Moment of a force about a point

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In defining the moment of a force about a point as "the tendency of one or more applied forces to rotate an object about an axis [going through a point, hence also about a point]", I see it logical to infer that it is somehow related to the angle between the force and the moment arm, but why put it directly proportional to the sine of that angle? Why not ##\ln^n(1+\theta)## for example? This will become zero as the angle tends to zero.

Furthermore, why have the magnitude of the moment arm and the force in the definition? Maybe so that to underline that for two forces of different magnitudes this tendency to rotate is different? And I guess the same for the moment arm's length?

Perhaps there is no aim to describe how this tendency "changes" as a function of the angle but we just want some formula that would tell us that for a given two forces and two moment arms, one is bigger than the other?
 
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archaic said:
I see it logical to infer that it is somehow related to the angle between the force and the moment arm, but why put it directly proportional to the sine of that angle?
Because it matches observation.
 
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A.T. said:
Because it matches observation.
You observe a tendency to rotate? It seems to me that it is more of a definition, not a natural "law".
 
archaic said:
You observe a tendency to rotate?

We observe that using sine in the definition of torque gives us quantity that is very usefull in predicting behaviour of zilions of mechanical systems and using other functions of angle doesn't. I remember that Feynman in his lectures gave somewhat alternative (with respect to most of textbooks I have seen) discussion and justification for the formula we use, you can give it a try.
 
A.T. said:
That's describing how fast the angular velocity is changing. I am sorry, I don't get your point.
If you want to say the presence of an angular acceleration tells us that there is rotation, then I can point out constant angular velocity, so that if you interpret the angular acceleration as a tendency to rotation in the same sense as moment of force, then it fails in this case.
 
weirdoguy said:
We observe that using sine in the definition of torque gives us quantity that is very usefull in predicting behaviour of zilions of mechanical systems and using other functions of angle doesn't. I remember that Feynman in his lectures gave somewhat alternative (with respect to most of textbooks I have seen) discussion and justification for the formula we use, you can give it a try.
I'll try to find it in the book, thank you. But then it is as I said, just a useful definition (i.e what I have said in the last question is true?)?
 
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archaic said:
You observe a tendency to rotate? It seems to me that it is more of a definition, not a natural "law".
There's much more to it than that. The 'observations' involve measurements (many of them) and the notion of Equilibrium (balance).
it makes me twitchy when the phrase "natural law" is used, as if there has to be some design involved. Many things in Science follow predictable and consistent behaviour and that's all that a Law implies. That word is still used but it is really only used as a shorthand term for repeatable behaviour, these days.
The 'definition' of a moment is used to compare two or more torques, acting on a body and they follow the same sort of 'observed rule' / 'Law' that Newton's First and Second Laws of Motion describe in translational dynamics.
I have great sympathy with students dealing with the meaning of "Perpendicular Distance" when they start with Moments. They nearly all get it wrong to start with. But I don't think there is a better term for it - except perhaps 'effective radius'. Too late to change now, though.
 
archaic said:
That's describing how fast the angular velocity is changing.
Yes, and that is proportional to torque, as commonly defined. That's why torque is defined this way, not some other way. Wasn't that your question?
 
It's worth pointing out that the Principle of Moments concerns the equilibrium condition where there is no actual rotation or change of rotation. So MI doesn't come into it under that condition. Angular acceleration would be zero.
Students start off with the see-saw for good reasons.
 
sophiecentaur said:
It's worth pointing out that the Principle of Moments concerns the equilibrium condition where there is no actual rotation or change of rotation. So MI doesn't come into it under that condition. Angular acceleration would be zero.
Students start off with the see-saw for good reasons.
Yes, one can deduct the same useful definition of torque from static experiments as well.
 
archaic said:
I see it logical to infer that it is somehow related to the angle between the force and the moment arm, but why put it directly proportional to the sine of that angle? Why not ##\ln^n(1+\theta)## for example? This will become zero as the angle tends to zero.
It has to give the correct answer, not only for an angle of zero but also for right angles and all other angles. That is what the sine does.
 
One can put a torque wrench on a lug nut and measure the torque that arises from pushing the end of the wrench with a variety of forces at a variety of angles.

It is hard to imagine much confusion arising from someone who has actually used a wrench.
 
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jbriggs444 said:
It is hard to imagine much confusion arising from someone who has actually used a wrench.
I'm not sure that everyone who uses wrenches and levers is actually aware, in a formal sense, of the effect of the actual angle of the force they are applying. They 'feel' what they are doing but do they appreciate the details, I wonder.

There is also the very confusing evidence of the geometry of hydraulic crane jibs and 'digger' arms. Those angles can be so acute that it's hard to comprehend just how extreme the forces must be, that the rams are producing. The 'perpendicular distance' is there, in front of you, but intuition can get in the way.
 
archaic said:
but why put it directly proportional to the sine of that angle?

Two reasons that are equivalent to each other:

1. The lever arm equals ##r \sin \theta##.
2. The component of the force that's perpendicular to the lever arm equals ##F \sin \theta##.
 
archaic said:
why put it directly proportional to the sine of that angle? Why not lnn(1+θ)lnn⁡(1+θ)\ln^n(1+\theta) for example? This will become zero as the angle tends to zero.
Arm waving arguments don't apply very well in Physics - especially when you want an actual answer and a measurable quantity from the exercise.
How about this? Levers and wheels behave the same way. You can replace a lever with a (rigid) disc or pulley. Mechanisms with levers behave the same as trains of gears and pulleys and the same principle of moments applies. If you pull a string wrapped round a pulley, the force will always be tangential to the pulley and the moment will always be rF, where r is the radius of the pulley. A lever doesn't automatically do this and you can choose any angle to pull the string at. But the moment will be the same as for a pulley with a radius which just touches the string. The r of the disc is pulley length sinθ (by simple geometry). The r of the disc is the 'perpendicular distance' in the lever formula. Only the sine function gives the right answer. Many other functions go through the origin but none of the others will give the right answer.

PS You have to pull a lever with a string if you want to be sure of the direction of th force you are applying. Turning a spanner / wrench usually involves the wrist and arm and you cannot be sure exactly what force or torque you are applying - so things can be counter-intuitive.

EDIT: PPS the perpendicular distance will actually change as the lever moves - unlike the result for a round pulley.
 
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