Inverse property of differential calculus of F=Ma

[MolecularEnergy]

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I know how to differentiate F=Ma, so since integration is the inverse process to differentiation, how would one perform integration on the equation F=Ma?

 

Thanks in advance

 

Does no one know?

[Kyrisch]

The same way you differentiate with respect to a specific variable, so you integrate with respect to a particular variable. Integrate with respect to what?

[the tree]

I'm unsure about what you're asking, precisely what are you trying to integrate, and with respect to what? (and between which limits, if that's relevant)

 

edit ninja'd.

[MolecularEnergy]

The same way you differentiate with respect to a specific variable, so you integrate with respect to a particular variable. Integrate with respect to what?

 

 

I really need to be shown. This is how i generally learn. But what if i show the differentiation of F=Ma, so you can see the operations i am using.

[mooeypoo]

F=ma is just a way of representing the actual [math]F=m \frac{dv}{dt}[/math], or [math]F=m \frac{d^2x}{dt^2}[/math]

 

So.. you need to be more specific. Are you integrating with respect to dv/dt ? or dx/dt? or are you integrating with respect to mass (with a system of particles) ?

[MolecularEnergy]

The mass, i do believe

[mooeypoo]

... you have to be more specific.

 

Mass in what volume? In what coordinates? Spherical? Cylindrical? What aer the limits?

 

Can you just post the full question with *all* the details so we could help?

[MolecularEnergy]

The volume is constant. As in, non-relativistic...

 

 

Sorry, if this is very abstract. geometry, however, has very little to do with it...

[mooeypoo]

Why do I have a feeling we're being set up towards a specific answer which then you could talk about?

 

Look. You don't just differentiate or integrate in physics, it's meaningless. You need to take a specific problem and then pick the right mathematical tools of analysis.

 

Give us an example.. where did you encounter the problem where you need to integrate a force/mass?

[MolecularEnergy]

Thank you, mooey, i am not right tonight. But, your help makes feel a bit better.

 

I thought the differential analysis of F=Ma had an inverse relationship, according to my math book...

[mooeypoo]

Inverse relationship, as in

[math]F=ma \dashrightarrow m=F/a \dashrightarrow a=F/m[/math]

?

 

Or

 

[math]F=m\frac{dv}{dt} \dashrightarrow \int \frac{F}{m} dt = \int dv[/math]

 

Or... what? See, there are so many things you can do with this concept.. we need a bit more information to answer your question.

[MolecularEnergy]

So it is basic algebraic manipulation, it speaks of?

 

Sorry, i know not much more, i am affraid... i wish there was some texbook, that simply took me through it.

[mooeypoo]

So it is basic algebraic manipulation, it speaks of?

 

Sorry, i know not much more, i am affraid... i wish there was some texbook, that simply took me through it.

It's not necessarily a basic manipulation... If I don't know more about what you intend on doing, i can't really answer.

 

 

There are textbooks about this, I personally have 3 just from previous courses.. look online, I'm sure there are online resources too.

[Reaper]

F=ma is also a way of expressing F=dp/dt, the change of momentum with respect to time.

[mooeypoo]

Indeed, thanks for that.

[Kyrisch]

Wait, I think I know what he's talking about (why is this so much like catch-phrase? >.<) Newton's Second Law: The acceleration an object undergoes when a constant force is applied thereto is directly proportional to the magnitude of the force and inversely proportional to the mass of the object. So the acceleration has an "inverse relationship" with the force and vice-versa. It is simple algebraic manipulation.

[mooeypoo]

Wait, I think I know what he's talking about (why is this so much like catch-phrase? >.<) Newton's Second Law: The acceleration an object undergoes when a constant force is applied thereto is directly proportional to the magnitude of the force and inversely proportional to the mass of the object. So the acceleration has an "inverse relationship" with the force and vice-versa. It is simple algebraic manipulation.

Well, those are the f=ma -> m=f/a -> a=f/m I wrote above.. he seems to refer to integration/differentiation, though, and we need more details for that.