Timeline for How do levers amplify forces?
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17 events
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Jan 3, 2020 at 21:02 | comment | added | bruno | @Joshua Ronis yes it does. In my recent answer I have tried to express that notion of spinning counterclockwise in terms of bending stresses, but ultimately as you have said if one side is deformed more than the other there is no equilibrium and it will tilt | |
Jan 3, 2020 at 20:51 | comment | added | joshuaronis | ...touching, the right clamp feels a greater force. That is, a force needs to be applied on the right side if we placed the entire rod on that little triangle, and wanted to stop it from spinning counterclockwise. Does that make sense? | |
Jan 3, 2020 at 20:50 | comment | added | joshuaronis | @21Brunoh I think you're close, but with one caveat...all the springs are the same length - they represent the forces only between adjacent molecules, which are all the same distance apart from one another, and must remain around the same distance apart (which is why its a "rigid" body). There are just more of them on the left side, so each one deforms by more due to the same overall vertical displacement in the right side than in the left side. Since the springs connected to the molecules that the right clamp is touching have been deformed by more than those the left one is... | |
Jan 3, 2020 at 20:37 | comment | added | bruno | Now I see what you are saying, it makes sense. It is similar to how it is harder to pull a short spring compared to the longer one. You need to exert more force in the shorter side. | |
Jan 3, 2020 at 20:27 | comment | added | joshuaronis | @21Brunoh I agree that the longer part in the vertical direction stretched the same. However, there are more molecules in the longer part. So, if the longer part stretched the same in total than the shorter part, the displacement between individual adjacent molecules in the shorter part must be greater, and thus they must be feeling a greater force from the particles around them, and exerting a greater force on the clamp. | |
Jan 3, 2020 at 20:23 | comment | added | bruno | @Joshua Ronis I'm sorry but you lost me when you started talking about the space part. Either way thank you for the time you've spent trying to explain the perspective, maybe you should consider making an answer to this post as comments are harder to follow. I would just disagree again that the deformation in one side is twice as much as the deformation on the other side. Deformation (in terms of strain) is how much the material as stretched over its original dimensions. In my opinion the longer portion of the rod also stretched a lot. At least in the vertical direction it stretched the same. | |
Nov 18, 2019 at 8:08 | comment | added | DS. | This picture made sense qualitatively, but to explain the actual relationship between forces, my attempt is in this answer. | |
Nov 6, 2019 at 3:41 | comment | added | joshuaronis | ...location that we must apply the force so that the vertical deformation between adjacent particles on the right side is twice as much as the vertical deformation between adjacent particles on the left side is 2/3 the length of the rod from the left side! That is, if we apply the force at this location (the center of mass) the entire thing accelerates in space with constant acceleration and no rotation, since the 2M is getting twice the force than the M is!!! | |
Nov 6, 2019 at 3:40 | comment | added | joshuaronis | ...vertical distance between horizontally adjacent particles on the right side would basically cause a linear increase in the total distance between horizontally adjacent particles on the right side, and thus a proportional increase in the vertical force delivered to any particle by the one next to it). Then, if each molecule on the right side feels twice the force as a particle on the left side from the one next to it, since it has twice the vertical displacement from it it'll deliver twice the force to the mass 2M as is being delivered to M! And, it shouldn't be hard to see that the... | |
Nov 6, 2019 at 3:33 | comment | added | joshuaronis | ....as the deformation of the rod on the left side. That is, we want the to apply the force in such a place that the vertical distance between any two adjacent (horizontally) atoms on the right side increases (for a brief moment, before the entire thing starts accelerating, since it's in space) by twice the amount that the bonds between adjacent atoms on the on the left side increase by, so that twice the force is exerted (vertically) on each particle on the right side than on each particle on the left side. (Note that since the x distances between particles are so small, increasing the... | |
Nov 6, 2019 at 3:30 | comment | added | joshuaronis | @21Brunoh from the point of view of the mass on the right, the only "springs" it would care about are those attached directly to the "balls" its in contact with. Say that the mass on the right has a mass of 2M, and the mass on the left has a mass of M. The rod is floating in space. I think (I'm almost sure) that at small enough distances, the force between bonded atoms in a rigid body is proportional to the distance between them. We want to apply the force (put the triangle) in such a place that the deformation of the rod (between any two particles) on the right side is twice as much... | |
Jul 12, 2017 at 17:49 | comment | added | John Alexiou | I don't think this answer answers the question sufficiently. | |
Jun 6, 2016 at 14:05 | comment | added | rmhleo | This answer does not explain the phenomenon. For an explanation see my answer of the following question | |
Nov 5, 2014 at 10:59 | comment | added | bruno | Although the springs from the longer side are less distorted there is a higher quantity of them, so why wouldn't the net distortion equal or be greater than the distortion from the short side? What's the reason? | |
May 6, 2012 at 22:40 | vote | accept | fiftyeight | ||
Mar 30, 2012 at 3:41 | history | edited | Noah | CC BY-SA 3.0 |
Tried to improve the readability of the sentences.
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Mar 29, 2012 at 17:43 | history | answered | Noah | CC BY-SA 3.0 |