Spring block system on inclined plane

[randomgamernerd]

1. Homework Statement :
The block of mass m₁ as shown in the figure (pic attached)is fastened to the spring and the block of mass m₂ is placed against it.
a)Find the compression of the spring in equilibrium position.
b)The blocks are pushed a further distance (2/k)(m₁ + m₂)g sinθ
against the spring and released. Find the position where the two blocks seperate.

Homework Equations

: [/B]
F=kx (Hookes law)

The Attempt at a Solution

:[/B]
a)For the time being...we are replacing the two blocks with a block of equivalent mass(M)
i.e M = m₁ + m₂.
At equilibrium, net force on this new block = 0.
MgSinθ= kx
⇒x=MgSinθ/k = (m₁ + m₂)gSinθ/k
b) The blocks separate when normal force between them is zero. So m₂gSinθ stops acting on m₁
Let the stretch in the spring be y at this instant.
So,
At this instant, on m₂, no N is acting.
So,
m₂a=m₂gSinθ
⇒a=gSinθ
⇒F_s/(m₁+m₂)=gSinθ
⇒ky = (m₁+m₂)gSinθ
⇒y=(m₁+m₂)gSinθ/k
so, the spring should by stretch by (m₁+m₂)gSinθ/k
Why is the answer given "when the spring attains its normal length"...? where am I going wrong?

 

[TomHart]

⇒Fs/(m1+m2)=gSinθ

Can you explain where this came from?

 

[TomHart]

Let me try this. Let's say you have a spring vertically oriented that has a platform on top of it and m1 and m2 sitting on top of the platform (m2 on top of m1). You compress the spring a great amount and then release it. In a perfect world, the two masses would be launched into the air from the platform and stay together. They would then continue on from there with an acceleration of -9.8 m/s².

Next, let's say that you glue m1 to the platform and repeat the process. m2 gets launched into the air and continues from there with an acceleration of -9.8 m/s². What can you say about m1's acceleration relative to m2's acceleration at the point they separate? What about after they separate? What condition causes that?

 

[randomgamernerd]

Let me try this. Let's say you have a spring vertically oriented that has a platform on top of it and m1 and m2 sitting on top of the platform (m2 on top of m1). You compress the spring a great amount and then release it. In a perfect world, the two masses would be launched into the air from the platform and stay together. They would then continue on from there with an acceleration of -9.8 m/s².

Next, let's say that you glue m1 to the platform and repeat the process. m2 gets launched into the air and continues from there with an acceleration of -9.8 m/s². What can you say about m1's acceleration relative to m2's acceleration at the point they separate? What about after they separate? What condition causes that?

Look man..I'm already ample confused...I appreciate your attempt to help me out...but the way you are giving me "hints" to the sum...it seems you are confused yourself and asking me a question rather than giving me hint...It will be helpful if you clarify what you want to say...A little bit of rephrasing will be better...

 

[randomgamernerd]

Someone please help...

 

[TJGilb]

Quick question, are you familiar with integral calculus at this point in time or expected to know it for this class?

 

[TomHart]

I think the main goal here is to try to help people understand. That is why I asked you the first question that I asked in post #2. I wanted you to think about that equation to consider if it really made sense. If you thought about post #3 and it did not help, then throw it away.

You wrote the following equation:

⇒a=gSinθ

This one made sense to me. It is the acceleration of block m2 at the instant the two blocks separate. The reason you came up with that is because you said:
"The blocks separate when normal force between them is zero." I agree with that.

And then your next equation was:

⇒Fs/(m1+m2)=gSinθ

So this equation substituted Fs/(m1 + m2) for a. So now you are basing the acceleration on Fs acting upon both masses. Why would you do that if you know the blocks are separated?

Basically what you found is how far the spring has to be compressed to produce a force that would give 2 blocks the same acceleration as one block would have if it was sliding freely on the ramp.

Here is the main point I was trying to get you to see:
Up until separation, blocks m1 and m2 have the exact same acceleration. When separation occurs, the only force acting on block m2 along the ramp is its weight component in that direction. Thus its acceleration is mgsinθ - like you found. The only reason m1 could have a greater acceleration in the down-the-ramp direction is if it had another force, in addition to its own weight component, acting in that direction. What additional force could possibly show up that would begin to pull m1 down the ramp, resulting in a greater down-the-ramp acceleration than m2 has? I'll give you a hint. It is something that is attached to m1 that, up until this separation time, was pushing it rather than pulling it.

I hope that helps.

 

[John Park]

Why is the answer given "when the spring attains its normal length"...? where am I going wrong?

Reference https://www.physicsforums.com/threads/spring-block-system-on-inclined-plane.907142/#post-5713716

It looks as though you're thinking of part (b) as another statics problem; you don't seem to have used the fact that the masses are moved a further distance and then released. Have you visualised what will happen then?

 

[randomgamernerd]

Quick question, are you familiar with integral calculus at this point in time or expected to know it for this class?

yup

 

[randomgamernerd]

I think the main goal here is to try to help people understand. That is why I asked you the first question that I asked in post #2. I wanted you to think about that equation to consider if it really made sense. If you thought about post #3 and it did not help, then throw it away.

yeah..post 3 confused me...Sorry if I sounded rude...I'm under huge pressure actually...
"So this equation substituted Fs/(m1 + m2) for a. So now you are basing the acceleration on Fs acting upon both masses. Why would you do that if you know the blocks are separated?"

I did not think about that...sorry..

Here is the main point I was trying to get you to see:
Up until separation, blocks m1 and m2 have the exact same acceleration. When separation occurs, the only force acting on block m2 along the ramp is its weight component in that direction. Thus its acceleration is mgsinθ - like you found. The only reason m1 could have a greater acceleration in the down-the-ramp direction is if it had another force, in addition to its own weight component, acting in that direction. What additional force could possibly show up that would begin to pull m1 down the ramp, resulting in a greater down-the-ramp acceleration than m2 has? I'll give you a hint. It is something that is attached to m1 that, up until this separation time, was pushing it rather than pulling it.

I hope that helps.

The additional force that shows up is the pull force due to the spring.
So forces on m₁ are m₁gSinθ and spring force...
So if we assume acceleration of m₁ to be a, then
m₁a = m₁gSinθ + kx
what is a equal to here??
Is it that common acceleration??
And what is the use of the other information("if the blocks are pushed a further distance (2/k)...")?

 

[TomHart]

yeah..post 3 confused me...Sorry if I sounded rude...I'm under huge pressure actually...

I have been there and fully remember the frustration. No problem.

So forces on m1 are m1gSinθ and spring force...

Yes. And I would say that the separation occurs just as the spring passes the zero-force point and just begins to pull. Or mathematically, it occurs when both blocks have acceleration mgsinθ. That is the point where the force of the spring is 0. For a particular spring, its zero-force point is ALWAYS at its normal, unloaded length.

And what is the use of the other information("if the blocks are pushed a further distance (2/k)...")?

I haven't verified it, but I'm fairly certain that the purpose of that information is to ensure that the spring gets compressed far enough for the 2 blocks to separate. If you imagine the 2 blocks resting on the spring in equilibrium and then you apply a gentle pressure compressing the spring slightly and releasing, the blocks would oscillate back and forth with a very small amplitude. It would not be enough spring compression that, when released, would cause the 2 blocks to separate. If you applied a very large pressure to compress the spring a large amount and then released, that would result in a much larger spring force - enough so that the blocks would separate.

 

[haruspex]

m₁a = m₁gSinθ + kx

What about the corresponding equation for the other mass? At the point where they separate, what will be the relationship between the two accelerations?
(This problem is much simpler than it appears.)

 

[TomHart]

(This problem is much simpler than it appears.)

Agreed. The solution is almost trivial, but not immediately obvious that it is almost trivial.

 

[randomgamernerd]

I have been there and fully remember the frustration. No problem.

Yes. And I would say that the separation occurs just as the spring passes the zero-force point and just begins to pull. Or mathematically, it occurs when both blocks have acceleration mgsinθ. That is the point where the force of the spring is 0. For a particular spring, its zero-force point is ALWAYS at its normal, unloaded length.

I haven't verified it, but I'm fairly certain that the purpose of that information is to ensure that the spring gets compressed far enough for the 2 blocks to separate. If you imagine the 2 blocks resting on the spring in equilibrium and then you apply a gentle pressure compressing the spring slightly and releasing, the blocks would oscillate back and forth with a very small amplitude. It would not be enough spring compression that, when released, would cause the 2 blocks to separate. If you applied a very large pressure to compress the spring a large amount and then released, that would result in a much larger spring force - enough so that the blocks would separate.

okay..now I got it...thanks for the help...

 

[randomgamernerd]

Okay..so I'm just describing what's happening here...
First, we placed the block m₂..or it was already placed...whatever..
Now the spring is not in its natural length..because the two blocks are exerting a force on it...
The spring is now compressed until its compressed to a length (m₁+m₂)gSinθ/k...where it attains equilibrium since net force on it is zero..
Now the spring is further compressed to 2(m1+m2...
The system will try to regain equilibrium...
So the restoring spring force acts...pushes both the block m1...which is again pushing block m2..
After the spring is again back to equilibrium length i.e (m₁+m₂)gSinθ/k..it overshoots itself because of inertia...so length of the spring is now approaching its natural length...now the forces on m₁ are spring force which is pushing it up and m₁gSinθ which is ushing it down..The normal force from m₁ on m₂ is pushing it up..forces on m2 are this normal force and m2gSinθ..As the spring approaches its natural length, spring force is decreasing..finally when the spring attains its natural length no spring force acts..So both are having same acceleration gsinθ...but now spring force comes back again to restore equilibrium position(which we calculated above) and hence now two forces acts on m1: one is m1gsinθ pushing down, another is spring force pulling it down..this spring force causes an additional downward acceleration and the two seperate..so they get separated when spring attains its natural length..

Am I right??
But I am still having trouble in expressing it mathematically I know I'm overthinking..but..

 

[randomgamernerd]

someone please confirm..

 

[randomgamernerd]

What about the corresponding equation for the other mass? At the point where they separate, what will be the relationship between the two accelerations?
(This problem is much simpler than it appears.)

That will also be a?? Then it becomes m₂a=m₂gSinθ
So we have two equations:
m₁a = m₁gSinθ + kx ...equation A
m₂a=m₂gSinθ...equation B
From these two I get:
m1m2gSinθ=m1m2gSinθ + m2kx
⇒x=0
Oh! I got it..

 

[haruspex]

That will also be a?? Then it becomes m₂a=m₂gSinθ
So we have two equations:
m₁a = m₁gSinθ + kx ...equation A
m₂a=m₂gSinθ...equation B
From these two I get:
m1m2gSinθ=m1m2gSinθ + m2kx
⇒x=0
Oh! I got it..

Great.
As TomHart noted in post #11, the information about the extra compression is to ensure that the spring has enough stored energy that the masses reach the point where the spring is relaxed. Otherwise, it could simply have said "assuming the masses later separate..."