# Tension in the same string between different masses

In the given diagram:

If the string in the entire system is the same and is assumed to be massless, then why is the tension in the string between objects $M_1$ and $M_2$ different from the tension in the string between objects $M_2$ and $M_3$? $M_2$ and $M_3$ have equal masses and $M_1$ is assumed to be moving downwards with a uniform velocity.

I am assuming the reason is because , even though the string is the same yet the force by which $M_1$ pulls onto $M_2$ is not the same as the force by which $M_2$ pulls onto $M_3$ because the weights of $M_1$ and $M_2$ are not equal. Is this right?

• Think of the two strings as two arms, where one arm carries the light box $M_1$ and the other carries the heavier box $M_2$. Naturally the tension to hold the heavier box is larger. – Steeven Nov 17 '17 at 15:12
• Edited to get the picture inline. – Gert Nov 17 '17 at 15:14
• "$M_1$ is assumed to be moving downwards with a uniform velocity." Why is this assumed? If there's no friction and $M_1$ is large enough there will be acceleration. – Gert Nov 17 '17 at 15:16
• @Gert it was given in the question that I was supposed to solve. Thanks for editing. – Aditi Nov 17 '17 at 15:18
• @Steeven but in a massless string , isn’t the tension the same throughout the string ? – Aditi Nov 17 '17 at 15:19

Now, let's consider an element in the string and let's say that the tension on either side of it is $T_1$ and $T_2$. Now recall Newton's second law which states that summation of all external forces on the body is equal to its mass times its acceleration. Hence we write that $T_2 - T_1 = m•a$. As mass of string is zero the mass of each element is also zero. Hence,we get $T_2-T_1 =0$ or $T_1 = T_2$ proving that the tensions are equal in a massless string. This equality of tensions can only happen if the mass of the body (under analysis) is zero. Hence, as the block has mass the tensions will be different. Also, question specific: the block is under influence of gravity also which must be taken account.(Although the acceleration is zero it's mass isn't).