Tension in a massive string How tension is different in a string with mass compared to a massless string .
I have referred many books and really does not intuitively understand the problem .
An intituitive explanation would be a great help .
 A: Think of a string (or rope) with mass holding a stone from the ceiling.


*

*The bottom part of the rope holds the stone's weight with a tension force $T$: $$T_{bottom}=w_{stone}$$

*The middle part of the rope holds the stone's weight and the weight of all rope below it:$$T_{middle}=w_{stone}+w_{rope\;bottom\;half\;}$$

*The top part of the rope holds the stone's weight and the weight of the whole rope:$$T_{top}=w_{stone}+w_{rope\;bottom\;half\;}+w_{rope\;top\;half\;}$$


The pattern is an increasing tension $T$ along the rope, because the next piece of the rope has to carry the load as well as all the rope before that point.
A: The difference appears when the string is subject to forces and accelerations or gravitational fields. The easiest way to see this is to consider just a string of total mass $m$ and length $L$, with uniformly distributed mass, accelerated by a force $F$ with an acceleration of, of course, $a=F/m$. Now let's draw a free-body diagram of, say, a length $x$ of the string measured from the end that has force $F$ acting on it. That piece has mass $m\,x/L$ and is acted on by $F$ on one side and tension $T(x)$ on the other. Its acceleration is $a$, just like the rest of the string. The total force $F-T(x)$ mus still satisfy Newton's Second Law, so we have
$$F-T(x)=m\,a-T(x)=m \frac{x}{L} a\quad\Rightarrow\quad T(x)=\frac{L-x}{L}\,m\,a=\frac{L-x}{L}\,F$$
Thus, at the front of the string the tension is (of course) just $F$, and tension decreases linearly towards its end, where it is zero.
A: Ill put it in simple words..
Find the acceleration of the whole system by dividing the NET external force by the total mass of the system.
Now break the string in two parts from the point where you want to find the tension at.
Now apply Newton's second law of motion to any part (you will get the same value of tension in both the parts because tension forms an action-reaction pair on the two separate strings) and find the required tension.
Example,
Suppose a box of mass 5kg is pulled by external force of 10N by a string of mass 5kg.
The acceleration of the system will be 10N/(5+5)kg = 1 ms^-2
Now consider a point in the middle of the string (the string has uniform mass per unit length).
We have two parts one with the block, the string and the tension force and the other part with the tension force and the external force.
Taking the first part.
a = 1ms^-2
Net force = Tension
So, T = Total mass times the acceleration 
  = (5+2.5)kg x 1 ms^-2

  = 7.5 N

Now taking the second part.
a = 1 ms^-2
Net force = External force - tension force
      = 10 - T

So, 10 - T = 2.5kg x 1 ms^-2
     T = 7.5 N

