- #1

prosteve037

- 110

- 3

**I was under the impression that, by experiment, Newton deduced**

[itex]\textit{F}\propto{m}[/itex] [itex]\rightarrow[/itex] [itex]\textit{F = k}_{1}\textit{m}[/itex]

(where [itex]\textit{k}_{1}[/itex] is some constant)

**and**

[itex]\textit{F}\propto{a}[/itex] [itex]\rightarrow[/itex] [itex]\textit{F = k}_{2}\textit{a}[/itex]

(where [itex]\textit{k}_{2}[/itex] is some constant)

**and then found that either/both**

[itex]\textit{k}_{1}\propto{a}[/itex] [itex]\rightarrow[/itex] [itex]\textit{k}_{1}\textit{ = c}_{1}\textit{a}[/itex]

(where [itex]\textit{c}_{1}[/itex] is some constant)

**and/or**

[itex]\textit{k}_{2}\propto{m}[/itex] [itex]\rightarrow[/itex] [itex]\textit{k}_{2}\textit{ = c}_{2}\textit{m}[/itex]

(where [itex]\textit{c}_{2}[/itex] is some constant)

**thus creating**

[itex]\textit{F = c}_{1}\textit{ma}[/itex]

and/or

[itex]\textit{F = c}_{2}\textit{ma}[/itex]

**where in SI Units they would be in the form**

[itex]\textit{F = ma}[/itex]

**However, I've read in some other forums how Newton actually meant**

[itex]\textit{F}\propto{ma}[/itex] [itex]\rightarrow[/itex] [itex]\textit{F = kma}[/itex]

**Which is the case? Did he use the first method or did he simply state the second?**

If he did use the first method, how did he resolve that[itex]\textit{k}_{1}[/itex]

If he did use the first method, how did he resolve that

**is dependent on acceleration and/or that**[itex]\textit{k}_{2}[/itex]

**is dependent on mass?**