91Ó°ÊÓ

An angler hangs a 4.50-kg fish from a vertical steel wire 1.50 m long and 5.00 \(\times\) 10\(^{-3}\) cm\(^2\) in cross-sectional area. The upper end of the wire is securely fastened to a support. (a) Calculate the amount the wire is stretched by the hanging fish. The angler now applies a varying force \(\overrightarrow{F}\) at the lower end of the wire, pulling it very slowly downward by 0.500 mm from its equilibrium position. For this downward motion, calculate (b) the work done by gravity; (c) the work done by the force \(\overrightarrow{F}\), (d) the work done by the force the wire exerts on the fish; and (e) the change in the elastic potential energy (the potential energy associated with the tensile stress in the wire). Compare the answers in parts (d) and (e). \(\textbf{Torques and Tug-of-War.}\) In a study of the biomechanics of the tug-of-war, a 2.0-m-tall, 80.0-kg competitor in the middle of the line is considered to be a rigid body leaning back at an angle of 30.0\(^\circ\) to the vertical. The competitor is pulling on a rope that is held horizontal a distance of 1.5 m from his feet (as measured along the line of the body). At the moment shown in the figure, the man is stationary and the tension in the rope in front of him is \({T_1} =\) 1160 N. Since there is friction between the rope and his hands, the tension in the rope behind him, \({T_2}\) is not equal to \({T_1}\). His center of mass is halfway between his feet and the top of his head. The coefficient of static friction between his feet and the ground is 0.65.

Step by step solution

01

Determine the force exerted by the hanging fish

First, calculate the force exerted by the fish due to gravity using the equation: \( F = mg \). Here, \( m = 4.50 \) kg is the mass of the fish, and \( g = 9.81 \) m/s² is the acceleration due to gravity.

02

Calculate the stress on the wire

Stress is defined as the force per unit area. To find the stress on the wire, use the equation: \( \text{Stress} = \frac{F}{A} \), where \( A = 5.00 \times 10^{-7} \) m² is the cross-sectional area of the wire.

03

Find the extension of the wire using Young's Modulus

To find the extension, use the formula associated with Young's modulus: \( \text{Strain} = \frac{\Delta L}{L} \) and \( \text{Stress} = Y \times \text{Strain} \), where \( Y \) is the Young's modulus for steel, \( Y = 2.1 \times 10^{11} \) N/m², \( L = 1.50 \) m is the initial length of the wire, and \( \Delta L \) is the extension.

04

Calculate the work done by gravity

Work done by gravity is calculated using: \( W_{gravity} = F_{gravity} \times d \), where \( d = 0.500 \) mm = \( 0.0005 \) m is the displacement caused by the downward motion.

05

Calculate the work done by the force \( \overrightarrow{F} \)

Since the force \( \overrightarrow{F} \) is applied very slowly and directly corresponds to the work done by the force on the wire, the work done is equal and opposite to the work done by gravity plus the change in elastic potential energy.

06

Calculate the work done by the force the wire exerts on the fish

For a non-accelerating system, this work is essentially equal to the work done by the force \( \overrightarrow{F} \) minus the work done against the elastic potential stored in the wire.

07

Calculate the change in elastic potential energy

The change in elastic potential energy of the wire is calculated using the formula: \( \Delta U = \frac{1}{2} \times \text{Stress} \times \text{Strain} \times V \), where \( V \) is the volume of the wire.

08

Compare the answers in parts (d) and (e)

The work done by the force the wire exerts on the fish should be equal to the change in the elastic potential energy with opposite signs, due to the conservation of energy principles in a conservative force field like elastic potential.

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Key Concepts

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

Tensile Stress

Tensile stress is a key concept in understanding how materials react when they are subject to forces. In this exercise, a steel wire is subjected to a force by hanging a 4.50 kg fish.
This force results in tensile stress within the wire. Tensile stress is defined as the response of a material to a force that attempts to stretch it. Mathematically, it is the force applied per unit area of the material:

• Formula: \( \text{Tensile Stress} = \frac{F}{A} \)
• Here, \( F = mg = 4.50 \times 9.81 \, \text{N} \ \) is the force due to the weight of the fish, and \( A = 5.00 \times 10^{-7} \, \text{m}^2 \ \) is the cross-sectional area of the wire.

Tensile stress helps us understand how much force the wire can handle before potential deformation. This concept is crucial in designing bridges, buildings, and even simple structures where material strength and stability matter.

Elastic Potential Energy

Elastic potential energy comes into play when a material like steel is either stretched or compressed. Just like a spring, the steel wire in this problem stores energy as it's being stretched by the fish and further stretched downward by the applied force.

• Change in elastic potential energy can be expressed as: \( \Delta U = \frac{1}{2} \times \text{Stress} \times \text{Strain} \times V \)
• The strain here relates to the extension or elongation experienced by the wire, and V is the volume of the wire.

The stored elastic potential energy will return as kinetic energy when the force is removed, causing the wire to return to its original state. This conversion of energy forms the basics of what keeps structures like suspension bridges stable and resilient, efficiently absorbing and releasing energy.

Mechanics of Materials

The mechanics of materials field lets us dig deeper into how different materials respond to different forces. In this problem, we see the principles like Young's modulus and stress-strain relationships applied. Young's modulus, a property of materials such as the steel in the wire, quantifies the ratio of tensile stress to tensile strain in the elastic deformation phase:

• Formula: \( Y = \frac{\text{Stress}}{\text{Strain}} \)
• In the exercise, Young’s modulus helps us calculate the elongation \( \Delta L \), showing how much a 1.50 m wire stretches under the force of the fish.

Understanding these mechanics ensures that architects, engineers, and designers can predict how materials behave in structural applications. It also aids in safely designing devices and components, where specific forces and loads are predicted and accounted for in advance.