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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TopNewest\(Let\quad the\quad displacement\quad of\quad mass\quad B\quad =\quad H\\ Let\quad the\quad tension\quad at\quad mass\quad B\quad =\quad T\\ So,\quad if\quad we\quad continue\quad finding\quad the\quad tension\quad upto\\ A,\quad we\quad end\quad up\quad with\quad 2T\quad at\quad mass\quad A.\\ \\ Now,\quad by\quad Virtual\quad work\quad done,\quad \\ Net\quad work\quad done\quad by\quad the\quad internal\quad forces\quad =\quad 0\\ { Force }_{ Int.A }\times { Displacement }_{ A }+{ Force }_{ Int.B }\times { Displacement }_{ B }=\quad 0\quad \\ 2T\quad \times \quad h\quad +\quad T\quad \times \quad H\quad =\quad 0\\ 2hT\quad =\quad -TH\\ Cancelling\quad T\quad from\quad both\quad sides,\\ H\quad =\quad -2h\\ (-)\quad sign\quad shows\quad that\quad displacement\quad is\quad in\quad opposite\quad direction\\ to\quad h\\ \\ Considering\quad only\quad magnitude,\\ H\quad =\quad 2h\\ H\quad =\quad 2\times 20\\ H\quad =\quad 40cm\)

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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Yes you are right bro. The answer should be

40cmfig

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=40cmBut 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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@Abhineet Nayyar missed that too. :P

Yes Yogesh you are right. I completely missed that and I thinkThank you Yogesh for pointing it.

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here

You can find the solutions of this bookLog in to reply

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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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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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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Is the answer 40 cm.?????

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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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No its 60 cm

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Yes its 60cm

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Could you prove your statement?

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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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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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this note. I think there is a similar problem on brilliant also.

Please refer toLog in to reply

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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

aand \(a_{1}\)Log in to reply

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F=4mg.But you wrote

F-2T=0. Block A is not in equilibrium.Log in to reply

@satvik pandey You in class 11th right?

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Can you find time period in which block A covered distance 'h'.

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