How did we get the formula for torque?

[rokuog]

I know torque is defined as r x f , but how did we come to define it in this way? Sorry if question seems a bit noobish but i haven't been able to find the answer

 

[BvU]

Hello roku, welcome to PF :)

Not exactly a homework posting, this is, eh ? You can read up http://www.physics.uoguelph.ca/tutorials/torque/Q.torque.intro.html or here or google around a bit.

Part of your answer is that it fits in nicely in the analogy between linear motion and angular motion.

 

[Stephen Tashi]

I know torque is defined as r x f , but how did we come to define it in this way? Sorry if question seems a bit noobish but i haven't been able to find the answer

In 2-D, imagine you are turning a bolt with a wrench with a handle of length L by pulling on a cord tied to the end fo the handle with a force F. If the string is not perpendicular to the handle of the wrench there are two ways of calculating the torque you are exerting. One way is to visualize an imaginary handle that is perpendicular to the line of the force and compute F times the length of the imaginary handle. Another way is to resolve F into two components and multiply L times the component of F perpendicular to the actual handle.

Both methods give the answer L F sin(theta) where theta is the angle between the string and the handle (or the line of the handle). That's the appropriate magnitude for the cross product.

To that 2-D explanation of the magnitude, you could add representing the torque as a vector that points in the direction that a right hand threaded bolt will move in such a situation. That would be straight up out of the 2-D plane or straight down. If you use the convention that the vector representing L has its tail at the bolt head and the vector representing F has its tail at the handle. then L cross F gives you the correct magnitude and the correct direction.

In a real life situation, the head of the bolt lies in a plane but you often don't have room to swing the handle of a wrench so it stays in that plane or parallel to it. (For example, you might have to use a ratchet wrench with a pivoting head.) Imagine a force F being exerted on the end of a wrench of length L in that situation. The force isn't perpendicular to the handle of the wrench, the handle of the wrench isn't perpendicular to the shaft of the bolt. How do you figure out the torque being exerted that is effective for turning the bolt?

Working in detail, you could resolve the force F into components so that one of the components Fp is in a plane parallel to the bolt head. You could calculate the length of an imaginary wrench handle Lp that is perpendicular to the shaft of the bolt and in that same plane. The torque would be Lp times Fp.

I've never worked out in detail whether using the cross product simplifies (or at least organizes) that calculation. The suggested reorganization is to let L be the vector pointing from the bolt head to the end of the handle. Represent the force F as vector with its tail at the end of the handle. Compute L cross F. Then project the vector (L cross F) on a unit vector pointing along the the shaft of the bolt. Let the unit vector have the direction that a bolt with a right hand thread will move when the head is turned counterclockwise.

 

[Brian T]

Think about opening a door. If you exert force parallel to the surface of the door (slide your hand across), the door won't rotate meaning it won't push open. The closer your angle gets to perpendicular to the surface, the more of your force goes into turning the door. When you are fully perpendicular, all of your force is used to rotate. rFsinA or (r) x(F) matches such phenomenon