Why is torque perpendicular to the force applied?

[MRzNone]

I understand the right hand rule and the equition, but why is the direction of the torque perpendicular to the force on the counter clockwise side of the force and crossing the rotational point?

 

[Orodruin]

Because this is how torque is defined and it turns out to be a useful definition. You could just as well have defined it with a minus sign, but this would have had other consequences in other formulas.

 

[MRzNone]

But, definitions are derived right? Can you explain me how? They are seemingly irrelevant to me.

 

[Orodruin]

But, definitions are derived right?

No. Definitions are not derived by construction. If it is possible to derive something, you do not need to define it.

 

[person_random_normal]

its a convention but i think you can feel that intuitive having thought over this -
look,
if you a screw and over it you have a nut ,tightly fitted ; and you are to undo it, using a spanner !
So, you will be rightly placing the spanner and forcing it to rotate it in counter clockwise sense about the screw
doing that you will be undoing the tightening , making the nut move towards you
so now you can think that your force on the spanner materialized into the outward movement of the nut (the direction of torque) !

How is it now ?!

 

[Orodruin]

So, you will be rightly placing the spanner and forcing it to rotate it in counter clockwise sense about the screw
doing that you will be undoing the tightening , making the nut move towards you
so now you can think that your force on the spanner materialized into the outward movement of the nut (the direction of torque) !

Note that this intuition is solely based on the convention of making nuts/screws with a particular handedness.

 

[Fredrik]

I second everything that Orodruin said, and I would like to add that the definition ensures that torque is to angular momentum what force is to momentum. What I mean by that is that force is the time derivative of momentum, and torque is the time derivative of angular momentum:
$$\dot{\mathbf L} =\frac{d}{dt}\left(\mathbf r\times(m\dot{\mathbf r})\right) =\mathbf r\times (m\ddot{\mathbf r}) =\mathbf r\times\mathbf F=\mathbf\tau.$$

 

[Orodruin]

And just to add to that, the relation would still hold if we defined both torque and angular momentum with a minus sign relative to the usual definition.

 

[andresB]

Is the question about the fact that the torque is perpendicular to the force or about the "clockwise" part?

 

[person_random_normal]

have
MRzNone got that
and how you felt about my reply