### 6. Class 11th Physics | Constrained Motion | Simple Pulley Constrained Motion | by Ashish Arora

Now we’ll study simple pulley block constrained
relations. you can simply say in pulley systems. these are developed. By analysing. Loosing
and. tightening. of strings . due to . motion of blocks . so just be careful about , a pulley
block system like, if we just talk about the simplest system which we can say , there is
a pulley, on the two sides of which there two blocks hanging. So these are ay and b.
We can simply state if ay is going up by a distance x, so when ay goes up by distance
x , the string on the left hand side of the pulley will, get a slackened by distance x.
Than obviously due to the weight of b it’ll fall down by the same distance x. So we can
say, if this is displace by distance y , the value of y=x. Velocity of b=velocity of
ay and Acceleration of b also equal to acceleration of ay. In the similar situation say we are
having a pulley block system like this , where there is a pulley . on one side of which there
is a block ay hanging. On other side there is movable pulley which is connected to, the
ceiling wire the same string , and another block b is hanging from this movable pulley.
And say, if we are required to develop constrained relation between ay and b, we can say b goes
down by x. And say ay will move up by y. Whatever relation exist between x and y the same relation
is exist between there velocities and acceleration. So we can say if b goes down, by distance
x. This pulley will also come down by the same distance x, as b is directly attached
to this pulley. Say if this pulley will come down by x , this string will be , pulled.
Twice of the length , i.e. if pulley goes down by x , the string will be pulled x on
this side x on this side . as on this end the pulley tight with the ceiling . the whole
string which is increased over here , i.e. 2 x will be provided by, slackened of string
due to motion of ay , so if ay goes up by y, y is the string which is transferred from
left side of this pulley to the right side . as the total length is(same) as the string
is inextensible we can simply say, the while length of string which comes on this side
is providing 2 x length, for tightening this string. So we can simply state here y=2
x. So we differentiated we can say velocity of block ay=twice the velocity of block
b, once again if you differentiate you will get acceleration of block ay is twice the
acceleration of block b. In this manner we develop constrained relation in pulley block
systems. Let us consider an example based on the same
concept . here, we are given that a system , system shown in figure ay moves upward , at
3 meters per second , we are required to find the speed of block b. Now just to find the
speed of block b first we relate the displacements of block ay and b. Say block ay goes up by
distance x, and b comes down by distance y. This we just analyze and check. What part
of length of string is getting loose . and it is transferred to the other part of the
pulley. Now in this situation, (if) block a is going up by distance x you can see this
string which is directly connected to block ay will move up by x and it is transferred
to the other part of, the other side of pulley i.e. x. So if it is coming toward right by
x, this pulley must move down , by the same distance x. Due to this, due to this motion
of pulley , on the two sides of the pulley , the string will get loose by a distance
2 x. And x is also coming due to motion of block ay. That means b will fall down by distance
x +-x +-x it will be 3-x. So total distance block b is coming down is y and this y can
be written as 3-x. Once again lets analyze how this y=3-x, when block ay goes up by
x, this first string is transferred to the other side of pulley i.e. by length x. Due
to which this pulley will come down by distance x. Due to this, the string was passing over
the second pulley will get loose by distance 2-x, x on this side x on this side. Along
with this as block ay goes up . x distance of string will also loose due to the upward
motion of ay. The total, loosing in this string will be 3-x, because of which we can say block
d will fall down by distance 3 x. Hence we can write y=3 x. If we differentiate it
we can directly state velocity of block b can be written as 3 times the velocity of
block ay, in opposite direction . so a block ay is going up at 3 meters per second velocity
of block b can be written as, 9 meters per second in, downward direction. This is the