How to solve this problem by using Newton's equations?

[FrancisClinton]

They have solved this problem by energy method but i want to solve this by normal force equations but when i try to solve by this method suddenly a i had a doubt whether i have to solve the problem assuming that the point x₂ always move in a straight line (this is easy to solve) or point x₂ can move in a curve for solving this way , i think there should be more data available for this.

 

[paisiello2]

No more data required.

It does say when the system is in equilibrium so that means that nothing is moving.

It also defines the length of string on either side to be equal, therefore x₂ is always located on a straight line when the system is in equilibrium. However, you are right that when the system is not in equilibrium then x₂ would move on a curve.

 

[FrancisClinton]

so how to solve this one

 

[paisiello2]

Didn't you say it was easy to solve?

 

[haruspex]

so how to solve this one

Can you show the tension should be the same thoughout the string? Can you then show that the middle pulley must be half way between A and B?

 

[FrancisClinton]

Didn't you say it was easy to solve?

Yes i can solve , but i am not getting the same answer as them , so can you help me .

 

[FrancisClinton]

Can you show the tension should be the same thoughout the string? Can you then show that the middle pulley must be half way between A and B?

Tension won't be same through out the string , it will be m₁g between mass m₁ and m₂ and it will be m₁g + m₂g in between mass m₂ and the point A .
I think it depends on the m₁, m₂, x₁ and x₂ for certain values of this the middle pulley will always be half way between A and B. So that horizontal components acting on mass m₂ cancel each other.
Correct me if am wrong.

 

[paisiello2]

Yes i can solve , but i am not getting the same answer as them , so can you help me .

Ah, maybe not so easy then. You'll need to show your work before anyone can help you.

 

[haruspex]

Tension won't be same through out the string , it will be m₁g between mass m₁ and m₂ and it will be m₁g + m₂g in between mass m₂ and the point A .

if the tension is different in the string on one side of a pulley from what it is on the other, consider the torques acting on the pulley. What will happen?

 

[FrancisClinton]

Ah, maybe not so easy then. You'll need to show your work before anyone can help you.

So far the point x₂ to move in a straight line , there should be no horizontal component acting on it ,
So at equilibrium force acting on the point x₂ in the vertical direction is m₁gcosθ₁ + Tcosθ₂= m₂g where T= m₁g+m₂g , and θ₁ is the angle between the line segments of point B and mass m1 and point B and mass m2 and θ₂ is angle between the line segments point A and the mass m2 and vertical line drawn from A. here θ₁ and θ₂ should be equal since x2 moves vertically downwards , so answer i got for x₁ is =(bm₂- 2d(2m₁ +m₂ ))/m₂
but the answer they got is

 

[FrancisClinton]

if the tension is different in the string on one side of a pulley from what it is on the other, consider the torques acting on the pulley. What will happen?

rotate towards the higher force side

 

[haruspex]

rotate towards the higher force side

Right, but in this set up we are told it is in static equilibrium, so what does that tell you about the string tension?

 

[FrancisClinton]

Right, but in this set up we are told it is in static equilibrium, so what does that tell you about the string tension?

So far the point x2 to move in a straight line , there should be no horizontal component acting on it ,
So at equilibrium force acting on the point x2 in the vertical direction is m1gcosθ1 + Tcosθ2= m2g where T= m1g+m2g , and θ1 is the angle between the line segments of point B and mass m1 and point B and mass m2 and θ2 is angle between the line segments point A and the mass m2 and vertical line drawn from A. here θ1 and θ2 should be equal since x2 moves vertically downwards , so answer i got for x1 is =(bm2- 2d(2m1 +m2 ))/m2
but the answer they got is

whether this is correct

 

[haruspex]

So far the point x2 to move in a straight line , there should be no horizontal component acting on it ,

I thought you were trying to use the force method, and this is in equilibrium, so nothing is moving. Why do you keep mentioning x2 moving in a straight line?
But, yes, there cannot be any net horizontal force on the central pulley. The correct answer to my last post was "the tension is constant throughout the string". You denied that before but seem to be accepting it now, so let's move on.
Since there is no net horizontal force at the central pulley, but there is a tension T each side, what does that tell you about the angles of the strings there?

So at equilibrium force acting on the point x2 in the vertical direction is m1gcosθ1 + Tcosθ2= m2g where T= m1g+m2g , and θ1 is the angle between the line segments of point B and mass m1 and point B and mass m2 and θ2 is angle between the line segments point A and the mass m2 and vertical line drawn from A.

I don't understand how you get either of those equations. T most certainly is not equal to m1g+m2g.

 

[FrancisClinton]

I

I thought you were trying to use the force method, and this is in equilibrium, so nothing is moving. Why do you keep mentioning x2 moving in a straight line?
But, yes, there cannot be any net horizontal force on the central pulley. The correct answer to my last post was "the tension is constant throughout the string". You denied that before but seem to be accepting it now, so let's move on.
Since there is no net horizontal force at the central pulley, but there is a tension T each side, what does that tell you about the angles of the strings there?

I don't understand how you get either of those equations. T most certainly is not equal to m1g+m2g.

I can't able to understand your statement can you please elaborate on that.

 

[haruspex]

I

I can't able to understand your statement can you please elaborate on that.

Which statement don't you understand?
Do you agree now that the tension must be the same throughout the string?
Consider the forces acting on the hanging pulley. There are three. What are their horizontal components? What equation does that give you?

 

[FrancisClinton]

Which statement don't you understand?
Do you agree now that the tension must be the same throughout the string?
Consider the forces acting on the hanging pulley. There are three. What are their horizontal components? What equation does that give you?

I don't understood how the tension in the string is constant throughout ? ,
I will tell the idea behind how i got the value of my tension please tell whether this is correct ? , first i imagined the setup without the pulley M₂ so the tension on the string is same and = m₁g , now you are putting the pulley M₂ on the string , so at this point of the pulley M₂ will experience two forces acting on it one due to the mass m₁(acting from pulley M2 to point B to restore to its original position) and due to mass m₂ which is acting vertically downwards , so the tension on this point will be vector sum of these two forces , so depending on the points in the string there will be different tensions .

 

[haruspex]

I don't understood how the tension in the string is constant throughout ? ,
I will tell the idea behind how i got the value of my tension please tell whether this is correct ? , first i imagined the setup without the pulley M₂ so the tension on the string is same and = m₁g , now you are putting the pulley M₂ on the string , so at this point of the pulley M₂ will experience two forces acting on it one due to the mass m₁(acting from pulley M2 to point B to restore to its original position) and due to mass m₂ which is acting vertically downwards , so the tension on this point will be vector sum of these two forces , so depending on the points in the string there will be different tensions .

The flaw in your reasoning is that adding m2 won't increase the tension. It will merely redirect the tension. Instead of the tension being horizontal, it is now at an angle each side. That gives it a vertical component. Equilibrium is achieved when the vertical components of the tension each side balance the weight of m2. (If the weight is too great, equilibrium will not be achieved.)
The moral is this: don't trust ad hoc arguments that seem logical. Go through the standard steps of drawing free body diagrams and writing out the force and torque balance equations.

 

[FrancisClinton]

The flaw in your reasoning is that adding m2 won't increase the tension. It will merely redirect the tension. Instead of the tension being horizontal, it is now at an angle each side. That gives it a vertical component. Equilibrium is achieved when the vertical components of the tension each side balance the weight of m2. (If the weight is too great, equilibrium will not be achieved.)
The moral is this: don't trust ad hoc arguments that seem logical. Go through the standard steps of drawing free body diagrams and writing out the force and torque balance equations.

Thank you so much for your explanation , I got the same answer as them . Finally I am Happy.