0
$\begingroup$

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?

$\endgroup$
3
  • 3
    $\begingroup$ Your magical massless string takes no force to accelerate. $\endgroup$ – M. Enns Apr 24 '16 at 21:00
  • $\begingroup$ and how's that ? $\endgroup$ – Shashaank Apr 24 '16 at 21:04
  • $\begingroup$ we are basically discussing the same issue over here: physics.stackexchange.com/questions/251724/… $\endgroup$ – Ilja Apr 24 '16 at 21:05
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.