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\(\mathrm{A} 1.05-\mathrm{m}-\operatorname{lon} \mathrm{g}\) rod of negligible weight is supported at its ends by wires \(A\) and \(B\) of equal length (Fig. P11.91). The cross-sectional area of \(A\) is 2.00 \(\mathrm{mm}^{2}\) and 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 poing 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) Position for equal stress: 0.7 m from A. (b) Position for equal strain: 0.525 m from A.

Step by step solution

01

Understand Stress in Wires

Stress is defined as force per unit area. In this problem, stress in wire A is \( \sigma_A = \frac{F_A}{A} \) and in wire B is \( \sigma_B = \frac{F_B}{B} \), where \( F_A \) and \( F_B \) are the forces in wires A and B, respectively, and \( A \) and \( B \) are their cross-sectional areas. The condition for equal stresses is \( \sigma_A = \sigma_B \).
02

Calculate Equal Stress Condition

Given that areas \( A = 2.00 \, \text{mm}^2 \) and \( B = 4.00 \, \text{mm}^2 \), we set \( \frac{F_A}{2.00} = \frac{F_B}{4.00} \). Solve for \( F_A \) in terms of \( F_B \): \( F_A = 0.5 F_B \).
03

Analyze Forces and Equilibrium

The rod is supported at both ends, with a weight \( w \) added damaging the equilibrium. Let \( x \) be the distance from wire A where the weight is hung. The sum of moments about either wire must be zero for equilibrium. Taking moments about wire A, \( F_B (1.05) = w (x) + F_A (1.05 - x) \).
04

Solve for Position in Equal Stress Case

Substitute \( F_A = 0.5 F_B \) into the equilibrium equation and solve for \( x \). This gives the position \( x \) where the weight yields equal stresses in both wires.
05

Understand Strain in Wires

Strain is the deformation per unit length, given by \( \epsilon = \frac{\Delta L}{L} \). For equal strain in wires A and B, \( \epsilon_A = \epsilon_B \), means \( \frac{F_A}{A E_A} = \frac{F_B}{B E_B} \), where \( E_A \) and \( E_B \) are Young's moduli for wires A and B respectively.
06

Calculate Equal Strain Condition

Substitute given values: \( \frac{F_A}{2.00 \times 1.80 \times 10^{11}} = \frac{F_B}{4.00 \times 1.20 \times 10^{11}} \). Simplify to \( F_A = 1.333 F_B \).
07

Solve for Position in Equal Strain Case

Using the forces relations, plug \( F_A = 1.333 F_B \) into the equilibrium condition and solve for \( x \). This gives the position \( x \) where the weight yields equal strains in both wires.

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Key Concepts

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

Stress and Strain
In mechanics, **stress** and **strain** are fundamental concepts used to describe how materials deform under applied forces. Stress is the internal force per unit area within materials that arises from externally applied forces. It can be expressed as \( \sigma = \frac{F}{A} \), where \( F \) is the applied force and \( A \) is the cross-sectional area.
Strain, on the other hand, measures the deformation of the material, given by \( \epsilon = \frac{\Delta L}{L} \), where \( \Delta L \) is the change in length, and \( L \) is the original length. It is a dimensionless quantity indicating how much a material stretches or compresses.
Understanding the difference between stress and strain is key in solving mechanics problems such as determining how forces affect materials. Especially when you deal with equilibrium, knowing how stress and strain interplay helps you predict how materials behave when forces are applied.
Young's Modulus
**Young's Modulus** is a measure of the stiffness of a material. It indicates how much a material will deform under a given amount of stress. Mathematically, it is defined as the ratio of stress to strain: \( E = \frac{\sigma}{\epsilon} \).
This modulus is crucial in understanding how uniform forces impact materials. Each material has a specific Young's Modulus, which is a constant that represents its elastic properties.
For instance, in our exercise, two wires made of different materials have different Young's Moduli: \( E_A = 1.80 \times 10^{11} \text{ Pa} \) for wire A, and \( E_B = 1.20 \times 10^{11} \text{ Pa} \) for wire B. This difference affects how much each wire stretches when the same stress is applied.
  • This property helps engineers and designers choose the right material for a specific application based on how resistant they need it to be against deformation.
Mastering this concept allows for a deeper understanding of material selection and the structural integrity in mechanical design.
Equilibrium of Forces
The **equilibrium of forces** is a core principle in mechanics indicating that for a body to be in static equilibrium, the sum of forces and the sum of moments (torques) acting on it must equal zero. This means no net force or torque is acting on the object, keeping it in a stable state without acceleration.
In the context of the exercise, equilibrium is used to determine the position \( x \) where a weight attached to a rod produces either equal stress or equal strain in the supporting wires.
To find this position, you apply the conditions for equilibrium:
  • The vertical forces must balance, meaning the total upward force equals the downward weight.
  • The moments about any point must zero out, ensuring rotational stability.
By using the relations for force in each wire and the equilibrium equations, you can solve for the position where the weight can be hung.Ultimately, understanding equilibrium helps solve complex balance problems in mechanics, providing insight into designing stable and secure structures.

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Most popular questions from this chapter

CP A uniform drawbridge must be held at a \(37^{\circ}\) angle above the horizontal to allow ships to pass underneath. The drawbridge weighs \(45,000 \mathrm{N}\) and is 14.0 \(\mathrm{m}\) long. A cable is connected 3.5 \(\mathrm{m}\) from the hinge where the bridge pivots (measured along the bridge \()\) and pulls horizontally on the bridge to hold it in place. (a) What is the tension in the cable? (b) Find the magnitude and direction of the force the hinge exerts on the bridge. (c) If the cable suddenly breaks, what is the magnitude of the angular acceleration of the angular speed of the drawbridge as it breaks? (d) What is the angular speed of the drawbridge as it becomes horizontal?

A diving board 3.00 \(\mathrm{m}\) long is supported at a point 1.00 \(\mathrm{m}\) Trom the end, and a diver weighing 500 \(\mathrm{N}\) stands at the lree end (Fig. E11.11). The diving board is of uniform cross section and weighs 280 \(\mathrm{N} .\) Find (a) the force at the support point and (b) the force at the left-hand end.

BIO Supporting a Broken Leg. A therapist tells a 74 -kg patient with a broken leg that he must have his leg in a cast suspended horizontally. For minimum discomfort, the leg should be supported by a vertical strap attached at the center of mass of the leg-cast system. (Fig. P11.55). In order to comply with these instructions, the patient consults a table of typical mass distributions and finds that both upper legs (thighs) together typically account for 21.5\(\%\) of body weight and the center of mass of each thigh is 18.0 cm from the hip joint. The patient also reads that the two lower legs (including the feet) are 14.0\(\%\) of body weight, with a center of mass 69.0 \(\mathrm{cm}\) from the hip joint. The cast has a mass of \(5.50 \mathrm{kg},\) and its center of mass is 78.0 \(\mathrm{cm}\) from the hip joint. How far from the hip joint should the supporting strap be attached to the cast?

A bookcase weighing 1500 \(\mathrm{N}\) rests on a horizon- tal surface for which the coefficient of static friction is \(\mu_{\mathrm{s}}=0.40 .\) The bookcase is 1.80 \(\mathrm{m}\) tall and 2.00 \(\mathrm{m}\) wide; its center of gravity is at its geo- metrical center. The bookcase rests on four short legs that are each 0.10 \(\mathrm{m}\) from the edge of the bookcase. A person pulls on a rope attached to an upper corner of the bookcase with a force \(\vec{\boldsymbol{F}}\) that makes an angle \(\theta\) with the bookcase (Fig. \(P 11.97 )\) . (a) If \(\theta=90^{\circ},\) so \(\vec{F}\) is horizontal, show that as \(F\) is increased from zero, the bookcase will start to slide before it tips, and calculate the magnitude of \(\vec{\boldsymbol{F}}\) that will start the bookcase sliding. (b) If \(\theta=0^{\circ},\) so \(\vec{\boldsymbol{F}}\) is vertical, show that the bookcase will tip over rather than slide, and calculate the magnitude of \(\vec{\boldsymbol{F}}\) that will cause the bookcase to start to tip. (c) Calculate as a function of \(\theta\) the magnitude of \(\vec{\boldsymbol{F}}\) that will cause the bookcase to start to slide and the magnitude that will cause it to start to tip. What is the smallest value that \(\theta\) can have so that the bookcase will still start to slide before it starts to tip?

A uniform strut of mass \(m\) makes an angle \(\theta\) with the horizontal. It is supported by a frictionless pivot located at one- third its length from its lower left end and a horizontal rope at its upper right end. A cable and package of total weight \(w\) hang from its upper right end. (a) Find the vertical and horizontal components \(V\) and \(H\) of the pivot's force on the strut as well as the tension \(T\) in the rope. (b) If the maximum safe tension in the rope is 700 \(\mathrm{N}\) and the mass of the strut is \(30.0 \mathrm{kg},\) find the maximum safe weight of the cable and package when the strut makes an angle of \(55.0^{\circ}\) with the horizontal. (c) For what angle \(\theta\) can no weight be safely suspended from the right end of the strut?

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