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A metal rod that is \(4.00 \mathrm{~m}\) long and \(0.50 \mathrm{~cm}^{2}\) in crosssectional area is found to stretch \(0.20 \mathrm{~cm}\) under a tension of \(5000 \mathrm{~N}\). What is Young's modulus for this metal?

Short Answer

Expert verified
Young's modulus for the given metal is \(2 * 10^{12} \mathrm{N/m^2}\)

Step by step solution

01

Convert units

Convert all units for consistent calculation. Here we need to convert the cross-sectional area and the change in length from cm to m, so A = 0.5cm^2 *10^-4 m^2/cm^2 = 5*10^-6m^2, \u0394L = 0.20cm*10^-2 m/cm = 2*10^-3m.
02

Calculate the Stress

Calculate the stress (\u03C3). Its formula is given by \u03C3 = F/A, substituting the values, we get \u03C3 = 5000N/5*10^-6m^2 = 1*10^9 N/m^2.
03

Calculate the Strain

Calculate the strain (\u03B5). Its formula is given by \u03B5 = \u0394L/L, substituting the values, we get \u03B5 = 2*10^-3m/4m = 5*10^-4.
04

Calculate Young's Modulus

Finally, calculate Young's modulus (Y). It's given by the formula Y = \u03C3/ \u03B5. Substitute the values from step 2 and 3, we get Y = 1*10^9 N/m^2 / 5*10^-4 = 2 * 10^12 N/m^2.

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Key Concepts

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

Stress and Strain
Understanding the relationship between stress and strain is essential in the field of materials science and engineering. When a material is subjected to force, it experiences a deformation; this phenomenon is analyzed with two key concepts: stress and strain.

Stress, denoted as \( \(\sigma\) \) in physics, is the force applied per unit area within materials that arises from externally applied forces, uneven heating, or permanent deformation. It is calculated using the formula \( \(\sigma = \frac{F}{A}\) \), where \(F\) is the force applied perpendicular to the area \(A\). In the exercise, we calculate stress by dividing the tension force (5000 N) by the cross-sectional area of the rod (5 x 10^-6 m^2).

Strain, denoted as \( \(\varepsilon\) \) is a measure of deformation representing the displacement between particles in the material body relative to a reference length. Strain is dimensionless and calculated as \( \(\varepsilon = \frac{\Delta L}{L}\) \), where \( \Delta L\) is the change in length and \(L\) is the original length. The exercise presents a strain calculation as the change in the rod's length divided by its original length.

Together, stress and strain define the rigidity or flexibility of materials. The greater the stress applied for a given strain, the stiffer the material is considered to be.
Mechanical Properties of Materials
The mechanical properties of materials describe their behavior under the influence of external forces. These properties determine how a material reacts when it is pulled, compressed, twisted, or bent. They are of utmost importance in selecting materials for engineering applications, ensuring structures and components can withstand the forces they will encounter without failure.

Some of the primary mechanical properties include
  • Hardness
  • Tensile strength
  • Ductility
  • Toughness
  • Elasticity
  • Plasticity
Each property reflects different aspects of a material’s behavior under stress. Elasticity, for example, refers to a material's ability to return to its original shape after the force is removed. In our exercise, when determining Young's modulus, we are essentially exploring the elastic property of the metal rod.

By understanding these mechanical properties, aspects such as safety, durability, and maintenance can be predicted and controlled. When engineers know the properties of the materials they work with, they can design structures and products that are both efficient and safe.
Elasticity in Solids
The concept of elasticity in solids refers to their ability to return to their original shape after being deformed by an external force. This property is described quantitatively by Young's modulus, a measure of the stiffness of a solid material. It relates stress and strain in the elastic region of the material’s stress-strain curve.

Young's modulus, often represented as \(Y\), is defined by the formula \(Y = \frac{\sigma}{\varepsilon}\) where \(\sigma\) is the stress and \(\varepsilon\) is the strain. In our example, Young's modulus for the metal was found by dividing the stress by strain, resulting in a value that reflects the metal's resistance to elastic deformation.

Materials with high Young's modulus values are stiffer, meaning they deform less under load, which is a desirable characteristic in many construction and industrial materials. On the other hand, materials with low Young's modulus values are more flexible. Elasticity plays a crucial role in designing systems where energy absorption and rebound are important, such as in vehicle suspension systems or shock absorbers.

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Most popular questions from this chapter

(a) In Fig. P11.64 a \(6.00-\mathrm{m}\) -long, uniform beam is hanging from a point \(1.00 \mathrm{~m}\) to the right of its center. The beam weighs \(140 \mathrm{~N}\) and makes an angle of \(30.0^{\circ}\) with the vertical. At the right-hand end of the beam a \(100.0 \mathrm{~N}\) weight is hung; an unknown weight \(w\) hangs at the left end. If the system is in equilibrium, what is \(w ?\) You can ignore the thickness of the beam. (b) If the beam makes, instead, an angle of \(45.0^{\circ}\) with the vertical, what is \(w ?\)

A lead sphere has volume \(6.0 \mathrm{~cm}^{3}\) when it is resting on a lab table, where the pressure applied to the sphere is atmospheric pressure. The sphere is then placed in the fluid of a hydraulic press. What increase in the pressure above atmospheric pressure produces a \(0.50 \%\) decrease in the volume of the sphere?

A \(60.0 \mathrm{~cm}\), uniform, \(50.0 \mathrm{~N}\) shelf is supported horizontally by two vertical wires attached to the sloping ceiling (Fig. E11.10). A very small \(25.0 \mathrm{~N}\) tool is placed on the shelf midway between the points where the wires are attached to it. Find the tension in each wire. Begin by making a free-body diagram of the shelf.

The left-hand end of a light rod of length \(L\) is attached to a vertical wall by a frictionless hinge. An object of mass \(m\) is suspended from the rod at a point a distance \(\alpha L\) from the hinge, where \(0<\alpha \leq 1.00 .\) The rod is held in a horizontal position by a light wire that runs from the right- hand end of the rod to the wall. The wire makes an angle \(\theta\) with the rod. (a) What is the angle \(\beta\) that the net force exerted by the hinge on the rod makes with the horizontal? (b) What is the value of \(\alpha\) for which \(\beta=\theta ?\) (c) What is \(\beta\) when \(\alpha=1.00 ?\)

His body is again leaning back at \(30.0^{\circ}\) to the vertical, but now the height at which the rope is held above-but still parallel to - the ground is varied. The tension in the rope in front of the competitor \(\left(T_{1}\right)\) is measured as a function of the shortest distance between the rope and the ground (the holding height). Tension \(T_{1}\) is found to decrease as the holding height increases. What could explain this observation? As the holding height increases, (a) the moment arm of the rope about his feet decreases due to the angle that his body makes with the vertical; (b) the moment arm of the weight about his feet decreases due to the angle that his body makes with the vertical; (c) a smaller tension in the rope is needed to produce a torque sufficient to balance the torque of the weight about his feet; (d) his center of mass moves down to compensate, so less tension in the rope is required to maintain equilibrium.

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