Suppose we are just pulling a block with the help of a string which is massless.
Now since the mass of the string is 0 the force exerted by us is transmitted without being diminished. So the equation of block is $F=Ma$ (supposing its Mass be $M$ and measured acceleration (here a constant) be $a$
We are continuously applying the force and the string is never slack.
So the entire system accelerates with $a$.
Equation of string is then $F=ma$ but $m=0$ so $F=0$. But the string is accelerating (as the system is accelerating) so a non zero force has to act on it. So why is $F=0$ by the above equation?
-
3$\begingroup$ Your magical massless string takes no force to accelerate. $\endgroup$– M. EnnsCommented Apr 24, 2016 at 21:00
-
$\begingroup$ and how's that ? $\endgroup$– ShashaankCommented Apr 24, 2016 at 21:04
-
$\begingroup$ we are basically discussing the same issue over here: physics.stackexchange.com/questions/251724/… $\endgroup$– IljaCommented Apr 24, 2016 at 21:05
2 Answers
The net force on the string is not $F$. You are pulling the string forward with force $F$ but I think you are forgetting that the block is pulling the string backwards with a force that is almost equal to $F$. If the masses of the string and block are $m$ and $M$ then for the whole system (string plus block) $F=(M+m)a$. The net force on the string is $F '=ma=\large{\frac{mF}{M+m}}$. As $M \gg m$, $F ' \rightarrow 0$ even though $F$ remains finite and possibly quite large.
You wrote it yourself, $F=ma$, so with zero mass there is zero force needed for a finite acceleration. A finite force gives an infinite acceleration.
... Lol, I just reviewed your answer about the electrostatic field being normal on the surface of a conductor. That's exactly the same, here! With a non-zero force the acceleration would be infinite ($F/m$ with $m=0$), and it would move to an equillibrium position instantaneously.