What are the conditions for the tension to vary in the rope. I have read below conditions

1. rope has to have some mass
2. rope is accelerating

I get the 1st one, but I am not sure if I get the 2nd condition. If a $10\ \text{N}$ force accelerates a mass-less role, what will be the tension?

  • $\begingroup$ Thx. I corrected. what happens if rope is massless but accelerating. ? I think in that case tension wwont vara right.? $\endgroup$
    – user31058
    Commented Feb 1, 2016 at 23:14
  • $\begingroup$ If you pull on a rope that force must either go into accelerating the rope ($F=ma$) or into countering a force from the other end. Now figure what that means if $m=0$. $\endgroup$
    – BowlOfRed
    Commented Feb 1, 2016 at 23:35
  • $\begingroup$ You are confusing net force with tension. The net force is the vector sum of the tension applied at the two ends. Now figure out what that means if $m=0$. $\endgroup$
    – garyp
    Commented Feb 3, 2016 at 17:12

3 Answers 3


Clearly, if the tension in one part of a rope is different than in another part, there will be a gradient in tension - which in turn means that if you look at a particular part of the rope at location $x$ where there is a gradient in tension $\frac{dT}{dx}$, then there is a net force on an element of length $\Delta x$:

$$F = \frac{dT}{dx}\Delta x$$

For a mass per unit length of $\lambda$, the mass of the element is $\Delta m = \lambda \Delta x$, and we find an acceleration

$$a = \frac{F}{m} = \frac{dT}{dx}\frac{1}{\lambda}$$

This shows there is a relationship between the acceleration $a$, non-uniform tension (expressed by the gradient $\frac{dT}{dx}$), and the finite mass per unit length of the rope, $\lambda$. If the rope is massless, the finite force would lead to infinite acceleration; this means that the difference in tension would propagate infinitely fast.


It also depends on whether or not the rope is inextensible. Assume it isn't. You tie one end to a rigid wall and pull on the other end with a constant tension $T$. Then at some time $t_0$, you start to pull with a slightly higher tension $T_1 > T$. It will take some time $\tau$ of order $\frac{L}{c}$, $c$ being the typical celerity of mechanical waves inside the rope, for the rope to reach equilibrium. During that time the tension inside the rope will be inhomogenous and somewhere between $T$ and $T_1$. After equilibrium is reached the tension in homogenous and equal to $T_1$.

This is indeed linked to the fact that between $t_0$ and $t_0 + \tau$ the rope is accelerating.

If you repeat the same experiment with an inextensible rope, its rigidity is infinite thus the mechanical waves propagate at a celerity $c$ that tends to infinity. When you pull on it with a tension $T_1$, the rope immediately reaches equilibrium at tension $T_1$ everywhere inside the rope. This is linked to the fact that the rope don't accelerate in this case.

You also need the rope to have a lineic mass for this explanation to stand as the celerity $c$ of mechanical waves inside the rope is also inversely proportional to the square root of its lineic mass.


Since the LINE, a section of ROPE, has no attached mass or restraint and it seems it is accelerated as a whole, there is no tension. If one end is pulled to accelerate it then F=ma=0.0 N.

  • $\begingroup$ I don't understand what you are trying to say. Your analysis seems to apply to the net force on a segment of rope, not the tension. $\endgroup$
    – garyp
    Commented Feb 3, 2016 at 17:09

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