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$\bullet$ $\bullet$ Two circular rods, one steel and the other copper, are joined end to end. Each rod is 0.750 $\mathrm{m}$ long and 1.50 $\mathrm{cm}$ in diameter. The combination is subjected to a tensile force with magnitude 4000 N. For each rod, what are (a) the strain and (b) the elongation?

a) $2.1 \times 10^{-4}$b) $1.6 \times 10^{-4} \mathrm{m}$

Physics 101 Mechanics

Chapter 11

Elasticity and Periodic Motion

Equilibrium and Elasticity

Periodic Motion

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

Hope College

University of Winnipeg

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.

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In physics, an oscillation is the repetitive variation, typically in time, of some measure about a central value or between two or more different states. The oscillation may be periodic or aperiodic.

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Hi there, Troy G here for numerator solving a problem having to do that length, elasticity and our constant of proportionality here Young's Module. It's for a combination situation of copper and steel. Young's module ist relates, the stress of material undergoes and the D formation. The strain that it experiences stress. We think of force per unit of cross sectional area and on dhe strain is the amount that the length changes okay in proportion to the original length of the objects. So we're told here. Both materials have a length of 0.75 meters were also given in the problem that the diameter of the wire system is 150 centimetres. Let's communicate that in meters 1.5 times 10 to the negative second meters divided by ah 100. Of course, the tensile forces 4000 Newtons on this combination metal wire and not giving right the problem but easily findable in your textbook or online. Young's module is for steel is two times 10 to the 11th Pascal's and for copper is 1.1 times 10 to the 11 Pascal's. This fits with our experience, Um, where in we are familiar with the fact that it is easier to stretch and deform copper than it is steel. Hence, the lower Young's module is. So let's go here were asked to solve for the strain on each wire in this situation. So let's use our relationship for Young's module is to figure that out. If we take, do some quick algebra and move the strain, multiply both sides by the strain and divide both sides by Young's module lists. We can come up with an expression where we can say the strain. In other words, the change in length over length is equal to one. We write it out like this one over. Young's module ist multiplied by force, divided by area, so hopefully you can see the algebra happening. They're off course. We're given the diameter of the wire. The cross sectional area that we're concerned with is going to be pie times. Half of our diameter squared. All right, so that's going to go in there for a and again, if we can put it all into one expression, I think it's the best way to do both calculations. We can say this strain is then one over. Young's module ist multiplied by the force. Let's bring a factor of four from the bottom of our area. Equation there. Let's bring that up to the top. And that's gonna leave pi times the diameter squared there at the bottom. So that's gonna be the same, um, expression for both. So again, just for part a. Here, let's go for the steel. Strained first. OK, again, what we're looking for. Change in length, divided by length, is going to be won over. The module is for steel, which is two times 10 to the 11th times, four times the 4000 Newton tense. I'll force divided by pi times the diameter 1.5 times, 10 to the negative second and again, that's meters. All omit units that's gonna be squared if we just calculate that all out the strain on the steel underneath this force is gonna be 1.1 times 10 to the negative. Forth again, it feels wrong. But these are unit list numbers here. That's okay because it's a length divided by a length. So there's the strain for the steal. The strain for the copper. Okay, let's see if I could save some time here. It's gonna be the exact same thing as the math or the steel, with the only difference being we're gonna have a different, um, module is, of course, so it's going to be won over the module list of copper, which is the 1.1 times 10 to the 11th. And if you would, we're gonna multiply by this same factor here, so that's gonna be the same. And when we calculate that out, the strain on the copper wire no surprise is gonna be bigger than the steel. That's gonna be 2.1 times 10 to the negative forth. So there's part a strain on the steel strain on the copper. Okay, I could clear some space here on the white board. Number two part B simply then asks us. We know we've found the strain. Can we, um, can recalculate the actual eat elongation? That is to say, doubt L Well, if my strain is delta l over l strain. That's what I just solved for. Bye. Isolate Delta L. I can say that my elongation delta l is just gonna be equal to length. The original length, times the strain that I just calculated So again first for steel. Okay, The elongation of the steel wire is gonna be the original length this was given to us in the problem. That's the 0.75 meters times the strain that we calculated in part a 1.1 times 10 to the negative fourth. Okay, we work that through the actual stretch on the steel part of this wire is going to be 8.3 times 10 to the negative fifth meters and again for the copper. It's elongation will be the original length 0.75 meters. Both played by its strain a little bit less than twice the steals 2.1 times 10 to the fourth from part, eh? Of course, the steel's gonna The copper's gonna stretch further. It's elongation is gonna be 1.6 times 10 to the negative fourth meters. So actually elongation for steel elongation for the copper. We worked all of that out by knowing the force and knowing Young's module iss

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