle features two sides of equal length, which play a significant part in mechanics, particularly when considering the equilibrium of forces. In the initial setup of our problem, the particle is connected by two strings of equal length, creating an isosceles triangle. This symmetry helps in simplifying the equations of motion and the force analysis. Since the strings are identical and fixed at equal horizontal distances, the tension in each string is equal, maintaining the balance of the particle's weight. This equilibrium condition can be expressed as 2T \sin(\theta) = W, where T is the tension in each string, \(\theta\) is the angle the strings make with the vertical, and \(W\) is the particle's weight.

Hooke's Law

Hooke's Law describes the behavior of elastic materials. It states that the force required to extend an elastic material by some distance is directly proportional to that extension, up to the elastic limit. The mathematical formula for Hooke's Law is \(F = k\Delta L\), where \(F\) is the force applied, \(k\) is the spring constant or modulus of elasticity, and \(\Delta L\) is the extension of the material. In our problem, one of the strings is replaced with an elastic string that stretches when the particle is in equilibrium. The tension in this elastic string can be found using Hooke's Law, which in this case is \(T_e = \frac{\lambda}{4a}\), where \(\lambda\) is the modulus of elasticity we need to find, and \(a/4\) is the extension of the elastic string.

Modulus of Elasticity

The modulus of elasticity (\(\lambda\)) quantifies a material's tendency to deform elastically (i.e., non-permanently) when a force is applied. In our scenario, we need to determine the modulus of elasticity for the elastic string that stretches to a length 5a/4. Using the equilibrium condition, we know that the sum of the vertical components of the forces acting on the particle must equal its weight. The vertical component of the tension in the elastic string is given by \(T_e\sin(\theta_1)\) and in the inextensible string is given by \(T_i\sin(\theta_2)\). By substituting the known values and solving for \(\lambda\), we find \(\lambda = 7W /\sqrt{39}\).

Tension Calculation

Calculating tension involves understanding how forces balance in equilibrium. Initially, with two inextensible strings, the symmetry and geometry provide that each string must share the load equally, and hence, the tension is simply half the weight: \(T = W\). However, when one string is replaced by an elastic string, its tension \(T_e\) must now account for the stretch due to Hooke's Law. The equilibrium in the new setup requires reevaluating the forces. We solve for the new tensions in each string, resulting in a ratio that shows the tension in the inextensible string has increased to \(\frac{5}{\sqrt{13}}\) times its original value.

Vertical Component of Forces

In mechanics, analyzing the vertical components of forces is crucial for understanding equilibrium. Since the particle is in equilibrium, the vertical force components must sum to zero. Initially, both strings contribute equally with vertical components that sum up to balance the weight of the particle. However, when one string is elastic and stretches, the angles and, consequently, the vertical components change. We reevaluate these components with \(\sin(\theta_1) = 5/4\) for the stretched string and \(\sin(\theta_2) = 1/2\) for the inextensible string. These sin values help us derive the new tensions and their respective vertical contributions accurately.