Given there are three blocks of masses , and (ref image in attachment)
When all three masses move together at an acceleration a, the force F is given by
F = ( + + ) *a ................(equation 1)
Also it is given that does not move with respect to , which gives tension T is exerted on pulley by only, Hence tension T is
T = *a ..........(equation 2)
There is also also tension exerted by . There are two components here: horizontal due to acceleration a and vertical component due to gravity g. Thus tension is given by
T = ................(equation 3)
From equation 2 and 3, we get
*a =
Squaring both sides we get
* = * (+)
* = ( * )+ ( *)
( - ) * = *
= */( - )
Taking square root on both sides, we get acceleration a
a = *g/()
Hence substituting the value of a in equation 1, we get
It says it's a Long cylindrical charge distribution, So you can ignore the border effects on the ends of the cylinder.
Also by the gauss law we know that E¨*2*pi*r*L = Q/ε0
where Q is the total charge inside our gaussian surface, that will be a cylinder of radius r and heaight L.
So Q= rho*volume= pi*r*r*L*rho
so replacing : E = (1/2)*r*rho/ε0
you may ask, ¿why dont use R on the solution?
since you are calculating the field inside the cylinder, and the charge density is uniform inside of it, you don't see the charge that is outside, and in your calculation actuali doesn't matter how much charge is outside your gaussian surface, so R does not have an effect on the calculation.
R would matter if in the problem they give you the total charge of the cylinder, so when you only have the charge of a smaller r radius cylinder, you will have a relation between r and R that describes how much charge density you are enclosing.