The Definition of Torque - a proof

In summary, the 'effectiveness of rotation' of a force, taken as ##g(F,x)## is a function of the force F and the distance from the origin x. If the force be scaled by ##\lambda##, the effectiveness ##g(F,x)## must also be scaled by that factor. This implies that the change in effectiveness per change in F just depends on x (and not on the particular value of F). Therefore, effectiveness itself is a linearly dependent on f and some function of x.
  • #1
Shreya
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Homework Statement
Please refer the image below.
Relevant Equations
Newton's Laws
I have been trying to understand this proof from the book 'Introduction to classical mechanics' by David Morin. This proof comes up in the first chapter of statics and is a proof for the definition of torque.
I don't understand why the assumption taken in the beginning of the proof is reasonable. A note given at the end tries to give some clarification, but I can't relate these 2 points.
Please be kind to help. :)
1683691085008.png
 
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  • #2
How about this:
The "effectivity" of a force of magnitude F applied in a specified direction at x from the axis is ##g(F,x)##. Applying some multiple of F, ##\lambda F##, in the same direction and at the same point should have effectivity ##\lambda g(F,x)##, i.e. ##\lambda g(F,x)=g(\lambda F,x)##. Hence ##\frac{\partial g(F,x)}{\partial F}=\frac{\partial g(\lambda F, x)}{\partial F}##. Since every force in the specified direction can be represented by a suitable choice of ##\lambda##, this implies ##\frac{\partial g}{\partial F}## is a function of x only. Integrating, ##g=F f(x)+c## for some function ##f##.
Adding that a zero force should have zero effect leads to the result.
 
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  • #3
Sorry, but I don't quite understand the partial differential equation, @haruspex . Should'nt there be a ##\lambda## on the left.
 
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  • #4
Shreya said:
Sorry, but I don't quite understand the partial differential equation, @haruspex . Should'nt there be a ##\lambda## on the left.
Sorry, I didn't write the algebra correctly.
##\lambda g(F,x)=g(\lambda F,x)##. Hence ##\lambda\frac{\partial g}{\partial F}\vert_{F,x}=\lambda\frac{\partial g}{\partial F}\vert_{\lambda F, x}##.
Cancelling,
##\frac{\partial g}{\partial F}\vert_{F,x}=\frac{\partial g}{\partial F}\vert_{\lambda F, x}##.
Since every force in the specified direction can be represented by a suitable choice of ##\lambda##, this implies ##\frac{\partial g}{\partial F}## is a function of x only. Integrating, ##g=F f(x)+c## for some function ##f##.
 
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  • #5
Yes sir, I understood it now. Just to check if I got it properly, I'll try to summarise here. The 'effectiveness of rotation' of a force, taken as ##g(F,x)## is a function of the force F and the distance from the origin x. If the force be scaled by ##\lambda##, the effectiveness ##g(F,x)## must also be scaled by that factor. This implies that the change in effectiveness per change in F just depends on x (and not on the particular value of F). Therefore, effectiveness itself is a linearly dependent on f and some function of x. I might have lost some rigorousness here due to departure from mathematical notation, but I hope I have understood the idea properly.

I also wanted to correlate the assumption with the note 1 given at the end. From your explanation, ##\lambda g (F,x) = g (\lambda F, x)## means the same as the note, right?
 
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  • #6
Shreya said:
I hope I have understood the idea properly.
Yes, you get the idea.
Shreya said:
From your explanation, ##\lambda g (F,x) = g (\lambda F, x)## means the same as the note, right?
My explanation is an attempt to use the hint to arrive at the answer. Whether that's what the author had in mind I cannot be sure.
 
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  • #7
Thank you so much @haruspex for helping me out again. I had tried many resources to understand this question and all that had failed. You are doing a great help to all students in the world
 

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