What exactly is wrong here? (F=ma when Force and acceleration are zero)

[NTesla]

Homework Statement

In Newton's 2nd Law, F = m*a, which implies that m = F/a. Let's take mass as any finite quantity. Now if F = 0, then a = 0. Then m = 0/0 which is undefined, but we already knew m as a finite quantity. So what exactly is wrong in the argument, how come m is undefined and defined at the same time. Something is wrong in this argument, I'm not able to figure out what exactly it is.

Relevant Equations

F = m*a

If F = 0 then a = 0. When the equation is written in the form F = m*a, it appears ok, that whatever the mass be, LHS and RHS of the equation are equal so no problem. But when the same equation is written in the form m = F/a, then m becomes undefined when F = 0 and a = 0. It occurs to me that some definition of division is out of order or maybe something else I'm missing, but at present I can't seem to pinpoint what exactly is the error in the argument.

 

[anuttarasammyak]

In mathematics say
$$c=ab$$
holds for any a, b, c including they are zeros. However
$$a=\frac{c}{b}$$
does not hold for b=0. So the two formula are not equivalent and are equivalent only for the region of devisor { b|$b \neq 0$} .

 

[NTesla]

In mathematics say
$$c=ab$$
holds for any a, b, c including they are zeros. However
$$a=\frac{c}{b}$$
does not hold for b=0. So the two formula are not equivalent and are equivalent only for the region of devisor { b|$b \neq 0$} .

Yes, I agree and I was already clear on that point. But I'm not able to work through as to how is it applicable in indentifying the error in saying that m is finite and undefined at the same time.

 

[anuttarasammyak]

But I'm not able to work through as to how is it applicable in indentifying the error in saying that m is finite and undefined at the same time.

$$0=0 \cdot 1$$ 1 is finite of course. But from
$$1=\frac{0}{0}$$ We are not sure 1 is finite or have a well defined value. We do not have to mind it because the second equation is a false statement.

 

[NTesla]

We are not sure 1 is finite or have defined value. But we do not have to mind because it is a false statement.

This part I don't understand. We are very clear that 1 is a finite number, and since we are clear on that, that's the only reason that we can say that 1= 0/0 is a false statement. Likewise, mass m is a finite number, but the formula m = F/a gives it as undefined. Would it be correct to say that m =F/a =0/0 is a false statement only because F/a is defined only when a !=0 ?

 

[haruspex]

This part I don't understand. We are very clear that 1 is a finite number, and since we are clear on that, that's the only reason that we can say that 1= 0/0 is a false statement. Likewise, mass m is a finite number, but the formula m = F/a gives it as undefined. Would it be correct to say that m =F/a =0/0 is a false statement only because F/a is defined only when a !=0 ?

In manipulating equations, it is often possible to throw away information. You have started with some number, multiplied it by zero to get zero, and now you are asking what number you started with. There is no way to know because whatever number you started with you will now have zero.

 

[anuttarasammyak]

Would it be correct to say that m =F/a =0/0 is a false statement only because F/a is defined only when a !=0 ?

We use mathematics to express physics. In mathematics we cannot define "division by zero" in a logical way. Isn't that enough ?

 

[anuttarasammyak]

Say a free body of unknown mass m is at rest in a IFR. We just see it. Applied force F=0 and acceleration of the body a=0. Then how can we know how much m is ? This situation would correspond in mathematics that 0/0 is indefinite. m could be any value.

[EDIT]
@NTesla I try to restate what you think.
-----------------
Say there is an object. We put 1N force on it and it gets acceleration of 1 m/s^2. Since m=F/a we know m=1 kg. Then we put this object free at rest in an IFR. Since m=F/a, m=0/0 so m becomes indefinite.
Should the law m=F/a be universal so that it gives same value of m for any situation ?
-----------------
m=F/a is not universal because in mathematics it is not defined for a=0.
F=ma which covers the case a=0 is better.

 

[malawi_glenn]

F = m*a, which implies that m = F/a.

WRONG

F = ma and a ≠0 implies m = F/a

 

[nasu]

@NTesla Why did you pick this specific formula (Newton's second law) to start with? The same can be said about any formula of the type a=bc. Your questions has nothing to do with physics, it is just a matter of mathematical missunderstanding.

 

[Chestermiller]

Have you tried applying L'Hopital's rule to the limit at 0?$$\frac{F(a)}{a}|_{a=0}$$

 

[Malapine]

Say a free body of unknown mass m is at rest in a IFR. We just see it. Applied force F=0 and acceleration of the body a=0. Then how can we know how much m is ? This situation would correspond in mathematics that 0/0 is indefinite. m could be any value.

You can't know what m is in that situation: you would have to apply a known force to it and measure the acceleration.

 

[willem2]

Homework Statement:: In Newton's 2nd Law, F = m*a, which implies that m = F/a.

It does not imply that if a = 0. no need to go any further

 

[Lnewqban]

Homework Statement:: In Newton's 2nd Law, F = m*a, which implies that m = F/a. ....
at present I can't seem to pinpoint what exactly is the error in the argument.

It seems to me that the interpretation of the dependence is misleading.

The second law describes that dependence:
“The change of motion of an object is proportional to the force impressed; and is made in the direction of the straight line in which the force is impressed.”

The mass exists independently from the force, and is never zero, if a force can be applied on it.

The proper relation is $a=F/m$

Acceleration is result, a consequence, a function of mass and force, not the other way around.

 

[kuruman]

I should point out that $m=\dfrac{F_{net}}{a}$ becomes the mathematician's anathema $m=\dfrac{0}{0}$ when the net force on an object is zero and the object is not accelerating. As I am typing this. I have a mass regardless of whether I am accelerating as a result of $F_{net}$ acting on me or not. So does anyone sitting at rest reading this but with a mass different from mine.

There are a lot of equations in physics in the form $C=PE$ that link cause $C$ and effect $E$ in terms of physical property $P$. The physical property P exists independently of the cause C and the effect E. For example, a wire has resistance (P) regardless of whether a current (E) is established by a voltage (C); two conductors in space have capacitance (P) regardless of whether a potential difference (E) between them is established by charges (C) placed on them; a hunk of metal has heat capacity (P) regardless of whether its temperature (E) rises when heat (C) is added to it.

In short, if there is no cause, there is no effect; conversely if there is no effect, there can be no cause. In that case, the equation $C=PE$ cannot be used to determine the physical property P, because nothing happens. The physical property still exists, however it is undefined. That's what $F_{net}=ma$ with $F_{net}=0$ and $a=0$ says, at least to me.

 

[votingmachine]

You can't know what m is in that situation: you would have to apply a known force to it and measure the acceleration.

That is a little too extreme for me. If one knew the mass before, it remains the same, even if it is not being measured. If I step on a scale at 1g and then am launched into space and not being accelerated, my mass is the same as when I had a force to measure it.

I share the bewilderment that simply noting division by zero is undefined should be an adequate explanation as to why the division by zero doesn't lead to a defined answer.

 

[hmmm27]

I share the bewilderment that simply noting division by zero is undefined should be an adequate explanation as to why the division by zero doesn't lead to a defined answer

The "result" is undefined within the context of usability.

Draw the graph $y=1/x$ ; extend both axes into both negative and positive. What are your conclusions about the value of $y$ when $x=0$ ?

Wikipedia take.

 

[vela]

Would it be correct to say that m =F/a =0/0 is a false statement only because F/a is defined only when a !=0 ?

Not really, though it's a subtle point.

When you ask how can you have a defined quantity on one side of the equation equal to an undefined quantity on the other, the question assumes the equation is valid. That's your error. The equation isn't valid because to get there from $F=ma$, you have to have $a \ne 0$. You can't expect an equation derived by assuming $a\ne 0$ to make sense when $a=0$.

Or to put it another way, the implication $F=ma$ leads to $m=F/a$ is false when $a=0$. When $a=0$, $F=ma$ does not imply $m=F/a$. So there's no contradiction with $m$ being defined and $F/a$ being undefined since there's no logical reason to expect the two expressions to be equal.