Determine if the equation given is dimensionally correct:
$\sf F=m(v^2)/(d^2)$

Question

Determine if the equation given is dimensionally correct: 
$\sf F=m(v^2)/(d^2)$

anonymous · Answer

@dan815

anonymous · Answer

@Abhisar

abhisar · Answer

Ok, so first of all you know what is the dimension of force right?

anonymous · Answer

Not really.  This is my pre-physics summer work.

anonymous · Answer

Newtons?

anonymous · Answer

ML/T^2

abhisar · Answer

Yes, dimension of force = $\sf  M^1L^1T-2$

Do you understand how?

anonymous · Answer

F=ma, and acceleration=L/T^2

abhisar · Answer

$\huge \checkmark$

abhisar · Answer

So, in order to determine if the given equation is dimensionally correct we just need to make sure that same dimension occurs on both hand sides of the equation. So in RHS of equation the dimension should also be $\sf  M^1L^1T^{-2}$

anonymous · Answer

Ok.

abhisar · Answer

RHS = $\sf \Large \frac{mv^2}{d^2}= \frac{mass \times (metre^2)}{second^2\times distance^2}$

Agree?

anonymous · Answer

what?  Okay, I have:
$$\frac{ mv^2 }{ d^2 }$$ would equal $$\frac{ M(L/T^2) }{ L^2 }$$

abhisar · Answer

$\sf V=L/T \ \Rightarrow V^2 = (L/T)^2=L^2/T^2$

abhisar · Answer

So, that would give you for RHS 

$$\frac{ M(L^2/T^2) }{ L^2 }$$ = $\sf M/T^2$

anonymous · Answer

yeah sorry i meant $$\frac{ M \frac{ L^2 }{ T^2 } }{ L^2 }$$

abhisar · Answer

Yup, what can you say now, Is LHS = RHS?

anonymous · Answer

Yes?  They've got the same letters...

abhisar · Answer

LHS=Force= $\sf  M^1L^1T^{-2}$

RHS=$\huge \frac{ M(L^2/T^2) }{ L^2 }$ = $\sf M/T^2$

Are they both equal?

anonymous · Answer

No, there's no L in the RHS.  If there was, would they be equal?

abhisar · Answer

Yes.

abhisar · Answer

But since, RHS$\neq$LHS, the equation is dimensionally incorrect but if somehow they both were equal then the equation would have been dimensionally correct.

anonymous · Answer

so the letters just have to be there or does the division, etc. matter as well?

abhisar · Answer

Division matters, treat them like algebraic variables.

anonymous · Answer

thank you!

abhisar · Answer

Do you want me to give you some examples for practice?

anonymous · Answer

I have a few more to do in the packet.  Hopefully I'll get it before class tomorrow.  I think I understand better now.  Thanks again

abhisar · Answer

You're most welcome c:

anonymous · Answer

@Abhisar Would L=2.5L ?