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$\bullet$ A thin, light wire 75.0 $\mathrm{cm}$ long having a circular cross section 0.550 $\mathrm{mm}$ in diameter has a 25.0 $\mathrm{kg}$ weight attached to it, causing it to stretch by 1.10 $\mathrm{mm}$ . (a) What is the stress in this wire? (b) What is the strain of the wire? (c) Find Young's modulus for the material of the wire.

a) $1.03 \times 10^{9} \mathrm{Pa}$b) $1.47 \times 10^{-3}$c) $7.01 \times 10^{11} \mathrm{Pa}$

Physics 101 Mechanics

Chapter 11

Elasticity and Periodic Motion

Equilibrium and Elasticity

Periodic Motion

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Hi there, Troy G here with numerator, we're gonna solve a quick problem today having to do with linked elasticity and the constant of proportionality for different materials there, which we call Young's module lists will show that as why that is a ratio of the amount of stress the material experiences in the form of force per cross sectional area force divided by Harry on tops. We can think of that as a pressure unit and then that's divided by the amount of strain. How much does the does the material stretch? Delta L changing length, divided by its initial length. L So Young's module list. Things that are very that don't stretch a lot underneath a lot of strain. Have a very Hae Young's module list. Things that do stretch a lot, um, underneath comparable stress Have a low Young's module is in our problem. We're given a series of, uh, pieces of information. We're told the length of this wire in question is 75 centimeters. Let's keep everything in meters, kilograms and seconds s high standards so we can work with force and Newton's pressure in Pascal's things like that you can see I have communicated the link. The wires 0.75 meters. The diameter of the wire was is given 0.55 millimeters. Divide that by 1000 to convert it to meters. 5.5 times times 10 to the negative. Forth the mass that is hanging from the wire providing the forces 25 kilograms were good there. No conversion, and we're told the wire stretches 1.1 millimeters. Again. Divide by 1000. That's going to be 1.1 times 10 to the negative third meter. So part A. We want stress. Part B. We want strain, and we're gonna finish this up by putting them together to find it. Young's module is before we do part A to find stress. Let's just figure the force that's occurring on on this wire. Presumably it's a wire that's hanging vertically and the masses hanging from the end of it. The 25 kilograms. Of course, that's gonna provide a way to force the force its own force of gravity downward, and we know that we can calculate that by the mass times the acceleration due to gravity so the force provided is the 25 kilograms multiplied by the 9.8 meter per second acceleration due to gravity on or near earth. Therefore, the force on the wire comes out to be in even 245 Newton's. We will need that. And if we then want to calculate the cross sectional area of the wire, presumably it's a circle and were given the diameter value of that circle. Of course, the area of that circular shape is pi. Times are square. Since we're given D. Let's just make it easy. Of course, radius is half the diameter. Let's go pi times t over two square Buying the cross sectional area when we punched that throw, putting in the 5.5 times 10 to the negative work meters for the diameter. The cross sectional area of the wire comes out to be 2.38 times 10 to the negative. Seventh meters cubed meter square. Rather, pardon me. Okay, so then that allows us to go ahead and solve for what? We're asked for part A. We want the stress on the wire and that's just simply how much force is provided per unit cross sectional area. And so that's just gonna be our 245. Newton's divided by our calculated area 2.38 times 10 to the negative seventh meters square. When we calculate that we get 1.3 times 10 to the ninth and we're talking pressure here. So Newtons per square meter weaken, Just calculate, weaken. Just communicate that as he s a unit for pressure the past cow. So there's our stress then and be what's the strain ratio here? The wire stretched, divided by the wire link that just given that's given to us in the problem. That's the stretch of 1.1 times 10 to the negative third meters, divided by the initial length of the wire 0.75 meters. When we calculate that we get a strain of one point for seven times 10 to the negative third. Okay. And interestingly, strain is a unit list number because we're taking a length measurement and dividing it by a length measurement. And lastly, we want the Youngs module is for this wire, which is the stress divided by the strain we just solve for both of those. So we're gonna put 1.3 times 10 to the ninth. Pascal's are calculated stress divided by our calculated strain. One point for seven times, 10 to the negative third and just simple calculator pushing. Here we come out with a Young's Module list of seven point no one times tend to the 11th Pascal's awesome good elasticity analysis of this one.

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