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I have recently started studying the Tension force and came across this diagram{attached below} of tension forces on a string and am unable to comprehend that how the direction of the tension forces are put below the point A and below the point B(which is fixed and rigid).

Shouldn't the tension force just point upwards the whole time because on a deeper level it is just due to electromagnetic forces? The mass attached on one end pulls the rope molecules farther away to which the molecules should respond by applying a force upwards to counter the downward force by the weight and hence in theory there shouldn't be any fore acting downwards by the string

The explanation for putting the arrows this way was given that the tension force is a pulling force and hence pulls each point although I think it dosen't make any sense. How is the tension force by the rope/string acting downwards at points B and A if it should instead act upwards to counter the force by the weight? Is the diagram incorrect or is my understanding of Tension force just incorrect all along? If so please help by providing an intuitive understanding of the cause and reason of Tension force.

Tension force direction

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3 Answers 3

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Tension is not in itself a force. To get an actual force you must choose a division of the rope into two parts. If a horizontal rope is in "tension" and you choose a point P on the rope, then the part of the rope to the right of P pulls on the part of the rope to the left of P with a force T to the right. Similarly the part of the rope to the left of P pulls on the part of the rope to the right of P with a force T to the left.
If instead of a rope you have rod which is in compression then, after choosing a point P, the part of the rod to the left of P pushes on the part of the rod to the right of P with a force to the right, and so on....

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  • $\begingroup$ "If a horizontal rope is in "tension" and you choose a point P on the rope, then the part of the rope to the right of P pulls on the part of the rope to the left of P with a force T to the right. Similarly the part of the rope to the left of P pulls on the part of the rope to the right of P with a force T to the left." So, this pulling occurs due to electromagnetic forces? If so then please explain how? $\endgroup$ Commented Jul 13, 2023 at 14:10
  • $\begingroup$ @BhavyaJain , See this $\endgroup$
    – khaxan
    Commented Jul 13, 2023 at 15:06
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Let me try to explain from another point of view.

If you look at a system as a whole, all the (outer) forces acting on that system will add to zero, otherwise the system will accelerate.

Now the tension on the rope is something internal to the ceiling/rope/mass system. To talk about such internal forces, we can (virtually) cut our system into two pieces at some convenient place. In our "tension" example, that could be somewhere along the rope, e.g. at point A.

At this cut, each part will exert some force on the other part. How to deal with that?

  • We observed that the lower part does not fall down, it stays in place with zero acceleration. So we can conclude the sum of all forces on that part must be zero. One force is the weight force W of the mass doing downwards, and the only thing that can compensate this is some force at the cut that (probably) the upper part exerts on the lower part. This has to be equal to the weight force W, but directed upwards - otherwise it would accelerate instead of staying in place. So, somehow the molecules of the upper part of the rope pull on those of the lower part, with this force. (If they wouldn't, the weight force would win, and the mass would fall down.)
  • As "actio = reactio", if some body X exerts a force F on another body Y, then also Y exerts a force of -F (same value, opposite direction) on X. So, as the upper part pulls on the lower part, then the lower part also pulls on the upper part, with opposite direction, now being a downward force.

Alternatively, we can place two cuts in the rope, and look at the "cutout" piece of rope. As it does not move, the downward-pulling force it receives from the lower end must equal the upward-pulling force from the upper end.

So, at whatever tiny length of rope we look, there are always two forces of equal value and opposite direction, one downward, coming from the lower end, and one upward, coming from the upper end, and that's what we call tension.

Or, we can place a cut directly above the mass and look at the mass. Again, the rope (the partner at our cut) must exert a force on the mass to keep it from falling, and vice versa, the mass exerts a force on the rope.

And in the situation from your image, it's close to impossible to reason about the system without making some specific cut, as most probably the ceiling is part of some building which stands on the earth. So the smallest candidate no-cut system would already include the whole globe, and no longer be useful to our question.

The picture you show us is correct in that it shows both force directions, but it does not make clear which part causes the force and which one receives it.

If I were to draw that image, I'd reverse the four arrows, so they are located at the part that receives the force. Then e.g. below A we'd see an upwards force, one that operates on the lower part at that point short below A.

This concept of placing virtual cuts into a system, and then reasoning about the forces at the cut points, is simple yet extremely powerful.

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Keep in mind that the tension force can act in either direction depending on whether you are looking at it as an internal force or an external force and how you draw the free body diagram (FBD). To see how this occurs you need to draw free body diagrams individually for the ceiling, block, and string connecting the block to the ceiling.

FIG 1 below is a FBD of the ceiling, FIG 2 a FBD of the block, and FIG 3 a FBD of the string (section including point A). Note in FIG 1 the tension is shown downward, since that is the external force acting downward on the ceiling. In FIG 2 is is shown upward since it is an external force acting upward on the block. Internal to the string at any section (such as that including point A) the tension forces (assuming a massless inextensible string) are equal and opposite for any section of the string.

Hope this helps.

enter image description here

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