The tension in rope AB is T. The tension in rope CP is S. The accelerations of A, B and C are respectively a,b, and c.
Find T and c.
Figure below:

- ujjwal

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- ujjwal

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- anonymous

where is the picture?

- ujjwal

just above your reply.

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## More answers

- anonymous

what do you want ?

- anonymous

What are the T and C?

- ujjwal

READ QUESTION!!!!!

- anonymous

Is it a spool ?

- anonymous

C=R?

- anonymous

are the pulleys massless?!

- ujjwal

yes!

- anonymous

since the masses on both sides are equal (3m and 2m+m= 3m).. why would the system accelerate?

- anonymous

find m for T and c ?

- ujjwal

ok, may be no.. let me check..

- anonymous

@Mashy : 3m = 3m ?!!!! Its ridicule !!!
we dont have Mechanical advantage ?!

- ujjwal

2m accelerates downward, lowering the CG of the system of masses of m and 2m.

- anonymous

In the left we have 3m and in right we have 1.5 .

- anonymous

@: ujjwal :Have it Attraction from earth ?

- anonymous

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- Vincent-Lyon.Fr

Both pulleys will rotate anticlockwise.

- Vincent-Lyon.Fr

It is a double Atwood machine. I will derive the solution.

- Vincent-Lyon.Fr

Does your answer key give c = g/17 as an answer?

- ujjwal

yes!

- Vincent-Lyon.Fr

And T = 24/17 mg

- ujjwal

yes!

- Vincent-Lyon.Fr

Let's take all accelerations positive if downwards.
Can you write N's 2nd law for m, 2m, and 3m?

- Vincent-Lyon.Fr

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@ujjwal Can you do it from the 3 free-body diagrams?

- anonymous

ok .but we have Movable pulleys and ...

- ujjwal

ma=T-mg
2mb=T-2mg
3mc=S-3mg

- anonymous

No !

- Vincent-Lyon.Fr

@E.ali : never mind they are moving or not
@ujjwal : correct but your are taking up as +ve.

- ujjwal

yeah, i assume all masses to have an upward acceleration..

- Vincent-Lyon.Fr

Then you will get c = - g/17, because 3m is definitely moving downwards.
Anyway, now we have 5 unknowns: a, b, c, S, T
and only 3 equations, so we need 2 more equations.
Can you suggest which ones?

- ujjwal

S=2T

- Vincent-Lyon.Fr

Perfect! Because pully is massless, hence net force = 0.(-c) = 0
Can you find the nature of the 5th one? (It is the most difficult to write)

- Vincent-Lyon.Fr

@Mashy : are you with us now? Do you have the final equation too?

- anonymous

i was doing something.. lemme read :P

- anonymous

i have one question.. Since initially both sides have same total mass .. why does the system accelerate?!.

- Vincent-Lyon.Fr

Because the masses on the right side are globally accelerating down and S will be slightly less than 3mg as a consequence.

- anonymous

I just don't get intuition :( :(.. why would the right side masses accelerate down (along with the pulley)
my gut says this.. initially the 3m would be stationary.. that second pulley would be stationary, 2m would be accelerating down and the m would be accelerating up with same acceleration..

- Vincent-Lyon.Fr

So (m+2m) centre of mass is globally going down.

- anonymous

Ok.. ok.. its starting to make sense now
what i am saying is true only for a split second, the moment i let of the system.. the very moment 2m starts going down, the tension in big string S decreases.. and so 3m also starts accelerating upwards... correct?!

- anonymous

i mean 3m accelerates downwards

- Vincent-Lyon.Fr

Correct!

- anonymous

ok.. thanks a ton!!!!.. now lemme solve this!

- anonymous

the final equation connects the three accelerations?

- Vincent-Lyon.Fr

Yes it does. It is simply of kinematic nature. Velocities of m, 2m and centre of pulley cannot be independent since the string is inextensible.

- anonymous

considering a , b and c upwards i get 2c = a+b

- Vincent-Lyon.Fr

If a and b are up, then c must go down, so actually: a + b = –2c

- anonymous

but his above equations are based on considering c downwards right?

- anonymous

i mean c upwards!!!

- Vincent-Lyon.Fr

No, ujjawal chose up=+ve for all three masses. (I had chosen down=+ve)

- anonymous

ahh.. ok ok ok.. yea yea..

- ujjwal

How do we get a+b=-2c ?

- Vincent-Lyon.Fr

Use the centre of the pulley (let'c call it P) as a starting point. And always reason with velocities.
Each side will cumulate the velocity of P and the one induced by the rotation (Ω) of the pulley (radius R).
Hence:
v(2m) = v(P) + ΩR
v(m) = v(P) – ΩR
Adding the equatins yields:
v(m) + v(2m) = 2 v(P)
Since strings are inextensible, v(P) = - v(3m)
v(m) + v(2m) = -2 v(3m)
Derivation wrt time yields:
a + b = –2 c QED

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