6. Class 11th Physics | Constrained Motion | Simple Pulley Constrained Motion | by Ashish Arora
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
answer to this question.

28 thoughts on “6. Class 11th Physics | Constrained Motion | Simple Pulley Constrained Motion | by Ashish Arora”

  1. Talib Iqbal says:

    A very nice video ! it clears all your concepts.. the only possible improvement would be , if u add more examples which are tough to handle !!
    Thank you

  2. Physics Galaxy says:

    Thank you. For more solved examples you can refer to the playlists on our channel Physics Galaxy… or you can visit our website for all videos. Good luck!

  3. Animesh Kumar says:

    thank u sir i watched all of ur example videos and now i m comfotable with the topic.
    thank u again…

  4. shridhar kulkarni says:

    thank u sooo much sir ur  channel is very useful for the students who do nt get a gud level teaching !! i salute u!!!! nice job.

  5. karim khan says:

    VERY  GOOD

  6. Anoop Varma Gottumukkala says:

    very nice sir you realy made my doubts clear

  7. honey sharma says:

    excellent sir heads off to you

  8. Ayushi Mewar says:

    thnks sir very nice videos

  9. Raj Mehta says:

    Very helpful, thank you.

  10. solai nathan says:

    excellent teaching sir. great!!!!

  11. Shaswat Gupta says:

    thanks sir

  12. S.Kavin Balaji says:

    sir it methodology of teaching is just amazing

  13. Ali Hasan says:

    What if we pull block B. How will we analyse then? Block B is pulled then the string connecting block B and A is pulled by a distance x and also the pulley itself is pulled by x so in total 2x distance covered by a?????

  14. jay dani says:

    what about questions that involve the calculations of tension and acceralation

  15. Yashu Juneja says:

    Sir, I have not done Laws Of Motion till now!!
    Should I do these questions?

  16. Viraam Rao says:

    Amazing explanation sir. This was very difficult for me to understand initially. After watching ur playlists my doubts are cleared. Thank you sir.

  17. Color Sirens says:

    At first, I was like omg this guy writes so slow what is he gonna teach me and then at 1.25 speed its perfect. Subbed and liked

  18. didde sekhar says:

    sir your explanation is superb . sir can YOU explain me constraint relations form introduction onward please

  19. Vaibhav Sharma says:

    sir what happened after 4:33 as the pulley comes down by x why does the strring looses by 2x?

  20. Deepak Rani says:

    Nicely explained

  21. sonam khurana says:

    sir at 3:11 you said that if A goes up by y then B goes down by x as y=2x it means that A goes up by 2x then B comes down by x but at 4:30 you said that if block A goes up by x B comes down by x + 2x??? but it should come down by x + x/2

  22. VD PRODUCTIONS says:

    sir i tried to do it by drawing a perpendicular but i had to take another variable z in the direction of length of the string but i m stuck and dont know how to proceed further

  23. ASHUTOSH DIXIT says:

    It can also solved by tension relation method i.e Va/ Vb =T/3T i.e 3 Va = Vb

  24. Harshit Modi says:

    Thank you so much sir your videos are very helpful for students who can't afford coaching.

  25. Munim Ahmad says:

    Is this relation only valid for closed system

  26. NightcoRohak says:

    Sir is the acceleration that we are calculating in pulley problems, the instantaneous acceleration due to which the body gains velocity , or is that acceleration constant and continues to be applied in the bodies from the beginning of hanging let's say two masses on 1 pulley , till reaching equilibrium?

  27. Bigboss 13 Khabri says:

    Sir in the last qn i m getting ans as 1m/s as x = 3y =) Va = 3Vb

  28. Vaibhav Tambade says:

    At 4:34 why does the shorter string looses by 2x though there is just x distance travelled by longer string downward?
    Why should it not be x/2 and x/2 after traveling downward by x?

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