Timeline for Rope Tension and change in length of rope
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Apr 26, 2017 at 1:57 | comment | added | Chet Miller | As a guy who worked in rheology and material science, I only feel comfortable using the Pythagorean theorem (or more general kinematic form) in getting the changes in length. For cases where you know in advance that the strains are going to be small, it is valid to use the approach you used. However, I still feel that there is less margin for error if the large deformation version is used, and then taking the limit of small displacements or strains. I'm not a teacher. I'm a retired industry engineer. | |
| Apr 25, 2017 at 17:04 | comment | added | Sherlock Homies | Hi chester can i actuall use the ratio of the length of the sides of the original traingle with the stretched out one? Can you see if you get the same results as me above? Thanks. Are you a teacher? | |
| Apr 22, 2017 at 15:09 | comment | added | Chet Miller | Actually, by a little mathematical manipulation of the answer, I get $$T_{VR}=(2-\sqrt{2})mg$$ | |
| Apr 22, 2017 at 12:38 | comment | added | Chet Miller | I don't know. I would have to see your algebra. The first thing to do is to solve for the center rope tension from the vertical force balance, as you did. My equation for that was $$T_{VR}(1+\frac{1}{\sqrt{2}})=mg$$where the 2nd term in parenthesis comes from the combination of the two side ropes. | |
| Apr 22, 2017 at 12:25 | vote | accept | Sherlock Homies | ||
| Apr 22, 2017 at 12:20 | comment | added | Sherlock Homies | is it because i forgot the otherside somewhere in my calculations? I got the relation between delta l1 and l2 as follows: delta l1 = sqrt2*delta l2 by using the sides as ratios correspondingly. | |
| Apr 22, 2017 at 12:12 | comment | added | Chet Miller | I get twice your value for the tension in the middle rope. | |
| Apr 22, 2017 at 12:06 | comment | added | Chet Miller | You linearize by neglecting a term such as $(\frac{\Delta L}{L})^2$ as being insignificantly small compared to unity and compared to $(\frac{\Delta L}{L})$ | |
| Apr 22, 2017 at 11:55 | comment | added | Sherlock Homies | i did the calculations and came up with the middle rope tension as mg/(sqrt2+2) | |
| Apr 22, 2017 at 8:42 | vote | accept | Sherlock Homies | ||
| Apr 22, 2017 at 8:43 | |||||
| Apr 22, 2017 at 8:22 | comment | added | Sherlock Homies | what does linearize mean? how do you do that? | |
| Apr 21, 2017 at 22:50 | history | answered | Chet Miller | CC BY-SA 3.0 |