Timeline for Torque of upside-down T-shape
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 2, 2021 at 6:49 | comment | added | DJohnM | You also need to include the direction of the torque, either explicitly by adding "clockwise" and "counter-clockwise", or by making one of the torques negative. | |
| Oct 2, 2021 at 6:48 | history | edited | ACB | CC BY-SA 4.0 |
added 9 characters in body
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| Oct 1, 2021 at 19:43 | vote | accept | explodingfilms101 | ||
| Oct 1, 2021 at 19:14 | comment | added | Bob D | Yes that is correct | |
| Oct 1, 2021 at 19:13 | answer | added | John Alexiou | timeline score: 0 | |
| Oct 1, 2021 at 19:07 | comment | added | explodingfilms101 | So the torque due to $F$ is $F \cdot (L + h)$? @BobD | |
| Oct 1, 2021 at 18:55 | comment | added | Bob D | @explodingfilms101 Don't you realize that the moment arm of F is the perpendicular distance between the line of action of F and the orange dot, i.e. L+h? | |
| Oct 1, 2021 at 18:48 | comment | added | Michael Seifert | Note that $$\sin\left[180^\circ - \tan^{-1} \left( \frac{ L+h}{r} \right) \right] = \sin\left[\tan^{-1} \left( \frac{ L+h}{r} \right) \right] = \frac{(L+h)/r}{\sqrt{ 1 + (L+h)^2/r^2}} = \frac{L+h}{\sqrt{ r^2 + (L+h)^2}}$$where I've used one of the identities in this table in the second step. | |
| Oct 1, 2021 at 18:39 | comment | added | explodingfilms101 | Look at the diagram. I'm applying a force that's not along the axis of rotation about the orange point. | |
| Oct 1, 2021 at 18:36 | answer | added | R.W. Bird | timeline score: 0 | |
| Oct 1, 2021 at 18:28 | comment | added | Bob D | Seems odd that you are drawing the moment arm for the weight to the right of the center of mass, instead of to the left. Also, what do you mean by $F$ being "off axis"? | |
| Oct 1, 2021 at 17:41 | history | asked | explodingfilms101 | CC BY-SA 4.0 |