Non-constant tension in rope Could somebody explain to me intuitively how tension is not the same in a rope with mass? My physics teacher (when regarding a massless string) told me that the tension is always equal because if you pull at one side more then the other side has to pull just as much to keep it in equilibrium resulting in the tensions being equal. This fits really well intuitively with me, but if the string has a non-negligible mass how can I adapt this idea? Or replace it if necessary?
 A: Imagine a load hanging in a vertical rope:


*

*The bottom particle carries the load.

*The next particle carries that particle plus the load.

*The next-next particle carries both below particles plus the load. 

*In general, a particle carries all particles below it plus the load. 


Clearly, the top particle carries the most whereas the bottom particle carries the least. 
Tension increases up through the rope since the particles gradually carry more total weight. Only in the special case of massless particles - a massless rope - is this not the case, since more particles don't add extra weight. 
A: Hopefully you can understand it this way: See, if an object is massless ($m=0$), by Newton's second law of motion, if we apply a force on the object, the acceleration cannot be defined. (Something divided by $0$ is not defined.) So, the net force on the object needs to be zero for the body to have a finite, defined acceleration. So, in case of a massless rope, the net force on the rope needs to be zero, and the net force on each small part (element) of the rope should be zero, so everywhere the tension would be the same, say $T$. 
Now if you have a rope with mass(as in real conditions), the small element is experiencing a gravitational force by the earth as well! So the tension on each point would be different, usually increasing as we go upwards. 
A: Non constant tension would be developed in a rope that hangs from a fixed support if it has a "non-negligible" mass. First let's discuss why such a variation is found in the tension. Say you take bottom of the rope as the origin and consider an element of length $ dx $ at a distance $ x $ from the bottom. Now analysing its free body diagram :

Now let's analyse the relevant mathematics involved. As you can see that the rope element is in equilibrium hence we should have a "ZERO" net force on it.  Hence we have $ T+dT-T = \lambda g dx $  which leads us to
$$ \int_{0}^{T} dT = \int_{0}^{x_{0}} \lambda gdx $$
The above expression on simplifying gives us  $ T(x) = \lambda gx_{0}  $ which is a linear function of $ x $(assuming linear mass density to be constant).
A similar intuition can help us a lot. For the same element divide the rope in two parts one above it and one below it. We can see that the upper mass has to support a weight equal to that of the lower part (as mass of element is negligible) and hence we can arrive at the same result.
A: Other answers here are adequate, but I like the following mental exercise to demonstrate tension in a rope due to self-weight.
Start with what you know; a weightless rope of a physics exercise; same tension throughout, lets say it's suspending a load of 10N, from a fixed point directly above and is at rest, in equilibrium.
We want to turn this into a real rope, but slowly, one small part at a time; like turning a square into a circle by adding more sides.
Now a real rope has mass $M_r$ distributed along it's length $L_r$ such that any length of rope $dL$ will have an equivalent mass $dM$
So in the first instance, to work out the tension at the top of the rope, you just add the weight of the rope $F_w=M_rG$ at the base of the rope, along with the 10N force, this helps, you now know how much the rope and load weigh together.
But then you know the mass of the rope is distributed along the length of the rope, not all at the end, so that's not very realistic.
So you break the mass up, into 2 sections both $\frac{M_r}{2}$, and place one at the end, one at the mid-point; getting better slowly.
Now; the tension at the top of the rope is still correct, so that's good, but the tension between the two masses is something different. It has to be, the lower part of the string is no longer supporting half the mass of the string, so it has to be less.
You can then take this concept and run with it, every time you break that mass up into smaller chunks and distribute along the string, the tension changes depending on how many chunks are above or below the point you are measuring the tension at.
I hope this has given you a more intuitive feel for the effect of the continuous mass distribution along a string and it's effect on the resultant string tension.
As always, happy to answer any further questions.
A: If you pull on a massless rope from both ends, the tension at every point must be the same. Why?
Because the net force on any segment of the string must be zero, because it's a massless object and hence, tension must be same at every point.
In a string with mass, we can apply two different forces at the two ends of the string because of which it has a net acceleration. 
Now, the Tension at any point must be such that it can account for the acceleration of the mass of that segment of the string. 
Suppose string has length L and constant mass density g. Now, you apply a force F on right. The acceleration of the entire string is 
(F/gL).
If you want to find the tension at a distance L/4 from the right. What will it be?
F - T = m.a
i. e T = F - (gL/4)*(F/gL) = 3F/4.
You take a different distance and the Tension would be different because it is dragging with it a different length of the string.
Think of the string as many many blocks with small links attached in between.
A: Imagine a rope having a finite mass and is placed in gravity free space and having a constant velocity.Now take the rope as your system and exert an external force on one end of the rope then the rope will bear some tension at the point of application of force.As rope is the system tension is the internal force.The whole rope will now be having same acceleration.At the point of application of force imagine an infinitesimal section of rope,by Newton's third law we know that the section will exert an equal and opposite force on the agent applying the force.As the whole rope is constrained to move with that infinitesimal section, then that section would apply a force on the adjacent infinitesimal section of the rope but this force would be less than the external force itself as the section of point of application of external force is itself accelerated and the acceleration of whole rope is same.In the same manner force applied by these infinitesimal sections would decrease in a defined order which can calculated using basics of calculus.Hence,tension in a rope having finite mass is not the same everywhere.
