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A \(1.05-\mathrm{m}\) -long rod of negligible weight is supported at its ends by wires \(A\) and \(B\) of equal length (Fig. 11.62\() .\) The crosssectional arca of \(A\) is 2.00 \(\mathrm{mm}^{2}\) und that of \(B\) is 4.00 \(\mathrm{mm}^{2} .\) Young's modulus for wire \(A\) is \(1.80 \times 10^{11} \mathrm{Pa}\) ; that for \(B\) is \(1.20 \times 10^{11} \mathrm{Pa}\) At what point along the rod should a weight \(w\) be suspended to produce (a) equal stresses in \(A\) and \(B,\) and \((b)\) equal strains in \(A\) and \(B ?\)

Short Answer

Expert verified
(a) Suspend the weight 0.35 m from wire A for equal stress; (b) suspend it 0.26 m from wire A for equal strain.

Step by step solution

01

Understanding Stress Equality

To achieve equal stresses in wires A and B, the forces on them must equal the ratio of their cross-sectional areas. Stress is defined as force per unit area, so if \(F_A\) is the force on wire A, \(F_B\) is the force on wire B, \(A_A\) is the area of wire A, and \(A_B\) is the area of wire B, then we have \(\frac{F_A}{A_A} = \frac{F_B}{A_B}\).
02

Calculating Forces for Equal Stress

The area ratio is \(\frac{A_A}{A_B} = \frac{2.00 \, \text{mm}^2}{4.00 \, \text{mm}^2} = \frac{1}{2}\). Therefore, \(F_A = \frac{1}{2}F_B\). If the rod length is \(L = 1.05\) m and the weight \(w\) is suspended at a distance \(x\) from the end attached to wire A, then from equilibrium, \(w = F_A + F_B\) and \(F_A \cdot (1-x) = F_B \cdot x\). Substitute \(F_A = \frac{1}{2}F_B\) into the equilibrium equations to solve for \(x\).
03

Solving for Position of Weight with Equal Stress

Using the substitution \(F_A = \frac{1}{2}F_B\) in \(F_A \cdot (1-x) = F_B \cdot x\), we have \(\frac{1}{2}F_B \cdot (1-x) = F_B \cdot x\). Simplify to get \(\frac{1}{2}(1-x) = x\). Solving \(\frac{1}{2} - \frac{1}{2}x = x\) gives \(\frac{1}{2} = \frac{3}{2}x\). Therefore, \(x = \frac{1}{3}\). So, the weight should be suspended at a distance of \(1.05 \text{ m} \times \frac{1}{3} = 0.35\) m from the end attached to wire A.
04

Understanding Strain Equality

Strain is stress divided by Young's modulus (E), hence equal strain means \(\frac{F_A}{A_A E_A} = \frac{F_B}{A_B E_B}\). Given that Young’s Modulus for wire A is \(1.80 \times 10^{11} \, \text{Pa}\) and for B is \(1.20 \times 10^{11} \, \text{Pa}\), the ratio becomes \(\frac{F_A}{F_B} = \frac{A_A E_B}{A_B E_A}\).
05

Calculating Forces for Equal Strain

The ratio of \(\frac{A_A E_B}{A_B E_A} = \frac{2.00 \, \text{mm}^2 \times 1.20 \times 10^{11} \, \text{Pa}}{4.00 \, \text{mm}^2 \times 1.80 \times 10^{11} \, \text{Pa}} = \frac{2.40}{7.20} = \frac{1}{3}\). Therefore, \(F_A = \frac{1}{3}F_B\). Use this in the equilibrium equations \(w = F_A + F_B\) and \(F_A \cdot (1-x) = F_B \cdot x\) to solve for the new \(x\).
06

Solving for Position of Weight with Equal Strain

With \(F_A = \frac{1}{3}F_B\) in \(F_A \cdot (1-x) = F_B \cdot x\), we have \(\frac{1}{3}F_B \cdot (1-x) = F_B \cdot x\). Simplify this equation to \(\frac{1}{3}(1-x) = x\). Solving \(\frac{1}{3} - \frac{1}{3}x = x\) gives \(\frac{1}{3} = \frac{4}{3}x\). Therefore, \(x = \frac{1}{4}\). So, the weight should be suspended at a distance of \(1.05 \text{ m} \times \frac{1}{4} = 0.2625\) m from the end attached to wire A.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stress and Strain
Stress and strain are foundational concepts in the study of mechanics of materials. Stress is the internal force experienced by an object per unit area, expressed mathematically as \( \sigma = \frac{F}{A} \), where \( F \) is the force applied and \( A \) is the cross-sectional area. Strain, on the other hand, is a measure of deformation representing the elongation or compression of a material relative to its original length. It is defined as \( \varepsilon = \frac{\Delta L}{L} \), where \( \Delta L \) is the change in length and \( L \) is the original length.

In problems involving rods or wires, achieving equal stress in different sections often involves ensuring that the forces applied to them are distributed according to their areas. If you want equal stress in two wires using a single weight, the weight should be hung such that the product of the force and the cross-section area is equal for both wires. This often requires leveraging the equilibrium position to ensure both sides hold proportionately to their strengths.

These principles extend to strain, where equal strain would mean that, despite having different stiffer materials, both wires should stretch proportionally to their original dimensions when forces are applied. Identifying a balance point is essential to solve practical problems involving complex structures or systems.
Young's Modulus
Young's Modulus, often denoted by \( E \), is a material property that describes its elasticity, specifically the stiffness of the material. It is the ratio of stress \( \sigma \) over strain \( \varepsilon \), expressed as \( E = \frac{\sigma}{\varepsilon} \).

This modulus offers insight into how much a material will deform under a given load. A high Young's modulus indicates that a material is stiff, meaning it resists deformation under stress. Conversely, a low Young's modulus suggests that the material will deform more easily.

In the provided exercise, Young's modulus plays a crucial role in determining the position of the weight so that equal strain is achieved in both wires. The distinct Young's moduli for wires A and B (\( 1.80 \times 10^{11} \text{ Pa} \) and \( 1.20 \times 10^{11} \text{ Pa} \) respectively) directly influence how each wire behaves under the applied force, requiring different calculations when addressing strain versus stress. Understanding Young's modulus can significantly aid in designing systems that rely on material deformation characteristics, ensuring that materials operate within their elastic limits.
Equilibrium of Forces
The concept of equilibrium is fundamental in mechanics. For an object or system to be in equilibrium, the sum of forces and moments acting upon it must equal zero. This ensures no resultant force or torque, thereby maintaining a stable system.

In the problem of suspending a weight on a rod supported by two wires, equilibrium principles are used to find the optimal point along the rod for hanging the weight. Both vertical and rotational equilibria come into play:
  • **Vertical Equilibrium:** Ensures the total upward force from the tension in wires equals the downward weight force; mathematically expressed as \( w = F_A + F_B \).
  • **Rotational Equilibrium:** Ensures there's no rotation of the rod around any point, determined by the moment condition \( F_A \times (1-x) = F_B \times x \).

By applying these equilibrium equations, it becomes possible to precisely calculate how a load should be shared between supports, and predict how it influences the system layout. This balance is critical not just theoretically, but in practical engineering applications to prevent failure and optimize structural design.

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Most popular questions from this chapter

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