本帖最後由 【YU】 於 19-2-2009 20:14 編輯
Newton's 2nd law
The rate of change of momentum of a body is proportional to and in the same direction as the resultance force that acts to it.
F = k d(mv)/dt = ma, where k = 1 <~ do you mean it is the special case when Lorentz factor = 1? or the special case when m is constant? or both? or no formula can be equivalent to the law itself?
F = ma is the general case for applying 2nd law ( i have not said that it is the origin, and i don't think it is important to distinguish between origin and implied result, when both of them can be used directly)
F = d (mv)/dt = m dv/dt = ma
it can be directly applied, and is the result explained by the 2nd law.
when answering mechanic question, we just wrote:
By Newton's 2nd law, F = ma....
Nobody will write the word definition first, and then prove from F = m(v-u)/t = ma everytime.
Even you will not do like this la? so don't condemn siuwing for his/her answer.
so what case do you think"F=m(dv/dt)+v(dm/dt)" is?
本帖最後由 Puppet 於 19-2-2009 20:08 編輯
21# 【YU】
From 2nd law, mathematically,
F = k(dp/dt) or F = dp/dt (take k = 1) (general case directly from 2nd law)
F = d(mv)/dt = m(dv/dt) + v(dm/dt) (general case)
N.B. k = 1 is just for the definition of the Newton unit.
For constant m (a certain condition), F = m(dv/dt) = ma
Do you know that there are cases in which dm/dt is not zero and 2nd law needs to be applied?
Still F = ma is the general case?
special case: F = ma
most commonly used case: F = ma
I just want to clarify which the general case is.
This does not matter with convenience or practicality.
"Most of the time m is constant" does not mean anything.
Also, I am clarifying that "2nd law states that F=ma" is improper because it does not state that.
"By 2nd law, F=ma" is okay because F=ma is derived from 2nd law (as long as it is obviously known that m is constant in that case).
By 2nd law, we know that/arrive at F=ma for m is constant. However, 2nd law does not state that F=ma.
c.f. by Avogadro's law, 2 moles of helium occupy 48 dm3 at r.t.p.
本帖最後由 【YU】 於 19-2-2009 20:23 編輯
When v -> c , dm/dt =/= 0
it is added after Albert's theory, newton 2nd law itself does not consider this case.
The dm/dt is an expansion from the original 2nd law.
general can also means most commonly used.[]
本帖最後由 Zend 於 19-2-2009 20:37 編輯
in the example , it is the mass changed due to addition/removal of mass from an moving object.
like rockets , when u give a upthrust force , if no gas is emitted, the F used is very big -.-
but when u emit the gas at the same time, with a rate of dm/dt(negative in value)
then the force used will be smaller, which is provided by the equation
F=mdv/dt+vdm/dt
and this is not related to Einstein's relativity...(the mass changed equation is also different)
it is the rate of mass emitted/added.
When v -> c , dm/dt =/= 0
it is added after Albert's theory, newton 2nd law itself does not consider this case.
The dm/dt is an expansion from the original 2nd law.
general can also means most commonl ...
【YU】 發表於 19-2-2009 20:20 http://www1.nakuz.com/bbs/images/common/back.gif
Too difficult for me la.
I just know we can have the rocket system where dm/dt is not 0, and empirical cases like pushing a piece of melting ice.
Where do you get dm/dt is an expansion from original 2nd law?
Does your "original 2nd law" actually mean F=ma?
歧義?
F=ma is not special case cuz constant m is not something special??[]
So is F=ma the general (通, applicable to all cases) case??
個MASS的分別
一個係本身物體自己會變， 當速度去到接近光速的運動
一個係比較上人為，係一個移動物體本身的質量減少
不過果個時代牛頓有無做過一路行一路抽的ＥＸＰＥＲＩＭＥＮＴ， 就唔知
但果式係數學推出來的公式，而唔係咩新發現
my orginal 2nd force is just from F = dp/dt with m is constant.
rocket system is ok, but have't calculate with dm/dt b4, so i am curious with his formula.
物理大師們激戰喎 我發現我原來唔識睇英文啦.....
樓主比小小時間去研究一下佢地講哥d野
不過後期哥d比較深 同埋d得太長哥條second law唔洗理住
用返f=ma夠用架啦 唔好諗得咁深
再深入=vector calculus
再深入=vector calculus
Zend 發表於 20-2-2009 02:01 http://www.nakuz.com/bbs/images/common/back.gif
我好似有個course讀過d到好變態架ma
一係分xyz 去partial d一係就 alpha 哥d用angle去d
yu哥有無take過1201