Three Pulley - two masses system.

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SciencyBoi
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Homework Statement


The given system is released from rest. Assuming no friction, mass-less pulleys and ideal strings; calculate the accelerations of the pulleys.
Pulley system: a fixed pulley hangs from a ceiling, below it a movable pulley labelled P1 and a lower movable pulley labelled P2, with a string over each; masses m1 and m2 hang from the strings below.


Homework Equations


Constraint equations.
F=ma

The Attempt at a Solution


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Diagram of a three-pulley, two-mass system: a fixed top pulley, a movable pulley labeled P1 with acceleration a0, and a lower movable pulley labeled P2, with masses m1 and m2 hanging below. Tension T labels the string segments and accelerations a1 and a2 mark the masses.
Taking the above assumptions;

And applying constraints to the longer string, we have;

-a1 + a0 + a0 + 2a2 - a0 = 0
=> a0 + 2a2 = a1 ~~~~~~~~~~~~~~~~~~ (1)

Now doing force balancing on m1;
m1a1 = T - m1g ~~~~~~~~~~~~~~~~~ (2)

similary on m2;
m2a2 = m2g - 2T ~~~~~~~~~~~~~~~~~~ (3)

For four variables a0, a1, a2 & T; we only have three equations, but we will neeed four. I'm not sure how to get the fourth.

I would greatly appreciate any hint/help.

Thank You...
 
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  • Pulley system: a fixed pulley hangs from a ceiling, below it a movable pulley labelled P1 and a lower movable pulley labelled P2, with a string over each; masses m1 and m2 hang from the strings below.
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  • Diagram of a three-pulley, two-mass system: a fixed top pulley, a movable pulley labeled P1 with acceleration a0, and a lower movable pulley labeled P2, with masses m1 and m2 hanging below. Tension T labels the string segments and accelerations a1 and a2 mark the masses.
    HAz4uxU.jpg
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Consider the relation between a0 and a2 regardful of the highest (the biggest) pulley
 
Sorry. Never mind! I am not sure any more about my previous comment
 
Diagram of a pulley system: a fixed ceiling pulley, a movable pulley P1 and a lower movable pulley P2, with masses m1 and m2 hanging below. Tension T labels mark each rope segment, and accelerations a1, a2 and a0 are shown with arrows.
How is it possible that 2T force acts on pulley P1 downward, and T acts upward? It is a massless pulley.
 
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  • Diagram of a pulley system: a fixed ceiling pulley, a movable pulley P1 and a lower movable pulley P2, with masses m1 and m2 hanging below. Tension T labels mark each rope segment, and accelerations a1, a2 and a0 are shown with arrows.
    haz4uxu-jpg.jpg
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ehild said:
View attachment 214889How is it possible that 2T force acts on pulley P1 downward, and T acts upward? It is a massless pulley.
I've thought of this;
The upward force is 2T.
Let the downward tension be T'.

Then T' - 2T = 0.a2
which gives us, 2T = T'
 
SciencyBoi said:
I've thought of this;
The upward force is 2T.

Why? It is the same string, the tension is the same everywhere.
 
srecko97 said:
It is not the same string
The roughness of the drawing may be misleading you. It is clearly intended that there is one string from m1, over P1, under P2, over the top pulley and down to the axle of P1. If not, P2 is not really a pulley.
 
Sorry, I know this is the same rope, but I meant the string between p2 and m2. I see that 2T is written next to this string (between p2 and m2) and I see no problem with this.
 
srecko97 said:
Sorry, I know this is the same rope, but I meant the string between p2 and m2. I see that 2T is written next to this string (between p2 and m2) and I see no problem with this.
There is no problem with that. Reread post #4. ehild is objecting to the forces shown on P1.
But of course, it does also appear that those forces shown are correct, since it is all one rope.
What is the resolution to this paradox?

(By the way, to answer the given question it is necessary to assume the pulleys do have some very small mass.)
 
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haruspex said:
There is no problem with that. Reread post #4. ehild is objecting to the forces shown on P1.
But of course, it does also appear that those forces shown are correct, since it is all one rope.
What is the resolution to this paradox?

(By the way, to answer the given question it is necessary to assume the pulleys do have some very small mass.)
Length of string is constant on the basis of this we can establish relation between acceleration.is there any other method to establish relation between acceleration?
 
Abhishek kumar said:
Length of string is constant on the basis of this we can establish relation between acceleration.is there any other method to establish relation between acceleration?
Not sure what you are asking. Are you saying the constant length of string gives one equation, but you do not see how to find a second equation?
 
haruspex said:
Not sure what you are asking. Are you saying the constant length of string gives one equation, but you do not see how to find a second equation?
No i am saying in constraint motion we can established relation between acceleration. If one displaced say xa other displaced say xb then we establish relation by taking length of string constant. My question is getting same is there amy easier method
 
Abhishek kumar said:
No i am saying in constraint motion we can established relation between acceleration. If one displaced say xa other displaced say xb then we establish relation by taking length of string constant. My question is getting same is there amy easier method
No, the constant length of string is a key fact, so you must write an equation expressing that.
Still not sure I understand your question.
 
I think I got it... When we talk about pulley P2;

Then assuming tension T in the longer string, we would have 2T tension in the string connecting m2 and P2.

However, when we do force balancing of P1;

We would have 2T - T = a0 x 0;

Which would give us T = 0;

hence the masses m1 and m2 both are effectively in freefall...

Using equation 1(which is derived from the length of string remaining constant) we will have a0 = 3g.

Thank you everyone... This discussion actually led me to the solution and better concepts...
 
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SciencyBoi said:
I think I got it... When we talk about pulley P2;

Then assuming tension T in the longer string, we would have 2T tension in the string connecting m2 and P2.

However, when we do force balancing of P1;

We would have 2T - T = a0 x 0;

Which would give us T = 0;

hence the masses m1 and m2 both are effectively in freefall...

Using equation 1(which is derived from the length of string remaining constant) we will have a0 = 3g.

Thank you everyone... This discussion actually led me to the solution and better concepts...
Well done. It was a tricky question.
 
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Yeah, I see that I was really confused. I wanted to help, but I was probably making this task even harder by writing false comments. Please accept my appologies.
 
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srecko97 said:
Yeah, I see that I was really confused. I wanted to help, but I was probably making this task even harder by writing false comments. Please accept my appologies.
No problem. I honestly appreciate that people took interest in this problem that in no way would benefit them. And I would keep appreciating anyone for their participation in any of the posts here or any of the forums posted by anyone as they intend to help. Intentions is what always matters. Thanks everyone again...:smile:
 
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