Acceleration of a spring - mass system

[songoku]

Homework Statement

The diagram given below.

Find the acceleration of each mass when the string connecting $m_4$ to the ground is cut.
You can assume that the system is at equilibrium before the string is cut, the string is massless and inextensible, the springs are massless, the pulley is smooth and massless.

Relevant Equations

Newton 2nd law

m₁ = top left
m₂ = bottom left
m₃ = top right
m₄ = bottom right

My questions:
1. Will all the object (m₁, m₂, m₃,and m₄) have same acceleration?

2. Should I assume initial extension of both spring is the same? (only based on the picture)

3. Will the extension of the spring change after the string is cut?

Thanks

 

[BvU]

https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/

Your attempt is appreciated (and required according to the guidelines): what do you think ?

 

[songoku]

https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/

Your attempt is appreciated (and required according to the guidelines): what do you think ?

I have tried to draw free body diagram for each object when they are in equilibrium.

For m₁, there are tension upwards, weight downwards and spring force downwards

For m₂, there are tension upwards and weight downwards

For m₃, there are tension upwards, weight downwards and spring force downwards

For m₄, there are tension downwards, weight downwards and spring force upwards

After the string is cut, I tried to set up equation of Newton 2nd law but not sure about all the things I asked in post#1

I do not know whether the acceleration of all the masses will be the same or not. Maybe the acceleration can be the same because they all move as one system but I also think the acceleration can be different if somehow the extension of the spring changes during the motion, so I can't proceed.

How to know whether the extension will change or not during the motion?

Thanks

 

[haruspex]

I do not know whether the acceleration of all the masses will be the same or not.

Then assume not. If they are, the equations will lead to that.

 

[songoku]

ΣF₁ = k.x₁ + m₁.g - T
m₁.a₁ = k.x₁ + m₁.g - T ...(1)

ΣF₂ = m₂.g- k.x₁
m₂.a₂ = m₂.g- k.x₁ ... (2)

ΣF₃ = T - k.x₂ - m₃.g
m₃.a₃ = T - k.x₂ - m₃.g ...(3)

ΣF₄ = k.x₂ - m₄.g
m₄.a₄ = k.x₂ - m₄.g ... (4)

Then how to continue? Thanks

 

[BvU]

Count the number of unknowns to check if you miss an equation ...

And: what is $T, x_1,$ etc ?
Do you describe equilibrium, or the situation after the wire is cut ?

 

[haruspex]

how to continue?

"string is massless and inextensible"

 

[vela]

ΣF₁ = k.x₁ + m₁.g - T
m₁.a₁ = k.x₁ + m₁.g - T ...(1)

ΣF₂ = m₂.g- k.x₁
m₂.a₂ = m₂.g- k.x₁ ... (2)

ΣF₃ = T - k.x₂ - m₃.g
m₃.a₃ = T - k.x₂ - m₃.g ...(3)

ΣF₄ = k.x₂ - m₄.g
m₄.a₄ = k.x₂ - m₄.g ... (4)

Then how to continue? Thanks

The amount the spring on the left is stretched depends on both $x_1$ and $x_2$. Same issue with the spring on the right. Your equations don't reflect this.

 

[songoku]

Count the number of unknowns to check if you miss an equation ...

I have 7 unknowns (a₁, a₂, a₃, a₄, T, x₁, x₂) and 4 equations so I am missing 3 equations and I do not know what those 3 equations are

And: what is $T, x_1,$ etc ?

T = tension on string connecting m₁ and pulley = tension on string connecting m₃ and pulley

x₁ = extension of spring connecting m₁ and m₂

x₂ = extension of spring connecting m₃ and m₄

Do you describe equilibrium, or the situation after the wire is cut ?

I describe the situation after the wire connecting m₄ and ground is cut.

I assume at equilibrium, both spring are extended by same amount so the springs produce equal forces on all the masses (let say k.x, where x is extension of both spring when the system is in equilibrium)

After the wire is cut, I assume the extension of both springs is not the same so on left spring is x₁ and right spring is x₂

"string is massless and inextensible"

Sorry I don't understand the hint. What I can do about "string is massless and inextensible" is assigning same value of tension on string on left and right side of the pulley but I don't think this is what you mean

The amount the spring on the left is stretched depends on both $x_1$ and $x_2$. Same issue with the spring on the right. Your equations don't reflect this.

I don't even realize this, that the amount the spring on the left is stretched depends on both $x_1$ and $x_2$

I can say I don't understand how the system will move. This is what I thought at first:

1.
m₃ and m₄ will move upwards and because I don't understand whether they will move with same acceleration or not, I assume they don't. The right spring will have less extension compared to when the system is equilibrium so the extension of spring changes from x to x₂, where x > x₂ (don't understand why, just pure guessing).

2.
m₁ and m₂ will move downwards and because I don't understand whether they will move with same acceleration or not, I assume they don't. The left spring will have more extension compared to when the system is equilibrium so the extension of spring changes from x to x₁, where x < x₁ (again, just pure guessing)

Based on vela's hint, it is obvious that my thinking is wrong. The extension of right spring does not change from x to x₂ and extension of left spring does not changes from x to x₁ but I don't know how to relate the extension to both x₁ and x₂

Thanks

 

[haruspex]

What I can do about "string is massless and inextensible" is assigning same value of tension on string on left and right side of the pulley but I don't think this is what you mean

If it is inextensible, what is the relationship between the acceleration of one end and the acceleration of the other end?

 

[songoku]

If it is inextensible, what is the relationship between the acceleration of one end and the acceleration of the other end?

Oh, the acceleration should be the same so a₁ = a₃

What will be the next step? What should I understand to continue? Thanks

 

[haruspex]

What will be the next step? What should I understand to continue? Thanks

We are only concerned with the instant after the string is cut. What will be the extensions at that time?

 

[songoku]

We are only concerned with the instant after the string is cut. What will be the extensions at that time?

Will be the same as initial extension of the spring when at equilibrium?

When at equilibrium, how will the extension of the spring be? I am trying to understand the direction of the restoring force. Let say I consider the spring on the right, is the extension of the spring is both upwards and downwards (I mean m₄ will cause downwards extension and m₃ causes upwards extension) so the restoring force on m₄ will be upwards and on m₃ will be downwards?

Thanks

 

[haruspex]

Will be the same as initial extension of the spring when at equilibrium?

When at equilibrium, how will the extension of the spring be? I am trying to understand the direction of the restoring force. Let say I consider the spring on the right, is the extension of the spring is both upwards and downwards (I mean m₄ will cause downwards extension and m₃ causes upwards extension) so the restoring force on m₄ will be upwards and on m₃ will be downwards?

Thanks

Extension is how much longer the spring is than when it is relaxed. There is no separate upwards and downwards. Likewise, it exerts the same pull at each end.

 

[songoku]

Extension is how much longer the spring is than when it is relaxed. There is no separate upwards and downwards. Likewise, it exerts the same pull at each end.

So the extension of the spring at the instant when the wire is cut is the same as the extension of the spring when at equilibrium? I think from vela's hint there will be change in the extension, or maybe I am misinterpreting something?

Thanks

 

[haruspex]

the extension of the spring at the instant when the wire is cut is the same as the extension of the spring when at equilibrium?

Yes. It can't change instantly.

vela's hint

@vela seems to have assumed your x variables refer to the heights of masses.
You didn't define them, but I assume they are the spring extensions.

 

[songoku]

Yes. It can't change instantly.

@vela seems to have assumed your x variables refer to the heights of masses.
You didn't define them, but I assume they are the spring extensions.

Ok, I think I am missing one more equation. Maybe equation relating a₁ to a₂ or a₃ to a₄? Will a₁ be the same as a₂ because at the instant the wire is cut, m₁, the left spring and m₂ will move downwards as one system and also the extension of the spring has not changed?

Thanks

 

[haruspex]

I think I am missing one more equation.

How so?
You know the tensions in the springs now. That leaves four accelerations (but two are the same), and the tension in the string. You have four equations. Solve.

 

[songoku]

How so?
You know the tensions in the springs now. That leaves four accelerations (but two are the same), and the tension in the string. You have four equations. Solve.

At equilibrium position:
1. for object m₁
∑F = 0
T = m₁.g + kx (where T = tension on string connecting m₁ and pulley = tension on string connecting m₃ and pulley and x is extension of the sping)

2. for object m₂
∑F = 0
kx = m₂.g

3. for object m₃
∑F = 0
T = m₃.g + kx

4. for object m₄
∑F = 0
kx = m₄.g + T_s (where T_s is the tension on string connecting m₄ to ground)After string connecting m₄ to ground is cut:
1. for object m₁
∑F₁ = m₁ . a₁
k.x + m₁.g - T' = m₁ . a₁ → I think the magnitude of T will change after T_s is cut because if not, m₁ won't have acceleration so I use T' to denote the magnitude of the tension connecting m₁ and pulley when m₁ is accelerating. Am I correct?

2. for object m₂
∑F₂ = m₂ . a₂
m₂ . g - kx = m₂ . a₂ → because m₂ . g = kx so a₂ = 0?

3. for object m₃
∑F₃ = m₃ . a₁
T' - kx - m₃.g = m₃ . a₁

4. for object m₄
∑F₄ = m₄ . a₄
kx - m₄.g = m₄ . a₄

Am I correct until this part?

Thanks

 

[vela]

@vela seems to have assumed your x variables refer to the heights of masses.
You didn't define them, but I assume they are the spring extensions.

Yes, I did. Sorry. Carry on!

 

[songoku]

Yes, I did. Sorry. Carry on!

It was on me because I didn't define the variables

 

[BvU]

At equilibrium position:

Counting variables is still a problem: you need to distinguish $x_1$ and $x_2$. The picture shows both springs have the same $k$.
"Knowns" $\qquad m_1, m_2, m_3, m_4, k$
Unknowns $\qquad T, T_s, x_1, x_2$
4 equations, so now the unknowns are determined. In particular $T_s$

After string connecting $m_4$ to ground is cut:

Counting variables is now a big problem: You can't make do with $x_1$ and $x_2$ any more: there are 4 masses and three accelerations, so you'll need to re-think the kx

My suggestion: have four $x$ (well defined) and four $a$ and try to find 8 equations ($a_1 = -a_3$ being one of them)

It might also help if you first do the exercise with $m_2 = m_4 = 0$ .

--------[edit: added:]---------

I think the magnitude of T will change after Ts is cut because if not, m1 won't have acceleration so I use T' to denote the magnitude of the tension connecting m1 and pulley when m1 is accelerating. Am I correct?

Yes.

 

[haruspex]

[after string is cut]
Counting variables is now a big problem: You can't make do with $x_1$ and $x_2$ any more:

But those have not changed. The only unknowns are the tension in the string and three accelerations.

 

[BvU]

My mistake --

 

[songoku]

At equilibrium position:
1. for object m₁
∑F = 0
T = m₁.g + kx (where T = tension on string connecting m₁ and pulley = tension on string connecting m₃ and pulley and x is extension of the sping)

2. for object m₂
∑F = 0
kx = m₂.g

3. for object m₃
∑F = 0
T = m₃.g + kx

4. for object m₄
∑F = 0
kx = m₄.g + T_s (where T_s is the tension on string connecting m₄ to ground)After string connecting m₄ to ground is cut:
1. for object m₁
∑F₁ = m₁ . a₁
k.x + m₁.g - T' = m₁ . a₁ → I think the magnitude of T will change after T_s is cut because if not, m₁ won't have acceleration so I use T' to denote the magnitude of the tension connecting m₁ and pulley when m₁ is accelerating. Am I correct?

2. for object m₂
∑F₂ = m₂ . a₂
m₂ . g - kx = m₂ . a₂ → because m₂ . g = kx so a₂ = 0?

3. for object m₃
∑F₃ = m₃ . a₁
T' - kx - m₃.g = m₃ . a₁

4. for object m₄
∑F₄ = m₄ . a₄
kx - m₄.g = m₄ . a₄

Am I correct until this part?

Thanks

Is my working correct? Acceleration of object 2 is zero?

Thanks

 

[haruspex]

Is my working correct? Acceleration of object 2 is zero?

Thanks

Yes, but you seem to have overlooked @BvU 's advice:

you need to distinguish x1 and x2.

 

[songoku]

Yes, but you seem to have overlooked @BvU 's advice:

Let me redo:

At equilibrium position:
1. for object m₁
∑F = 0
T = m₁.g + kx₁ ...(i) (where T = tension on string connecting m₁ and pulley = tension on string connecting m₃ and pulley and x is extension of the left sping)

2. for object m₂
∑F = 0
kx₁ = m₂.g ... (ii)

3. for object m₃
∑F = 0
T = m₃.g + kx₂...(iii) (where x₂ is the extension of the right spring)

4. for object m₄
∑F = 0
kx₂ = m₄.g + T_s ...(iv) (where T_s is the tension on string connecting m₄ to ground)

From (i) and (ii): T = m₁.g + m₂.g ...(v)

From (v) and (iii):
m₁.g + m₂.g = m₃.g + kx₂
kx₂ = m₁.g + m₂.g - m₃.g ... (vi)After string connecting m₄ to ground is cut:
1. for object m₁
∑F₁ = m₁ . a₁
k.x₁ + m₁.g - T' = m₁ . a₁, where T' is the tension of string after T_s is cut
m₂.g + m₁.g - T' = m₁ . a₁
T' = m₂.g + m₁.g - m₁ . a₁ ... (vii)

2. for object m₂
∑F₂ = m₂ . a₂
m₂ . g - kx₁ = m₂ . a₂
a₂ = 0

3. for object m₃
∑F₃ = m₃ . a₁
T' - kx₂ - m₃.g = m₃ . a₁ ... (viii)

From (vi), (vii) and (viii):
m₂.g + m₁.g - m₁ . a₁ - m₁.g - m₂.g + m₃.g - m₃.g = m₃ . a₁
a₁ = 0?

Thanks

 

[haruspex]

Let me redo:

At equilibrium position:
1. for object m₁
∑F = 0
T = m₁.g + kx₁ ...(i) (where T = tension on string connecting m₁ and pulley = tension on string connecting m₃ and pulley and x is extension of the left sping)

2. for object m₂
∑F = 0
kx₁ = m₂.g ... (ii)

3. for object m₃
∑F = 0
T = m₃.g + kx₂...(iii) (where x₂ is the extension of the right spring)

4. for object m₄
∑F = 0
kx₂ = m₄.g + T_s ...(iv) (where T_s is the tension on string connecting m₄ to ground)

From (i) and (ii): T = m₁.g + m₂.g ...(v)

From (v) and (iii):
m₁.g + m₂.g = m₃.g + kx₂
kx₂ = m₁.g + m₂.g - m₃.g ... (vi)After string connecting m₄ to ground is cut:
1. for object m₁
∑F₁ = m₁ . a₁
k.x₁ + m₁.g - T' = m₁ . a₁, where T' is the tension of string after T_s is cut
m₂.g + m₁.g - T' = m₁ . a₁
T' = m₂.g + m₁.g - m₁ . a₁ ... (vii)

2. for object m₂
∑F₂ = m₂ . a₂
m₂ . g - kx₁ = m₂ . a₂
a₂ = 0

3. for object m₃
∑F₃ = m₃ . a₁
T' - kx₂ - m₃.g = m₃ . a₁ ... (viii)

From (vi), (vii) and (viii):
m₂.g + m₁.g - m₁ . a₁ - m₁.g - m₂.g + m₃.g - m₃.g = m₃ . a₁
a₁ = 0?

Thanks

Right so far. What about m₄'s acceleration?

 

[songoku]

Right so far. What about m₄'s acceleration?

For object m₄:
k.x₂ - m₄.g = m₄ . a₄
m₁.g + m₂.g - m₃.g - m₄.g = m₄. a₄
a₄ = (m₁.g + m₂.g - m₃.g - m₄.g) / m₄

Only object 4 is accelerating, while object 1, object 2 and object 3 move with constant speed?

Thanks

 

[haruspex]

For object m₄:
k.x₂ - m₄.g = m₄ . a₄
m₁.g + m₂.g - m₃.g - m₄.g = m₄. a₄
a₄ = (m₁.g + m₂.g - m₃.g - m₄.g) / m₄

Only object 4 is accelerating, while object 1, object 2 and object 3 move with constant speed?

Thanks

Yes, but what "constant speed"?

 

[songoku]

Yes, but what "constant speed"?

hhmm... I think the value of the speed will be zero because initially object 1 to 3 does not move and no acceleration means that no resultant force acting on it.

So, at the instant the wire is cut, only object 4 will move?

Thanks

 

[haruspex]

hhmm... I think the value of the speed will be zero because initially object 1 to 3 does not move and no acceleration means that no resultant force acting on it.

So, at the instant the wire is cut, only object 4 will move?

Thanks

Yes.

 

[songoku]

Yes.

What if I modify the question a little bit, find the acceleration 2 seconds after the wire is cut?

This is what I have in my mind:
The m₄ will push the right spring so the new extension (x₃) will be less than x₂ and all the other masses will move.

The extension on left spring does not change (still x₁) because the left spring, m₁ and m₂ will move as one system so the acceleration of m₁ and m₂ will be the same.

This means that the left spring will behave just like an inelastic string.

The magnitude of a₁, a₂ and a₃ will always be the same but all the acceleration, including a₄ won't be constant because there is continuous change in the extension of right spring, therefore there is continuous change in restoring force and there is continuous change in resultant force.

Am I correct?

If yes, how to set up equation of m₄ to account for continuous change in acceleration?

Thanks

 

[haruspex]

What if I modify the question a little bit, find the acceleration 2 seconds after the wire is cut?

Then you will need to have unknowns for the positions of three of the four masses independently (m1 and m3 will have a fixed relationship) and write the differential equations relating them. It is safe to assume there will be SHM involved, but it might be quite complicated.

 

[songoku]

Then you will need to have unknowns for the positions of three of the four masses independently (m1 and m3 will have a fixed relationship) and write the differential equations relating them. It is safe to assume there will be SHM involved, but it might be quite complicated.

Ok then I will leave it for now

Thank you very much for all the help BvU, haruspex, vela