Timeline for Friction forces and sliding slabs
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 1, 2013 at 14:18 | comment | added | Frank | Staying stationary feels incorrect to me: first, in the case of 2 slabs, if the bottom slabs was moving in unison with the top block when F was not strong enough, why would the bottom slab suddenly stop? Second, even if an interface "breaks", it doesn't mean the friction coefficient there drops to zero. It changes to the kinetic friction coefficient, and there is still a force transmitted to the bottom slab via that friction. So, there are forces on the bottom slab, and hence possibly an acceleration. IMHO the bottom slab becomes stationary only if the forces on it precisely cancel. | |
| Oct 1, 2013 at 12:57 | comment | added | Brian Moths | @Frank, for your first comment about the case of two slabs, just reread the bottom paragraph with $n=2$, so the bottom slab doesn't move unless the bottom interface is the weaker one and the force is sufficiently high to break it. For your second comment, the forces at the interface decrease to $\frac{\mu_{nk}}{\mu_{ns}}F_n$ which is too weak to get any of the other interfaces to break, so only a maximum of one interface will break. | |
| Oct 1, 2013 at 3:36 | comment | added | Frank | I am not sure that your reasoning for the n slabs case is valid: if you continue increasing the force after the weakest interface "broke", why wouldn't you reach a point where the second weakest interface can break? | |
| Oct 1, 2013 at 3:35 | comment | added | Frank | I can't find any fault in what you are saying, but it doesn't IMHO answer the question: what direction does the bottom slab move, in the case of 2 slabs? | |
| Oct 1, 2013 at 0:54 | history | answered | Brian Moths | CC BY-SA 3.0 |