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Why do we need Newton's First law? And how the First law works? 
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#19
Jun2714, 01:43 PM

P: 120

EDIT: Provide the answer please. 


#20
Jun2714, 03:17 PM

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PF Gold
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#21
Jun2714, 03:59 PM

P: 462




#22
Jun2714, 04:05 PM

P: 120

I am sorry to say, but it seems like I don't know enough about inertial and noninertial frames and rotational motion to continue following this discussion properly (though I started this thread )
For example: I cant simply figure out How one person in a noninertial frame can measure force on an object properly? (I mean avoiding fictitious forces.) But thank you all for your replies. Meanwhile I should go and do some study! EDIT: ... thanks DrS! I just had your reply. To use the third law, I think you mean, I need to find if that accelerating object is giving any reaction force...? 


#23
Jun2714, 04:33 PM

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#24
Jun2714, 05:07 PM

P: 120

Suppose I put a ball on a table and the ball starts moving. It's certainly moving without a counter force. So I decide my frame is non inertial.
But what if there was actually a counter force. Suppose someone with an almost invisible string attached to the ball was moving it around? So unless I can be sure about all the forces acting on the ball and by the ball, how can I tell if the frame was inertial or not? And in a certain situation it might not be possible for me to make sure about all the forces acting on and by an object. As you said too  Also it seems like Newton's first law is not enough to check for an Inertial frame. Because to exclude the fictitious forces We also need the Third law. What's happening? 


#25
Jun2714, 05:27 PM

PF Gold
P: 1,145

How so? Let the position of the particle at time ##t_0## be ##x_0##. Now imagine the force ##F## is a function of position ##x## only and ##F(x_0) = 0##. What will happen to the particle put at ##x_0##? Based on the 1st law, we conclude the particle will stay at ##x_0## forever, because there is "no impressed force to compel the particle to change its state of rest". Based on the 2nd law only, we cannot make such conclusion. This is because in its modern formulation, the 2nd law is this statement: mass times second derivative of position equals net force impressed: $$ m\frac{d^2 x}{dt^2} = F. $$ This equation connects force only to the second derivative of the position and disregards any other characteristics of motion. Due to this, in this case it is not sufficient to determine what will happen to the position of the particle: depending on the function ##F(x)##, there may be infinity of solutions distinguished by the time ##t_1## the particle begins to move. 


#26
Jun2714, 05:30 PM

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#27
Jun2714, 05:38 PM

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#28
Jun2714, 05:57 PM

PF Gold
P: 1,145




#29
Jun2714, 06:13 PM

P: 462

To my knowledge the first law and the original wording of the second law differ in a single special case only: In contrast to the first law [itex]F = \dot p = m \cdot \dot v + v \cdot \dot m[/itex] allows accelerations without force for [itex]\dot v =  v \cdot \frac{{\dot m}}{m} \ne 0[/itex] I don't know if this was intended by Newton but this means that the use of forces for open systems might be problematic. 


#30
Jun2714, 07:25 PM

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PF Gold
P: 5,583

Newton's first law, in its original formulation, is a special case of the second law. Newton didn't intend his laws to be logically independent, and it's not possible to read them as being independent. He probably just wrote the first law separately in order to emphasize that he was making a break with Aristotelianism.
Ernst Mach wrote a book, The Science Of Mechanics, 1919, http://archive.org/details/scienceofmechani005860mbp , which critiqued the logical basis of Newton's laws. Most people hear about Mach only through Mach's principle, which was not formulated precisely until long after Mach's death, so they get the impression that Mach was some kind of fuzzyheaded philosopher. But his critique of Newton's laws was very logically sound, and was influential. Probably influenced by Mach, some modern textbook authors started presenting Newton's laws in a rewritten form, with the first law rewritten as a statement that inertial frames exist. (It's not just a definition, it's an existence claim.) It's only in this rewritten form of Newton's laws that the first law is not a special case of the second. 


#31
Jun2814, 02:58 AM

PF Gold
P: 1,145

For example, according to the 2nd law, particle put at rest on top of a sphere at instant ##t_0## may begin to change its velocity at any subsequent time ##t_1##. The 2nd law does not determine ##t_1##. According to the 1st law, the particle will stay put. $$ F = \dot{m}v + m\dot v. $$ It is $$ F = m\dot{v}. $$ In the last form, the 2nd law applies also to systems with variable mass. The form ##F=\frac{dp}{dt}## is correct only if in addition, ##m## is assumed constant. 


#32
Jun2814, 06:32 AM

P: 462

http://cudl.lib.cam.ac.uk/view/PRADVB0003900001/46 and the definition of momentum http://cudl.lib.cam.ac.uk/view/PRADVB0003900001/26 it is. 


#33
Jun2814, 07:28 AM

PF Gold
P: 1,145

http://www.pitt.edu/~jdnorton/Goodies/Dome/ Norton argues that actually the first law is valid even here, but he uses different "1st law" than the one Newton wrote. I only say that Newton's first law predicts something that Newton's 2nd law does not: that the particle will stay at the top. But it shows that Newton's first law is different kind of law: it is not a mathematical equation, but causality statement. Unfortunately, it is impossible to check this distinction experimentally. But Newton wrote a whole book on this. In the section on definitions you cited, he deals with "body", which most of the time means that mass is assumed constant. I did not read whole of Newton's book, but I am sure he did not meant actually to include the term ##\dot{m}v## from the derivative of momentum as a part of the equation of motion. That would lead him to incorrect results and I'm sure he would check before publishing them. His 2nd law is thus correct only if in addition mass of the body is assumed to be constant, something he forgot/did not care to say or he said it elsewhere. 


#34
Jun2814, 10:16 AM

P: 462

[itex]r = \frac{{t{}^4}}{{144}}[/itex] Edit: Now I got it. He mixed the timedependent solution for the initial condition ##r(T) = 0## with the trivial solution ##r(t) = 0## and spliced them at ##t = T##. The resulting hybrid is still a solution of the differential equation and meets the initial condition ##r(0) = 0##. There is either no force and no acceleration (for ##r=0##) or both force and acceleration (for ##r \ne 0##). Whether this is a violation of the first law or not seems to be a matter of interpretation. The second law does not care about causality but gives a quantitative definition of force. 


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