Timeline for How is torque transmitted between inclined surfaces?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 26, 2020 at 14:31 | comment | added | DKNguyen | Yes............ | |
| Oct 26, 2020 at 14:20 | comment | added | Modulus | F2, sorry. M3 is transmitted to K1 in the same way as F2 is exerted onto K1. If M3 / lever = F2, then a) is identical to b)? | |
| Oct 26, 2020 at 14:10 | comment | added | Modulus | Yes, exactly. But a) with a free force F1 and b) with M1 resulting in another F1 are identical in a static state for the result of only F1 - ignoring the fact that both somehow need to hold still? So by applying superposition and looking in a first step only at F1, a) and b) are identical and both result in a net toqrue applied in ccw direction - which torque is then countered by K2 blocking K1. Right? | |
| Oct 26, 2020 at 14:09 | comment | added | DKNguyen | If the F2 is being exerted as a free force and K2 is not there, whether or not it turns cw or ccw is dependent on which force component of F2 has the longer moment arm. | |
| Oct 26, 2020 at 14:08 | comment | added | Modulus | Or differently phrased, suppose I would be excerting a torque in clockwise direction to K1. This torque would have to be F2t * a + F2o*b*(friction coefficient) to be able to move K1, right? So the F2t part is transmitted to K1 and holds K1 in place? | |
| Oct 26, 2020 at 14:06 | comment | added | DKNguyen | Are you now asking about a free force being applied along F2 to K1, with K2 not being present? With K2 present, there would be an something akin to an internal force since the stress is there but nothing is moving because of binding. There can't be a CCW net torque in that case, by definition. | |
| Oct 26, 2020 at 14:03 | comment | added | Modulus | Yes. But there would be a resulting torque ccw (which would make the object spin if K2 would allow it)?. | |
| Oct 26, 2020 at 14:01 | comment | added | DKNguyen | There can't ever be spinning in the CCW direction because the two objects will bind. All it can do is reduce the net torque in the CW direction up until there is no slippage and all the CW net torque is completely nullified. | |
| Oct 26, 2020 at 14:00 | comment | added | Modulus | lets suppose infinite friction (or enough friction for no slippage to occur). In the case that F2 is a linear force, would there be ccw "spinning"? I edited that into the question. | |
| Oct 26, 2020 at 13:59 | history | answered | DKNguyen | CC BY-SA 4.0 |