Why is Torque perpendicular to the force?

[FallingMan]

Yes, I understand the equation for the cross product and the right hand rule.

People keep using this example with a wrench twisting a bolt and the right hand rule makes sense there. Okay fine, the force is perpendicular in that case...

But suppose there is no bolt grooves.

The bolt won't experience a perpendicular force anymore because it lacks grooves, no matter how much you turn. I.E. the bolt is completely smooth on the inside, where it contacts the nut.

I think that's part of what I'm misunderstanding here. Is there some inherent perpendicular force that's part of torque?

Let me know if I'm not explaining my confusion clearly.

 

[dauto]

No, a force is not part of the torque. The torque is a separate quantity defined to be perpendicular to the force as per the right hand rule. That direction is part of its definition.

 

[Khashishi]

Because that's the way that torque is defined. There's nothing fundamental about the direction. Torque isn't something you can measure, except indirectly through the force. So physicists can choose a direction for it to point in, as long as the mathematics all work out consistently.

If you want to get more theoretical about it, any value that is equal to a cross product is actually a different kind of vector. It shouldn't be imagined as an arrow with a direction and magnitude. Rather, it is a little 2D surface patch with an orientation and an area. In 3D, we can map from these surface patches (or 2-forms) to regular vectors by associating a perpendicular vector to every surface patch, which is called a Hodge dual. It is convenient to write the laws of physics to directly define the torque in terms of the Hodge dual rather than the 2-form. If written in terms of the 2-form, there is nothing perpendicular about the torque.

 

[rcgldr]

For angular velocity, angular acceleration, or torque (angular force), one reason a vector perpendicular to the plane of rotation or torque is used because the math for vectors is simpler than trying to create math that works with rotating planes. The normal convention for the direction of the vector is to use right hand rule, so torque can be calculated as the cross product of two vectors: radius x force.

 

[gmax137]

First, you're right, the torque direction doesn't have anything to do with the threads moving the bolt in or out as you turn it.

At the risk of causing more confusion, imagine the bolt head is at the center of a clock face. You get the same effect if your wrench handle is at 3 o’clock and you pull down, or if the handle is at 9 o’clock and you pull up, right? Or if the handle is at 12 o’clock and you pull sideways, or 6 o’clock and you push the other way. And so on. If all these are the same, then the torque *has* to be along the axis of the clock, doesn’t it?

And the right-hand-rule convention just says whether we call the torque pointing in or out.

But it is perpendicular either way, due to the symmetry described above.

That’s how I always saw it, anyway.