0
$\begingroup$

Let me first give you a brief idea of my understanding of Tension force: Assuming a string is made up of a line of molecules, we fix one end of the string and pull it at the other. Each molecule will be pulled in the direction of the applied force and since nothing is moving and assuming the string is somehow made unbreakable, we conclude ( using Newton' laws ) that each molecule experiences a reaction force in the direction opposite to the pull which we call Tension in the string. Note here that the direction of this reaction force is opposite to the pull at every point on the string.

Now consider a situation where we have a string pulled at both its ends with two equal forces. Applying the above reasoning, due to the force at the left end, a tension should develop towards the right and similarly due to the pull at right end a tension along the left.
Now the situation we have is that each molecule is experiencing two tension forces of equal magnitude and in opposite directions which seems to me very absurd. In this particular case of two equal forces at two ends of a string what will be the direction of tension force at any point of the string?

Improve this question
$\endgroup$
7
  • $\begingroup$ I don't see how you can have unequal forces on the two ends of a string without causing the string to move in the direction of the greater force until forces balance. $\endgroup$
    – Hot Licks
    Commented Jan 8, 2018 at 4:30
  • $\begingroup$ @HotLicks I've got it. But let's say we have two equal forces. What will be the direction of tension in this case? $\endgroup$
    – Deepak Joshi
    Commented Jan 8, 2018 at 4:34
  • 1
    $\begingroup$ Tension in a string/rope is always bi-directional. $\endgroup$
    – Hot Licks
    Commented Jan 8, 2018 at 13:51
  • 2
    $\begingroup$ Every molecule in the rope is hanging on for dear life to the molecules on either side. $\endgroup$
    – Hot Licks
    Commented Jan 8, 2018 at 14:00
  • 1
    $\begingroup$ It just makes the arithmetic simpler, in those cases where there is movement. $\endgroup$
    – Hot Licks
    Commented Jan 8, 2018 at 14:16

4 Answers 4

Reset to default
1
$\begingroup$

Now consider a situation where we have a string pulled at both its ends with two equal forces.

Unless it has equal forces on both ends, it will accelerate in one direction, so this is always true for a string at rest. There's no difference between this case and the case where one end is fixed to a wall.

Tension is normally dealt with as a scalar in a string, not a force with a specific direction. For a string at rest, the tension at any point in the string is equal to the forces at each end.

Improve this answer
$\endgroup$
0
$\begingroup$

The tension in the string will be leftwards, from the left, and rightwards from the right.

Consider a particle right in the middle of the string, that particle will experience zero force.

Each particle to the the right of this central particle will get displaced from it's mean position towards the right, by small length. The same applies for every particle to the left of the central particle.The particles on the left will get displaced leftwards by some length.

The extent of displacement will increase as we go leftwards or rightwards.

Every particle will try to come back to it's mean position. For this, it will exerta force in the opposite direction. This is tension.

I hope my answer is satisfactory.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Why only the particle in the middle? Don't you think the net force experienced by every particle is zero? $\endgroup$
    – Deepak Joshi
    Commented Jan 8, 2018 at 7:39
  • $\begingroup$ Only if the net force on that string is zero. Suppose you hold a string, and everything is in equilibrium. Suddenly, you let go, at this time, every particle will experience a finite net force towards mean position. When you are holding a string, say, which is tied to a wall. The net force on the string is zero only because the force that you are exerting on the string is greater than the force the particle exerts on you to drive itself towards the mean position. $\endgroup$
    – Sarvesh Sontakke
    Commented Jan 8, 2018 at 9:14
0
$\begingroup$

A mass-less string is a concept which allows a force to be transmitted from one point to another whilst also allowing the direction of a force to be changed.

Now the situation we have is that each molecule is experiencing two tension forces of equal magnitude and in opposite directions

So the net force on that molecule, and the forces on all other molecules which make up the string, is zero and the molecule(s)/string does not accelerate.

Just cut the string at the position of the molecule and see what you need to do to keep the two parts of the string from moving.

Now the situation we have is that each cut end of the string is experiencing external forces of equal magnitude and in opposite directions

Improve this answer
$\endgroup$
0
$\begingroup$

Tension is a name we invent for the force that holds the particles of a rope together.

You are pulling (right wards) with a pulling force in the outermost particle, and that particle is pulling (right wards) with a tension force in the next particle. This next particle is then holding on (leftwards) with a tension force, and it is pulling (right wards) in the next particle in the row with a tension force. And so on. All these tension forces are equal in equilibrium (if nothing else pulls as well, of course).

So in this sense, tension is always pulling both ways.

  • From your perspective, tension holds back against your pull, so it has a direction away from you.
  • From the walls perspective (or the person who pulls the other end), tension also pulls away.
  • But for every particle inside the rope, tension pulls equally in both direction.

In this particular case of two equal forces at two ends

Note here that it makes no difference if you are pulling in both ends or if a wall is pulling in one end. The effect is the same. The rope end is kept static. The force exerted by wall or pull would be the same.

So why do we require the assumption of massless string

I mentioned above that the tension forces on either side of every particle are equal. And they are equal to those at every other particle. This is a result of Newton's 3rd law, as you mentioned.

But it is only true if nothing else pulls as well. And if the string has mass - which means that the particles have mass - then there is weight pulling as well. If you tie a string to a ceiling and pull down in the other end, then

  • the bottom particle has to hold back against the pull as well as it's own weight. The tension equals both of these.
  • The next particle has to hold back against both this tension force as well as its own weight. So that particle is now carrying the pull and the weight of two particles.
  • And so on. Every particle will carry all the particles below, and therefore the tension will be larger and larger from bottom to top of the string.
Improve this answer
$\endgroup$

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.