I meant it is a new quantity. Why is it defined to be $\vec{r}\times \vec{F}$ ? Why not any other definition of the torque?
$\begingroup$
$\endgroup$
7
-
$\begingroup$ What do you mean by “a new quantity”? $\endgroup$– G. SmithCommented Jun 14, 2020 at 3:46
-
1$\begingroup$ Related: physics.stackexchange.com/q/155869/50583 $\endgroup$– ACuriousMind ♦Commented Jun 14, 2020 at 7:39
-
$\begingroup$ If we like to "combine" force and leverage, the standard definition of torque seams to be reasonable to me: It projects the force onto the distance vector, thereby only considering the orthogonal part of the distance vector (=leverage). Which other / better definition do you have in mind? $\endgroup$– NotMeCommented Jun 14, 2020 at 13:30
-
$\begingroup$ For example, to define force we knew that there's something that causes mass an acceleration. Hence F=ma. Likewise, how did that definition came in mind T=r cross F. From this we later came to know that T= Ia Did you get me? Would you mind guiding me? $\endgroup$– Rusan LamsalCommented Jun 15, 2020 at 1:32
-
$\begingroup$ From observations. Go through my answer carefully as I tried to solve your problem minutely . $\endgroup$– user262060Commented Jun 15, 2020 at 2:30
|
Show 2 more comments
2 Answers
$\begingroup$
$\endgroup$
The notion of torque probably comes from the most basic tools of mankind: the lever.
answered Jun 15, 2020 at 1:27
$\begingroup$
$\endgroup$
2
See the concept of Torque was introduced to deal with the cases in which different parts of the body moves differently or precisely for rotatory motions. Torque actually defines the amount of force from a point other than the point of application.
If any force is applied on an object (at a point other than it's center of mass ) then the body will show some rotational motion because the force is not distributed equally (refer to the example in the note) . So to find the amount of the force from the point of application we need to define the distance of the point from the point of application and also the direction of the angular acceleration is obtained by screw rule and so the cross product sets good here.
Also, the amount of force determines the amount of rotation.
So , the formula for Torque holds good in all scenarios.
NOTE: Suppose we have a rod made up of only 9 particles , each of mass m (refer to the image ) and hence total mass is 9m. If the force is applied at the fifth point then it is distributed equally to the points above and below the point and thus the behaviour is same as if the force was applied to a body of mass 9m. But if it is applied at the seventh point, then there will be only two particles below and six particles above. So upper part having more mass will show less motion than the lower part of the body and we say that the body rotates or tends to rotate .
From observations we can calculate the angular displacement and hence angular acceleration and then we can calculate the amount of force at any point using
r × f = I( alpha)
!(https://i.sstatic.net/zmqBo.jpg)
Sorry for the approximations.
Hope it helps.
-
$\begingroup$ Ankit Kumar, I would like you to answer my another question. Do you think you could help me? Link:physics.stackexchange.com/questions/559470/… Please focus on explanation rather than equations and I didn't mean that you can't include any equations $\endgroup$ Commented Jun 16, 2020 at 1:59
-
$\begingroup$ I tried to answer your question and if you have some doubt comment below. $\endgroup$– user262060Commented Jun 16, 2020 at 7:41