Three Pulley - two masses system.

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SUMMARY

The forum discussion centers on calculating the accelerations of a three-pulley system with two masses, m1 and m2, released from rest under ideal conditions (massless pulleys and frictionless strings). The participants derive equations based on force balancing and constraints, ultimately concluding that the acceleration a0 equals 3g, where g is the acceleration due to gravity. The discussion highlights the importance of understanding tension in the strings and the relationship between the accelerations of the masses and pulleys.

PREREQUISITES
  • Understanding of Newton's second law (F=ma)
  • Familiarity with constraint equations in mechanics
  • Basic knowledge of tension in strings and pulleys
  • Concept of acceleration due to gravity (g)
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  • Study the derivation of constraint equations in pulley systems
  • Learn about the dynamics of multi-body systems in classical mechanics
  • Explore the effects of friction and mass on pulley systems
  • Investigate advanced topics in tension and forces in non-ideal pulley systems
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Students in physics, mechanical engineers, and anyone interested in understanding the dynamics of pulley systems and forces in mechanics.

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.
VOekIsH.jpg


Homework Equations


Constraint equations.
F=ma

The Attempt at a Solution


b9e552fd-9e66-41fb-a8ca-ad8e1c9fb75f
HAz4uxU.jpg
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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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
 
haz4uxu-jpg.jpg
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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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.
 
It is not the same string
 
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.
 
  • #10
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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  • #11
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?
 
  • #12
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?
 
  • #13
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
 
  • #14
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.
 
  • #15
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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  • #16
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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  • #17
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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  • #18
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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