Deepak Kapur Posted June 9, 2015 Posted June 9, 2015 Newton's second law states (roughly), The rate of change of momentum of a body is equal to the force acting on the body and is in the same direction.... Why 'equal to' and not 'proportional to' In various other formulas we use 'proportional to' and not 'equal to'. What's the difference? 1
ajb Posted June 9, 2015 Posted June 9, 2015 I would more-or-less take that as the definition of a force \frac{dp}{dt} = F. Why add some constant here to make the definition more complicated?
Strange Posted June 9, 2015 Posted June 9, 2015 On 6/9/2015 at 8:41 AM, Deepak Kapur said: Why 'equal to' and not 'proportional to' How about: Quote Newton's second law of motion can be formally stated as follows: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object. (http://www.physicsclassroom.com/class/newtlaws/Lesson-3/Newton-s-Second-Law) It is "equal" when you use a system of units that makes the constants of proportionality equal to 1. 2
ajb Posted June 9, 2015 Posted June 9, 2015 On 6/9/2015 at 9:11 AM, Strange said: It is "equal" when you use a system of units that makes the constants of proportionality equal to 1. This is an important point; one can get rid of awkward constants by picking sensible units.
studiot Posted June 9, 2015 Posted June 9, 2015 (edited) Did you guys read my mind? I just finished a mammoth post on this subject in your last thread Deepak. +1 for encouragement. Edited June 9, 2015 by studiot
Deepak Kapur Posted June 9, 2015 Author Posted June 9, 2015 On 6/9/2015 at 9:06 AM, ajb said: I would more-or-less take that as the definition of a force \frac{dp}{dt} = F. Why add some constant here to make the definition more complicated? Then...i think....it shouldnt be called a LAW.. Why didnt Newton use proportionality for this because Studiot tells me that proportionality was much prevalent those days than the concept of an equation... On 6/9/2015 at 9:11 AM, Strange said: How about: (http://www.physicsclassroom.com/class/newtlaws/Lesson-3/Newton-s-Second-Law) It is "equal" when you use a system of units that makes the constants of proportionality equal to 1. Does it mean there is really no difference between proportinality and equality....its just a matter of convenience and perspective?
Strange Posted June 9, 2015 Posted June 9, 2015 On 6/9/2015 at 10:01 AM, Deepak Kapur said: Then...i think....it shouldnt be called a LAW.. Why didnt Newton use proportionality for this because Studiot tells me that proportionality was much prevalent those days than the concept of an equation... What he actually wrote was: Quote LAW II A change in motion is proportional to the motive force impressed and takes place along the straight line in which that force is impressed. Actually, what he wrote was in Latin. This translation is from I. Newton, The Principia, Mathematical Principles of Natural Philosophy, translated into English from the Latin of the third (1726) edition by I. Bernard Cohen and Anne Whitman, assisted by Julia Budenz (University of California Press, Berkeley, 1999) Quote Does it mean there is really no difference between proportinality and equality....its just a matter of convenience and perspective? We can use = rather than \propto when the constant of proportionality is 1.
swansont Posted June 9, 2015 Posted June 9, 2015 On 6/9/2015 at 10:01 AM, Deepak Kapur said: Does it mean there is really no difference between proportinality and equality....its just a matter of convenience and perspective? F=dP/dt is true regardless of the unit system you choose.
BR-549 Posted June 14, 2015 Posted June 14, 2015 What makes it confusing is the term P. The self P (angular momentum) is always there and can change without changing physical position. The translational P (linear momentum) may or may not be present. Also, if the translational P is at a high enough rate of change.......the translational P is transferred (thru resonance) to the self P. Self P is mass. In the formula, F = M X A, the velocity of the force determines which term, M or A, increases. For a slow F, A increases. For a fast, short F, M increases. If you have a fast long F, the self P (mass) and the translational P will alternate in steps. The reason that the F can be proportional or equal is because of the velocity of the force.
J.C.MacSwell Posted June 14, 2015 Posted June 14, 2015 On 6/14/2015 at 3:51 PM, BR-549 said: What makes it confusing is the term P. The self P (angular momentum) is always there and can change without changing physical position. The translational P (linear momentum) may or may not be present. Also, if the translational P is at a high enough rate of change.......the translational P is transferred (thru resonance) to the self P. Self P is mass. In the formula, F = M X A, the velocity of the force determines which term, M or A, increases. For a slow F, A increases. For a fast, short F, M increases. If you have a fast long F, the self P (mass) and the translational P will alternate in steps. The reason that the F can be proportional or equal is because of the velocity of the force. It's not confusing (at least…it wasn't). P is translational momentum.
swansont Posted June 14, 2015 Posted June 14, 2015 On 6/14/2015 at 3:51 PM, BR-549 said: What makes it confusing is the term P. The self P (angular momentum) is always there and can change without changing physical position. The translational P (linear momentum) may or may not be present. Also, if the translational P is at a high enough rate of change.......the translational P is transferred (thru resonance) to the self P. Self P is mass. In the formula, F = M X A, the velocity of the force determines which term, M or A, increases. For a slow F, A increases. For a fast, short F, M increases. If you have a fast long F, the self P (mass) and the translational P will alternate in steps. The reason that the F can be proportional or equal is because of the velocity of the force. Wrong stuff is often confusing. It's hard to find anything in there that's right.
Phi for All Posted June 14, 2015 Posted June 14, 2015 On 6/14/2015 at 3:51 PM, BR-549 said: What makes it confusing is the term P. The self P (angular momentum) is always there and can change without changing physical position. The translational P (linear momentum) may or may not be present. Also, if the translational P is at a high enough rate of change.......the translational P is transferred (thru resonance) to the self P. Self P is mass. In the formula, F = M X A, the velocity of the force determines which term, M or A, increases. For a slow F, A increases. For a fast, short F, M increases. If you have a fast long F, the self P (mass) and the translational P will alternate in steps. The reason that the F can be proportional or equal is because of the velocity of the force. ! Moderator Note BR-549, please don't introduce speculative ideas in mainstream sections of the site. Students and professionals look here for mainstream answers. We have a section for speculations, please use that instead. And please don't go further off-topic by responding to this modnote. Report it if you have a problem with it.
John Cuthber Posted June 14, 2015 Posted June 14, 2015 (edited) On 6/9/2015 at 12:29 PM, swansont said: F=dP/dt is true regardless of the unit system you choose. I doubt that. It's true for self-consistent units but if I choose to measure force in dynes; momentum in stone furlongs per fortnight and time in jiffies it's going to go to hell in a handbasket. None of that detracts from the fact that the best thing to do with BR-549's "contribution" is to ignore it Edited June 14, 2015 by John Cuthber 1
studiot Posted June 14, 2015 Posted June 14, 2015 Quote None of that detracts from the fact that the best thing to do with BR-549's "contribution" is to ignore it +1
J.C.MacSwell Posted June 14, 2015 Posted June 14, 2015 On 6/14/2015 at 9:04 PM, John Cuthber said: I doubt that. It's true for self-consistent units but if I choose to measure force in dynes; momentum in stone furlongs per fortnight and time in jiffies it's going to go to hell in a handbasket. None of that detracts from the fact that the best thing to do with BR-549's "contribution" is to ignore it Fair enough. You would need some conversion factors. But would that not mean that dynes, stone furlongs per fortnight, and jiffies were already more hell in a hand basket than a system?
swansont Posted June 15, 2015 Posted June 15, 2015 On 6/14/2015 at 9:04 PM, John Cuthber said: I doubt that. It's true for self-consistent units but if I choose to measure force in dynes; momentum in stone furlongs per fortnight and time in jiffies it's going to go to hell in a handbasket. That's not a unit system.
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now