In the arrangement shown in figure, the mass of the body A is 4 times that of body B. The height h=20 cm . At a certain instant, the body B is released and the system is set in motion. What is the maximum height (in cm), the body B will go up? Assume enough space above B and A sticks to the ground.
Can anyone help me to understand the concept behind pulley with the help of problem
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Yogesh Ghadge
4 years, 2 months ago
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Top NewestIs the answer 40 cm.?????
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No its 60 cm
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Could you prove your statement?
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Yes I can
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I and satvik has discussed it's solution below you can refer to that. And if you still have doubt after reference then ask me I will clear the doubt
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Yes its 60cm
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I agree with you.
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The answer is got to be 40...Wait i'm gonna post a solve!!
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Sorry, Abhineet. I was thinking about the static situation. If A is lowered 20 cm then B will end up at 40 cm higher. I wasn't thinking about the accelerations involved.
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\(Let\quad the\quad displacement\quad of\quad mass\quad B\quad =\quad H\\ Let\quad the\quad tension\quad at\quad mass\quad B\quad =\quad T\\ So,\quad if\quad we\quad continue\quad finding\quad the\quad tension\quad upto\\ A,\quad we\quad end\quad up\quad with\quad 2T\quad at\quad mass\quad A.\\ \\ Now,\quad by\quad Virtual\quad work\quad done,\quad \\ Net\quad work\quad done\quad by\quad the\quad internal\quad forces\quad =\quad 0\\ { Force }_{ Int.A }\times { Displacement }_{ A }+{ Force }_{ Int.B }\times { Displacement }_{ B }=\quad 0\quad \\ 2T\quad \times \quad h\quad +\quad T\quad \times \quad H\quad =\quad 0\\ 2hT\quad =\quad -TH\\ Cancelling\quad T\quad from\quad both\quad sides,\\ H\quad =\quad -2h\\ (-)\quad sign\quad shows\quad that\quad displacement\quad is\quad in\quad opposite\quad direction\\ to\quad h\\ \\ Considering\quad only\quad magnitude,\\ H\quad =\quad 2h\\ H\quad =\quad 2\times 20\\ H\quad =\quad 40cm\)
Here @Guiseppi Butel @Yogesh Ghadge @satvik pandey
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Hello Abhineet your answer is correct but you have missed a last point that makes your answer wrong
When block B travels 40 cm then it gains a final velocity v
v^2=2as
v^2=2×g/2×0.4....................40cm in 0.4m
v^2=4
v=2m/s
Then we have to find the distance traveled by block under gravity
u^2/2g=s
4/20=s
0.2=s
0.2m=20 cm
Total distance covered = distance covered by block due to pulley + distance covered by block under gravity
=40+20=60cm
Now see if you have any doubt about the ans.
And thank you for sharing you mechanical work method
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Your answer is wrong because in taking components of force you have made an mistake force on block b is 2T and 4mg in downward direction because of taking on T it will not make any sense because both the tension are in upward direction so if you T only the force acting on block it will be totally wrong
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You are in class 10, right? There is a method in class 11, "The virtual work done method". You may learn about it on internet.
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Yes you are right bro. The answer should be 40cm
fig
For A
\(4mg-2T=4ma_{1}\)...................(1)
For B
\(T-mg=ma_{2}\)..............................(2)
Also \(2a_{1}=a_{2}\).............................(3)
putting this value in equation (2)
\(T-mg=2ma_{1}\)
Multiplying this equation with 2 yields
\(2T-2mg=4ma_{1}\)............................(4)
Adding eq. 1 and 4 yields
\(2mg=8ma_{1}\)
So \(a_{1}=\frac{g}{4}\) and \(a_{2}=\frac{2g}{4}\)
As \(S=ut+0.5at^{2}\)
So for block A \( 0.2=\frac{g}{8}t^{2}\)
or \(t^{2}=\frac{8(0.2)}{g}\)
So for block B
\(S=0.5(\frac{2g}{4})(\frac{8(0.2)}{g})\)=0.4 m=40cm
But I must say @Abhineet Nayyar that your method was the shortest I think.
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Satvik but when block b travel 40 cm it's gains a final velocity v which is 2m/s then block b travels a distance till its velocity is decelerated so after calculation the distance traveled by block b under gravity is 20cm . So the total distance traveled by block b is 20+40=60cm
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Yes Yogesh you are right. I completely missed that and I think @Abhineet Nayyar missed that too. :P
Thank you Yogesh for pointing it.
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Satvik i have misconception about friction can you explain me the way to solve h c verma's book question no. 11 in friction chapter. please give a quick response
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Q.11 of exercise?
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yes!
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You just need to draw FBD of the three blocks separately then use \(F=ma\).
Do have problem regarding the direction frictional force?Suppose if the blocks are accelerating towards right. Then friction forces acting on the blocks are towards left.
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see i want to buy fundamentals of physics by halliday so which edition should i buy
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Every editions are basically same. So you can by any edition. I think the latest is \(9^{th}\) edition.
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i have bought fundamentals of physics 8th edition but there is a problem . There are no answers of problem written so how would we come to know that we have solved the problem correctly.
And i was going through 42 problem of chapter no 2 i got the ans 0.94 m/s^2 is it correct?
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In that book answers of all question in exercise are not given. Only answer of odd question numbers are given.
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OK
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i wanted to ask you a question that was really means centripetal acceleration? i defines that its acceleration is towards center but , as an example if earth rotates at the centripetal acceleration of 9.8 m/s^2 then will earth exert force to us towards space or towards earth itself ?
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You can find the solutions of this book here
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Simply find the accelerations of A and B. Then find the time (t) that A will require to cover distance 'h' then use equations of motion to find the distance covered by B in time 't'.
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The main problem is I don't know to find acceleration in like this arrangement of pulley
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fig
Always try to draw the FBDs first.
Can you setup two equations for block A and B using newton's second law?
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I got it thank you . Can you tell me how you post diagram in comment box
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Please refer to this note. I think there is a similar problem on brilliant also.
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I have got that wrong so I asked you. And see I have framed equipment like this F-2T=0 T=F/2 taking block b as system a=T/m a=F/2m Thus acceleration of block A a=F/4m because acceleration of block B is twice than A .is this correct
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Equipment is wrong word there should be eq
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I don't understand what is 'F' in your equation.
Refer to diagram:
For Block A
\( 4mg-2T=4ma\)............(1)
For block B
\(T-mg=ma_{1}\)................(2)
Also \(2a=a_{1}\) .................(3)
Now you have 3 equations and 3 variables. Just find the values of a and \(a_{1}\)
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F refers to the downward force of block A
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So F=4mg.
But you wrote F-2T=0. Block A is not in equilibrium.
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See in first condition before Block B is released they both are in equilibrium
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I have done by different point of view but our result is same 2a=a1
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This eq. is just the part of whole solution. I have made 3 eq earlier. By solving those equations you will get the exact values of the accelerations of the A and B.
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But try to understand I make an sequence of finding T first then the acceleration and the tension in the rope is equal in every period of time so it doesn't matter if I take the period before motion or after
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I think we are approaching to the problem in different ways.
Can you find time period in which block A covered distance 'h'.
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T=0.4s
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i got same too. :)
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We both think in different ways but have same direction
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Yeah! Vector having different magnitude but same direction. :D
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Satvik when do you get time to do all this stuff .when I was in class 10 in didn't get time to do this. What is the board of your school
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I am in CBSE board. Well I have started studying mechanics when I was in class 9. I have studied most part of the mechanics in summer vacations.
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Is CBSE board better than SCC board
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I don't know what is SCC board. What is the board of your school?
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SSC
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satvik i want to ask you about writing comments. In my comments all the words are one behind the other I use enter to start the sentence from new line the but when i click post the words again come one behind the other so how do you write your comment so nicely
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To start a new line you need to press enter key two times.
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Here we have to consider the period in which the blocks are in motion
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@satvik pandey You in class 11th right?
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a comes out to be g/4 and a1=g/2
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Correct. :)
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You may also use the Mechanical Energy Conservation Law and have in mind that displacement h of body A will correspond to displacement 2h of body B and that every moment B body speed is twice that of body A. Then you can find B body speed (v) at the instant body A sticks the ground. You will easily get that v^2=2gh. And the maximum height will be 2h+h=3h.
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