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(I) A sign (mass 1700 $\mathrm{kg}$ ) hangs from the end of a verticalsteel girder with a cross-sectional area of 0.012 $\mathrm{m}^{2} .(a)$ What isthe stress within the girder? (b) What is the strain on thegirder? (c) If the girder is 9.50 $\mathrm{m}$ long, how much is itlengthened? (Ignore the mass of the girder itself.)

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A. $1.39 \times 10^{6}$B. $6.94 \times 10^{-6}$C. $6.59 \times 10^{-5}$

Physics 101 Mechanics

Chapter 12

Static Equilibrium; Elasticity and Fracture

Equilibrium and Elasticity

Cornell University

University of Michigan - Ann Arbor

McMaster University

Lectures

04:12

In physics, potential energy is the energy possessed by a body by virtue of its position relative to others, stresses within itself, electric charge, and other factors. The unit for energy in the International System of Units is the joule (J). One joule can be defined as the work required to produce one newton of force, or one newton times one metre. Potential energy is the energy of an object. It is the energy by virtue of an object's position relative to other objects. Potential energy is associated with restoring forces such as a spring or the force of gravity. The action of stretching the spring or lifting the mass is performed by a force which works against the force field of the potential. The potential energy of an object is the energy it possesses due to its position relative to other objects. It is said to be stored in the field. For example, a book lying on a table has a large amount of potential energy (it is said to be at a high potential energy) relative to the ground, which has a much lower potential energy. The book will gain potential energy if it is lifted off the table and held above the ground. The same book has less potential energy when on the ground than it did while on the table. If the book is dropped from a height, it gains kinetic energy, but loses a larger amount of potential energy, as it is now at a lower potential energy than before it was dropped.

04:17

04:49

(I) A sign (mass 1700 kg) …

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(I) A sign (mass 2100 $\ma…

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One end of a heavy beam of…

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(II) A 20.0-m-long uniform…

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\bullet A uniform beam 4.0…

so for party to find the stress often denoted as lower case sigma. This would be equaling the force over the area. In this case, this would be the mass times acceleration due to gravity divided by the area. This is equaling 1700 kilograms multiplied by 9.8 meters per second squared. This would be divided by 0.12 meter squared and we find that the stress sigma would be equaling 1.39 times 10 to the sixth Newtons per square meter for part B to find this strain, this is often denoted as lower case sick lower case Absalon. Rather, this would be equaling these stress or sigma over the Youngs modulates the Youngs module ist is often denoted as e or more commonly we can say why. So the stress over the Youngs modules Why would be equaling 1.388 12 round at the very end times 10 to the sixth Newtons per square meter. This would be divided by the Youngs module is of 200 times 10 to the Ninth Newtons per square meter. And so the strain which we find to be unit lis, would be equaling 6.94 times 10 to the negative sixth. So this would be our answer for part B, our stress for part a and then for part C to find our change in length. This would simply be equaling the strain Absalon times the original length. And so this is equaling 6.94 times 10 to the negative. Sixth multiplied by the original length of nine point 50 meters. And this is equaling 6.59 times 10 to the negative fifth meters. This would be our answer for the change in length for part C. That is the end of the solution. Thank you for

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