中美之间的博弈,投资者从中会诞生哪些投资机会
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 3 at 2:01 | vote | accept | Stefan Karwasiecki | ||
| Feb 2 at 22:37 | vote | accept | Stefan Karwasiecki | ||
| Feb 3 at 2:01 | |||||
| Feb 2 at 22:26 | comment | added | Bob D | You are absolutely right. $\mu Mg$ is the maximum possible value of the static friction force to have angular and translational acceleration without slipping at which point the friction becomes kinetic. It is not necessarily the actual static friction force. The actual static friction force adjusts on the basis of the applied force and the other constraints of the problem, in particular equation (4) of my answer. | |
| Feb 2 at 22:19 | comment | added | Stefan Karwasiecki | Oh, I think I understand what is meant. When F = Mg(\mu), in other words F is the maximum friction value, the ACTUAL friction force is not yet at its maximum value. Hence, a_com is nonzero and the circles COM translates. The friction reaches its maximum value when F_max (7) is applied, at which point it begins to slip. Is my understanding correct? | |
| Feb 2 at 18:42 | history | edited | Bob D | CC BY-SA 4.0 |
added 14 characters in body
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| Feb 2 at 18:22 | history | answered | Bob D | CC BY-SA 4.0 |