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Anonymous Wed 24 Mar 12:02:07 2021 No.12876467 View Reply Original Report

Quoted By: >>12876531 >>12876536 >>12876598 >>12876604 >>12876768 >>12876922

I dare you to mathematically and rigorously prove that F=ma

Anonymous

Anonymous Wed 24 Mar 2021 12:27:53 No.12876525 Report

Quoted By:

Sheeet ino bruh shits just is

Anonymous

Anonymous Wed 24 Mar 2021 12:28:13 No.12876527 Report

Quoted By:

Lagrange already did that

Anonymous

Anonymous Wed 24 Mar 2021 12:28:31 No.12876531 Report

Quoted By:

>>12876467
>mathematically and rigorously prove that F=ma

You just did.

Anonymous

Anonymous Wed 24 Mar 2021 12:29:38 No.12876536 Report

Quoted By:

>>12876467
F=ma
There, i just proved it.

Anonymous

Anonymous Wed 24 Mar 2021 12:50:57 No.12876598 Report

Quoted By:

>>12876467
force is like watts, its a measurment. You don't have to prove f=ma because its just a simple equation. The same way we do math at all is the proof

Anonymous

Anonymous Wed 24 Mar 2021 12:52:26 No.12876604 Report

Quoted By:

>>12876467
are you retarded

you cant prove an imaginary force, its used because its comfortable

Anonymous

Anonymous Wed 24 Mar 2021 13:46:33 No.12876750 Report

Quoted By:

okay then prove, $F=G\frac{Mm}{r^2}$

Anonymous

Anonymous Wed 24 Mar 2021 13:49:44 No.12876756 Report

Quoted By:

Definition.
F=ma.

Proof.
True by definition.
QED.

Anonymous

Anonymous Wed 24 Mar 2021 13:54:43 No.12876768 Report

Quoted By:

>>12876467

It is a postulate

Anonymous

Anonymous Wed 24 Mar 2021 15:02:05 No.12876922 Report

Quoted By:

>>12876467
$kg m/s^2 = (kg) \times m/s^2$

Fucking idiot

Anonymous

Anonymous Wed 24 Mar 2021 15:15:54 No.12876965 Report

Quoted By:

We know that $\mathbf p\;=\;m\,\mathbf v$ by definition. Suppose $m$ is constant and $\mathbf v$ is differentiable with respect to time, with $\dot{\mathbf v}\;=\;\mathbf a$ . Then for all time $t$ and time difference $\Delta t\;\neq\;0\;\mathrm s$
$\frac1{\Delta t}\,\left(\mathbf p\left(t\,+\,\Delta t\right)\,-\,\mathbf p\left(t\right)\right)\;=\;\frac1{\Delta t}\,\left(m\,\mathbf v\left(t\,+\,\Delta t\right)\,-\,m\,\mathbf v\left(t\right)\right)\;=\;m\,\frac1{\Delta t}\,\left(\mathbf v\left(t\,+\,\Delta t\right)\,-\,\mathbf v\left(t\right)\right)\;\xrightarrow[\Delta t\,\to\,0]{}\;m\,\mathbf a\left(t\right).$
Therefore, $\dot{\mathbf p}\;=\;m\,\mathbf a$ . By Newton’s second law of motion, $\dot{\mathbf p}\;=\;\mathbf F$ . By symmetry and transitivity, $\mathbf F\;=\;m\,\mathbf a$ , QED.

Next time, do your own homework.

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