Timeline for If I hold a ladder at an incline to a frictionless surface and then release it, will it slide?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Jun 11, 2020 at 15:52 | comment | added | Bob D | Welcome. Fun working with you. | |
| Jun 11, 2020 at 15:33 | vote | accept | Vulgar Mechanick | ||
| Jun 11, 2020 at 15:33 | comment | added | Vulgar Mechanick | That explains it perfectly. Many thanks! | |
| Jun 11, 2020 at 13:47 | comment | added | Bob D | Although as the angle goes down the required horizontal friction goes down, the normal force also goes down but at a greater rate. In doing some crude experiments using pencils on a granite counter and slo-mo iphone video, sliding always occurs nearer the end of the fall. Try it and see. | |
| Jun 11, 2020 at 13:47 | comment | added | Bob D | @OVERWOOTCH Let’s just say your understanding and mine are the same and, hopefully, are correct. If my trig is correct, the reactions to mg sin θ are a normal force of N= mg sin$^2$θ and a required horizontal friction reaction of $F_{f} = \frac {mg}{2}$ sin 2θ. The interesting thing is if this is correct, then when θ is less than 45$^0$ the coefficient of static friction would need to be greater than 1 to prevent sliding. At 10$^0$ or less it would have to be greater than 5.67! | |
| Jun 10, 2020 at 16:28 | comment | added | Vulgar Mechanick | Oh ok so “N” and “F” are just the Mgsinx resolved back right? Is my understanding about the rest as mentioned in the comments correct? | |
| Jun 10, 2020 at 15:40 | comment | added | Bob D | @OVERWOOTCH Oops. I was working on a couple of drafts and used the wrong Fig 3. See updated figure and text. Sorry for the confusion. | |
| Jun 10, 2020 at 15:38 | history | edited | Bob D | CC BY-SA 4.0 |
clarification
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| Jun 10, 2020 at 4:44 | comment | added | Vulgar Mechanick | Where m is the infinitismal mass of that point. As a result, the only force on the POC is actually Mgsin θ, which is again resolved and results in two components, one along the x axis which produces friction, and one along the y axis, which produces the normal contact force. If this is correct, whats the origin of F| at the point of contact. Shouldnt it be the reaction force to that infinitismal mgcos θ and thus be effectively 0? | |
| Jun 10, 2020 at 4:38 | comment | added | Vulgar Mechanick | Thankyou for the answer. Cleared a lot of things up. i think my main problem here is that I am ignoring the forces actimg on each specific point and generalising he forces the force(s) acting on the centre of mass to every point. If understood this correctly, we can consider the gravitational force to act as a single downward force, Mg, which is resolved into 2 components. Now, since the line of action of Mgsin θ passes through the point of contact, it acts on it but not Mgcos θ. The force perpendicular to the ladder at the point of contact is actually mgcos θ | |
| Jun 9, 2020 at 22:54 | history | answered | Bob D | CC BY-SA 4.0 |