1
$\begingroup$

I am studying systems of pulleys and ropes. For simplicity, these pulleys and ropes are defined to be massless. But if they are massless, then how do they accelerate? According to Newton's first law, a body can accelerate only if there is a net force, but if the mass of a body is zero, then $F = 0(a) = 0.$ So in this case, the body accelerates even though the net force is zero. Isn't this a violation of Newton's first law? What is going on here?

$\endgroup$
1
  • 4
    $\begingroup$ Because the other masses that the ropes are holding together accelerate. It is a learning exercise. $\endgroup$
    – Jon Custer
    Commented Sep 10, 2016 at 21:07

4 Answers 4

3
$\begingroup$

Perhaps a better statement is "the mass of the ropes is very much less than the other masses involved in the problem"?
So to accelerate such a rope there must be a net force acting on the rope but that net force is much less than other forces which are acting.

In your studies you will find many such assumptions made to make the problem easer to handle.
Frictionless surface (frictional forces much less than the other forces which are acting), inextensible string (extension of string much less than the distances moved by the masses), point mass (mass small enough so that positions of application of forces does not matter very much) etc.

If such approximations are not made then the analysis of a problem becomes more difficult but it may be important that more accurate solutions are needed.

$\endgroup$
3
$\begingroup$

When we say "assume the rope is massless," it's really an idiom that's letting you know what sort of simplifying assumptions you will be permitted to make. A more rigorous definition is that we are exploring the behavior of the system as the mass of the rope becomes infinitesimally small. If you know calculus, we're taking the limit as the mass of the rope approaches zero.

In cases where you will be told to assume the rope is massless, you could be more rigorous and assign a mass to the rope. You could then look at the behavior of the system as that mass approaches 0. For example, if you're using a rope to pull a block, the force you are applying on the rope has to accelerate both the block and the rope itself. However, if the rope has an infintessimal mass, the acceleration of the system will be dominated by the mass of the block. In fact, you can show that the result of this would be the same as though you simply transfered the force directly to the block!

In the case of Newton's first law, it applies to objects which have mass. This object would clearly have to behave differently. One formal approach to this is to treat the behavior of the rope as a constraint on the system, rather than an actual member of the system. We might say "the block must accelerate at the same rate as the object on the other side of the rope." It just so happens that, if you do Newton's first law with a mass-ful rope, and take the limit as that mass goes to 0, you will get the exact same behavior.

In real life situations, all you really have to know is that the rope's mass is negligible. In a large number of situations, that assumption is actually quite reasonable. In some situations, you won't be so lucky. As an example, if you are a crane operator, lifting a heavy object, you must be aware that your cable stretches under load. You cannot pretend that it is a simple constraint.

$\endgroup$
0
$\begingroup$

you assume rope to be massless because its mass is very small in comparison of the masses attached with the rope,and if you have one equation in which one variable is very small in comparison of the other variables then you could neglect that.By following the argument only you could say that the rope is massless.

$\endgroup$
0
$\begingroup$

rope is not massless, its mass is too small that it can be neglected.As far as your question is concerned that why it have acceleration my answer is that t is part of a system which is moving with a certain acceleration so it acquire s acceleration of the system.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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