# pulley problem

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

Note by Yogesh Ghadge
5 years ago

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

- 4 years, 12 months ago

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

- 4 years, 12 months ago

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.

- 4 years, 12 months ago

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

- 4 years, 12 months ago

Yes Yogesh you are right. I completely missed that and I think @Abhineet Nayyar missed that too. :P

Thank you Yogesh for pointing it.

- 4 years, 12 months ago

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

- 4 years, 11 months ago

Q.11 of exercise?

- 4 years, 11 months ago

yes!

- 4 years, 11 months ago

You can find the solutions of this book here

- 4 years, 11 months ago

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.

- 4 years, 11 months ago

see i want to buy fundamentals of physics by halliday so which edition should i buy

- 4 years, 11 months ago

Every editions are basically same. So you can by any edition. I think the latest is $9^{th}$ edition.

- 4 years, 11 months ago

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?

- 4 years, 11 months ago

In that book answers of all question in exercise are not given. Only answer of odd question numbers are given.

- 4 years, 11 months ago

OK

- 4 years, 11 months ago

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 ?

- 4 years, 11 months ago

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

- 4 years, 12 months ago

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.

- 4 years, 12 months ago

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.

Is the answer 40 cm.?????

- 5 years ago

I agree with you.

- 4 years, 12 months ago

The answer is got to be 40...Wait i'm gonna post a solve!!

- 4 years, 12 months ago

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.

- 4 years, 12 months ago

No its 60 cm

- 5 years ago

Yes its 60cm

- 1 year, 3 months ago

Could you prove your statement?

- 4 years, 12 months ago

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

- 4 years, 12 months ago

Yes I can

- 4 years, 12 months ago

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'.

- 5 years ago

The main problem is I don't know to find acceleration in like this arrangement of pulley

- 5 years ago

fig

Always try to draw the FBDs first.

Can you setup two equations for block A and B using newton's second law?

- 5 years ago

I got it thank you . Can you tell me how you post diagram in comment box

- 5 years ago

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

- 5 years ago

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

- 5 years ago

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 $a_{1}$

- 5 years ago

a comes out to be g/4 and a1=g/2

- 5 years ago

Correct. :)

- 4 years, 12 months ago

F refers to the downward force of block A

- 5 years ago

So F=4mg.

But you wrote F-2T=0. Block A is not in equilibrium.

- 5 years ago

@satvik pandey You in class 11th right?

- 4 years, 12 months ago

See in first condition before Block B is released they both are in equilibrium

- 5 years ago

Here we have to consider the period in which the blocks are in motion

- 5 years ago

I have done by different point of view but our result is same 2a=a1

- 5 years ago

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.

- 5 years ago

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

- 5 years ago

I think we are approaching to the problem in different ways.

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

- 5 years ago

T=0.4s

- 5 years ago

i got same too. :)

- 5 years ago

We both think in different ways but have same direction

- 5 years ago

Yeah! Vector having different magnitude but same direction. :D

- 5 years ago

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

- 5 years ago

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.

- 5 years ago

Is CBSE board better than SCC board

- 5 years ago

I don't know what is SCC board. What is the board of your school?

- 5 years ago

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

- 4 years, 12 months ago

To start a new line you need to press enter key two times.

- 4 years, 12 months ago

SSC

- 5 years ago

Equipment is wrong word there should be eq

- 5 years ago