Mathematically Define Torque (1 Viewer)

[Sy123]

It is given that the torque is:



(disregarding the angle by which the force is applied, which just results in a trivial trigonometry problem)

How do you prove this formula? I have been told that torque by definition is the distance multiplied by the force. But I don't see how you can do that.
I don't see why it must randomly be a multiplication of 2 variables that have to do with turning force, or rather the operation between them has to be multiplication.

Why can't it be: for instance, I mean as d increases so does torque so that it follows the same pattern as multiplication.

So can you prove the formula for torque? And if its like that by definition why multiplication?

 

[Lieutenant_21]

My physics tutor said torque gets very complicated in university but we don't need to know that for the HSC.
You can read about it here: http://en.wikipedia.org/wiki/Torque

 

[Riproot]

Work it out by putting units in. Duh.

(clearly I know nothing about this)

 

[Sy123]

My physics tutor said torque gets very complicated in university but we don't need to know that for the HSC.
You can read about it here: http://en.wikipedia.org/wiki/Torque

So there really isn't a simple answer to this?
That page has a lot of vector calculus in it hah I'll look at it in my free time I think

 

[yasminee96]

well torque is also nBiacostheta
maybe working from there, and using F=Bilsintheta, you can maybe derive the formula?
But i don't see where you could pull distance from - maybe work done or something...

 

[bleakarcher]

It is given that the torque is:



(disregarding the angle by which the force is applied, which just results in a trivial trigonometry problem)

How do you prove this formula? I have been told that torque by definition is the distance multiplied by the force. But I don't see how you can do that.
I don't see why it must randomly be a multiplication of 2 variables that have to do with turning force, or rather the operation between them has to be multiplication.

Why can't it be: for instance, I mean as d increases so does torque so that it follows the same pattern as multiplication.

So can you prove the formula for torque? And if its like that by definition why multiplication?

Torque is the instantaneous rate of change in angular momentum, the rotational analogue of linear momentum. Angular momentum is well defined as r*p (cbf using vector notation) and so when you differentiate with respect to time you get τ=d/dt(rp)=r(dp/dt)=Fr (obviously I assumed that the radius of the circle stays constant).

 

[D94]

It's technically r x F, where x denotes the cross product (it is not a multiplication sign). When you are dealing with 2 vectors in 3D space, you need to use the cross product - it is clear that an object will not rotate about a pivotal axis if you apply a force in the same direction as the arm or the displacement vector (r).

So applying the cross product, you get rFsinØ, in this case, it's all simple multiplication.

NB: r and F are vectors.

 

[Sy123]

Torque is the instantaneous rate of change in angular momentum, the rotational analogue of linear momentum. Angular momentum is well defined as r*p (cbf using vector notation) and so when you differentiate with respect to time you get τ=d/dt(rp)=r(dp/dt)=Fr (obviously I assumed that the radius of the circle stays constant).

I am guessing r is radius and p momentum right?







This is a correct proof yeah?
Thanks man

 

[bleakarcher]

yeah, that's right.

 

[Sy123]

It's technically r x F, where x denotes the cross product (it is not a multiplication sign). When you are dealing with 2 vectors in 3D space, you need to use the cross product - it is clear that an object will not rotate about a pivotal axis if you apply a force in the same direction as the arm or the displacement vector (r).

So applying the cross product, you get rFsinØ, in this case, it's all simple multiplication.

NB: r and F are vectors.

I see, thank you for the clarification

 

[Trebla]

Work it out by putting units in. Duh.

This. Adding them doesn't make sense because the units don't match (Newtons and metres).

 

[Lieutenant_21]

I am guessing r is radius and p momentum right?







This is a correct proof yeah?
Thanks man

Torque is the rate of change in angular momentum. Makes sense.

 

[Sy123]

OP,

to answer this dot point in the syllabus, just memorise the equation (torque = F . d) .... the HSC physics exam will (hopefully) not ask for a mathematical proof of torque.


"Mathematically Define Torque"...

translation:
Identify the mathematical formula of Torque,
AND.... Identify the meaning of the symbols and units in the formula of Torque,
where F = force (newtons)......
d = perpendicular distance between the "point of action of the turning force" and the axis (metres)...
and so on.


EDIT: Don't for get your units!

I am very familiar with the HSC concept of torque since the questions that utilise it are all very similar and simple.
I am looking for a proper definition of torque, not the washed up HSC version.

 

[asianese]

You'll find that knowledge beyond what the HSC requires is kinda useless for now, esp wrt maths and physics.

 

[Riproot]

This. Adding them doesn't make sense because the units don't match (Newtons and metres).

It's the only reason I passed physics.

Like put everything in terms of units and you'll see what does and doesn't work equation-wise.

 

[Drongoski]

You know the seesaw we knew as kids. When you sit on it,the farther out from the fulcrum (the centre where it balances the 2 riders) the greater the moment. This moment (or torque) is proportional to applied force and the distance from the fulcrum . In fact with the appropriate units (e.g. Newton-metre) moment = force x distance (i.e. f x d). This assumes the force is applied at right angle to the board. But if the force is applied to at an angle @ to the board then only the component perpendicular to the board has effect (the turning "force" - or torque). In this case the moment = fsin@ x d. Now torque is a vector quantity. The cross-product F x d fully reflects this situation.

In fact: |F x d| = |Fsin@|x|d|

Note it is not the dot-product F . d which is a scalar quantity.

 

[Sy123]

This. Adding them doesn't make sense because the units don't match (Newtons and metres).

I don't see how that justifies the proof though, the only reason why torque is Nm is because t=Fd. So its like proving a definition using its own definition.
Doesn't make sense

 

[ProtoStar]

Well I can show you a derivation for torque on opposite sides of a lever using W=FS