1
$\begingroup$

I recently studied that Tension in a string is a kind of force originated from electrostatic attraction between the atoms of the string in which the force is originating. My doubt was that:

Assume that I am pulling a rope with a force $F$, and the rope will develop a tension $T$ in itself and will pull me with $T$ but what about the force $F$ with which I started pulling the rope? Where would the reaction pair of this force would be felt? I know that this force is not included in the free body diagram of me (the one who is pulling the rope) as FBD only incorporates the forces acting on a body and not the ones exerted by it and hence I am kind of confused where I can find the reaction pair of $F$ being felt.

|cite|improve this question
$\endgroup$
  • 2
    $\begingroup$ I'm not sure I understand the question. Is the rope attached to something? Are you pulling both ends of the rope? Perhaps a diagram would be necessary here. $\endgroup$ – Matt Jul 14 '16 at 9:06
  • 1
    $\begingroup$ The answer is that tension is an representation of the instantaneous acceleration per unit length in a string, interpreted in a wave equation. Because the tension is defined for every infinitesimal unit length in the string, we can say that the 'pairs' you are referencing are innate in the definition when you take the limit of the string. It was introduced by newton to analyse the motion of a mechanical wave. $\endgroup$ – Cppg Jul 14 '16 at 10:11
  • 1
    $\begingroup$ I don't understand the question either. But maybe this helps: if you grab a rope and pull with a force F, there will be a tension force in the rope of F at the point your hand touches it. The reaction force is the force on your hand due to the rope. It's value is F also. $\endgroup$ – garyp Jul 14 '16 at 11:11
  • 1
    $\begingroup$ @Cppg Tension is defined whether or not a wave is involved. $\endgroup$ – garyp Jul 14 '16 at 11:14
  • 2
    $\begingroup$ @sammygerbil We can use the definition of tension as a reactive force, but you have to be very careful to fully understand what that means. I personally think it's a confusing definition, especially for novices. It sounds like backwards reasoning to me. I prefer (being specific to ropes at the moment) "the force that a rope applies to something else", where that "something else" could be some object, or another section of the same rope. But keep in mind that tension forces arise in systems that don't involve rope, like two objects glued together. $\endgroup$ – garyp Jul 14 '16 at 12:27
0
$\begingroup$

To simplify this answer assume that the string is made up of a line of molecules so each molecule bar the end ones have only two nearest neighbours.

When there is no tension force in a string then on average the molecules which make up the string are at their equilibrium spacing and have no net force acting on them.

Imaging that you apply a force $F$ on the first molecule in the string.
That molecule will exert an equal but opposite force on you - Newton's third law.
That first molecule is attached via a bond (electrostatic interaction) to its neighbouring molecule.
The spacing between the molecules increases and the neighbouring molecule exerts a force on the first molecule.
In turn the first molecule exerts a force on the second molecule - Newton's third law..
Those forces between the molecules we call the tension in the string.

You can liken the situation as having a line of point masses initially being connected by unextended springs.
Pull the end mass and the springs extend thus producing forces between the masses.

If the string is held at a fixed point at the other end and nothing is moving (static equilibrium has been established) then all of the molecules are separated by a distance greater than their equilibrium separation and so have forces on them due to their nearest neighbours but the net force on each of those molecules is zero.

The first molecule which you are pulling with a force $F$ has a force of the same magnitude but opposite in direction acting on you which you call the tension $T$.
The first molecule has a force $F$ acting on it due to you and a force $T(=F)$ acting on it due to the second molecule.

|cite|improve this answer
$\endgroup$
0
$\begingroup$

If the rope is in static equilibrium then $T=F$. If $T\neq F$ then that section of the rope (where tension is $T$) must be accelerating, which may happen if the rope is slack or if it is extensible.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ It will also happen in a straight inextensible rope if the rope is accelerating. $\endgroup$ – garyp Jul 14 '16 at 11:13
  • $\begingroup$ @garyP You are right. I assumed that rope was tied at one end. $\endgroup$ – Deep Jul 14 '16 at 11:19
  • $\begingroup$ I must add that above explanation applies only to massless ropes (rather academic). Real ropes have mass, and things become little more complicated. $\endgroup$ – Deep Jul 16 '16 at 4:53
0
$\begingroup$

Tension $T$ is the reactive force in the string when you pull at both ends with force $F$. If the string has stopped stretching or accelerating then $T=F$.

The action-reaction pair of forces in Newton's 3rd Law are two equal but opposite forces with the same cause which act mutually on two different bodies : object A exerts force $F$ on object B, and B exerts the same force $F$ back on A.

So the two forces $F_L$ and $F_R$ which you apply at the left and right ends of the string are not an action-reaction pair : although they might be equal and opposite and both caused by your muscles, they both act on the same object - the string. The paired force to your pull $F_L$ with your left hand on the string is the pull $T=F_L$ of the string back on your left hand. And likewise for your right hand. Only one force in each pair acts on a particular Free Body ; the paired force acts on the paired Free Body.

Tension is transmitted through the string, from one particle to the next, in both directions. In ideal, massless strings tension $T$ is the same at all points between the applied forces $F$ at each end. If you divide the string into two parts (left and right) by an imaginary perpendicular line, creating two Free Bodies, then the force which each Free Body exerts on the other is the tension $T$. These two forces form an action-reaction pair. In real strings (which have mass), the tension can vary along the string - eg if it is being accelerated, or if it is suspended in a gravitational field.

The two applied forces $F_L$ and $F_R$ are not necessarily equal. The resultant (net) force accelerates the string. If the string has zero mass, this means infinite acceleration, so $F_L \ne F_R$ is inconsistent with a massless string.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Thanks a lot of sparing your time to guide me through the concept $\endgroup$ – user118752 Jul 15 '16 at 16:42

Your Answer

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