# Modelling implication of an inextensible string in a pulley system?

I am learning this in a highschool maths textbook btw, so I need answers at that level.

Before I got to the chapter involving pulleys, whenever there was a question involving the implications of an inextensible string holding together more than one particle moving in the same direction of the system; the book told me that this implied that the acceleration was constant for all particles.

I never got an explanation why so I assume that this is because if part of the string did in fact extend then from the moment it did extend until it stopped extending: the acceleration for the particle behind where this happened would be slower than ones in front. Is this correct?

but now for questions involving a light inextensible string over a smooth light pulley, whenever the question is for what's the effect of the strings inextensibility is the answer is that tension is implied to be equal at both weights.

basically I don't understand why the inextensibility used to imply equal acceleration and now it implies equal tension. as before if the string is light, even if it extends wont the tension still be the same because it has no mass. And if it extends won't the acceleration on one side be different from the other?

https://postimg.cc/R67GRDrB Example question. The answer to the relevant part D is the tension on both sides of the pulley will be the same.

Thankyou.

• If the string extends, it is not "inextensible" :) – alephzero Mar 5 at 20:02

The equal tension condition has nothing to do with extensibility, but is because the string is light (i.e. it has no mass). If you consider a short element of the string with tension $$T_1$$ and $$T_2$$ at each end, Newton's second law says that $$T_2 - T_1 = ma$$ where $$a$$ is the acceleration of the string and $$m$$ is its mass. When the string is "light", $$m = 0$$ and therefore $$T_2 = T_1$$.