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19. Two samples of bone, A and B, are modeled as cylinders fixed in place at one end_ Equal forces are applied to the other ends of the bone samples as shown: Sampl...

Question

19. Two samples of bone, A and B, are modeled as cylinders fixed in place at one end_ Equal forces are applied to the other ends of the bone samples as shown: Sample A has twice the radius of sample B, but they have the same length. Ifthe two samples stretch the same distance when the force is applied, what is true about the Young's modulus E of the two samples? EA EB EA 2EB EA = 4E8 EA = EB/2 EA = EB/46 F

19. Two samples of bone, A and B, are modeled as cylinders fixed in place at one end_ Equal forces are applied to the other ends of the bone samples as shown: Sample A has twice the radius of sample B, but they have the same length. Ifthe two samples stretch the same distance when the force is applied, what is true about the Young's modulus E of the two samples? EA EB EA 2EB EA = 4E8 EA = EB/2 EA = EB/4 6 F



Answers

(II) If a compressive force of $3.6 \times 10^{4} \mathrm{N}$ is exerted on the end of a 22 -cm-long bone of cross-sectional area $3.6 \mathrm{cm}^{2},(a)$ will the bone break, and $(b)$ if not, by how much does it shorten?

The equation for stress is force upon area. Yeah. Um, here the force applied on the bone is F an area area of cross section area of section is a So sigma is f upon it is by upon four the outer square minus d inner square. So we have Ah, sigma is and that is a stress. We have 3.8. Well, two forces 1200 newton upon by by four um and they diameter of the outer diameter is 25 in a diameter is to be calculated. So that is, um, by by four. 25 square minus in our demeter square is equal to 315.79 So 25 square minus the inner diameter is equal to four, 102.7 six. So inner diameter square is 2. 22.9243 So inner diameter is 14.93 millimeters. Therefore, the inner diameter of the bonus 14.93 and

First the stress by dividing force by area on. Given those two values, we find stress to be 9.167 times 10 to the past seven new drink for meager scraps. Now to see if the board will break or not. We compared this value off stress with the compressive strength off the bone, which is 1.7 times 10 to the part eight newtons per meter squared. As you see, if you compare these two values, the stress that we found is less than the compressive strength off the bone. Therefore, the bone will not break Now the changing land can be calculated by using a question. 12 months. The question right here on dhe for the elastic modelers we use the Let's plastic model is off the born with this phone, Toby 15 times 10 to the power nine Newton per meter squared. So use thes values and plug them in this a question and find the change in them to be equal to 1.3 times 10 to the minus three meter

This question we have to find the um threat in Rod A. B. And city which are the two identical rod when there is a farce P apply at the end of the beam. So um from figure A. It is the bravery diagram of the beam and we use the equilibrium sigma moment of white F. You go to 020 is here and we'll have that if baby time A plus if C D E. A. He's minus P. three a. equal to zero, we have that if A B plus f city, they are equal to you three peat. That is the first situation you get. And from the compatibility equation and figure T africa P you have that Dell A B is equal to D. L. C. D. Has shown in the figure B. So we have that maybe. Uh huh. What about E is it good to F C D L was about E now because two brought identical. So all the parameter except have identical. So we have had F A B is equal to f C D. No. If we substitute this in the first equation, you have that F A B. It is equal to F city, which is three P. Do I buy 2? So the straits A B is the total straight of CD, which is F forward area, and that is six P by pi D square. That is the answer.

So once again welcome to new problem. This time we have ah born that's sitting on this office and it houses office area, cross sectional area given by, um, 3.6 centimeters squared and we always want to transform the two middle squared by multiplying by, um bye. One meter squared of ah, 100 centimeters squared. And this is gonna give us 3.6 times 10 to the negative. Four. Be two squared. Also the heights off the bone is 22 centimeters. Original height off the bone is 22 centimeters. We want to change that to meters. So it becomes one meter 100 senator unit conversion and we end up getting zero point to two meters. This is the information that's given in the problem. Also, another thing that happens is these air force being applied in the course sectional area. That force f we call it in Newton's 3.3 times turned to the for mutants. That's the force being applied. Our goal, in part, is to determine if the born well, the born break. Um, if the stress sigma is greater than the compressive strength off the bone, then obviously the bone is gonna break? Absolutely. The compressive strength of the bone is 170. The mega Paschal's You can see that. So that's the first thing I wanna find out in the problem. The second thing we're looking for in the problem is that if the, um the bone does not break, what's the change in length that we have? It's the bone does not break so in, but a We have to compute the stress, which is the force by unit area. Well, given the forces 3.3 time stand to the four Newtons and we divide that by 3.6 time. Stand to the negative for minutes wed. When you simplify that, we get the stress. A cz being equivalent to 91.67 Newton Amita squared Then obviously, if you compare that 270 it's less than that. So the stress is less Thune, the compressive strength. Um so say the born mmm does not break, you know, because the forces bottoms out. I mean, you know, this this force that's being applied is less than the compressive strength that the bone can take. So the bone itself doesn't. It doesn't. Kevin doesn't break on the next page, we're gonna see by how much does the land change given that these air force being applied on the bone itself. Now, this is a relationship between stress and Young's modules that's connected to the change in land and original men. So sigma he this is the stress. And this is the Youngs modular. So of elasticity and the Youngs modular us off elasticity. It's given us The value of start is 15 times 10 to the nine Newton Mideast squared. This is gonna help us Compute Doubt L, which is Sigma over times l note and this becomes signifies simmers 91 0.67 Newton meter squared. And then, uh hell, not his Sam was a 0.22 meters and then Young's model asses 15 times 10 to the nine new committee squared. That gives us the change in length as being equivalent to one point 33 millimeters or rather, 1.3 millimeters. You know, we can round it off if you change it to millimeters. So I hope you enjoy the problem. Feel free to send any questions or comments and have a wonderful day. Okay, thanks. Bye.


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