0
$\begingroup$

Suppose I have a two masses attached with a spring on ei either side and one of the masses is kept in contact to the wall like in this diagram with friction neglected everywhere.

enter image description here

Now when the wall is removed after equilibrium is set up then instantaneous ly the spring force remains the same and it accelerates m2 towards right.What I don't understand is how the mass m2 is experiencing a spring force even thought it is not deforming the spring..is the reason just that we are assuming it as a counter force against normal and that too equal to F. Can objects experience spring force without deforming the spring ?

Like if I was holding a spring with it's end free and my friend starts pushing t until he isn't able to....will I move in this situation or more carefully will I be always moving in this situation

Improve this question
$\endgroup$

2 Answers 2

Reset to default
1
$\begingroup$

The spring force only shows up for linear springs like this when the spring is in expansion or compression.

Spring force is a resistance to deformation, so the spring has to be deformed from it's resting length for it to provide a spring force.

Consider the net force on the spring. If it didn't have two opposite forces acting on it, it wouldn't compress or extend, it would just be pushed by the force and move like every other object being moved by a single force.

Improve this answer
$\endgroup$
6
  • $\begingroup$ So the two forces have to be opposite to justify the compression of spring.....that makes a lot of sense and this thought will be major help in solving spring questions $\endgroup$
    – Abhinav
    Jul 19, 2018 at 10:30
  • $\begingroup$ But how can I relate between the magnitude of forces on the springs which is compressing it....Of course they will be equal in equilibrium but what can we say before equilibrium $\endgroup$
    – Abhinav
    Jul 19, 2018 at 10:33
  • $\begingroup$ Before equilibrium, you would have your spring force (which is like an internal force for the entire spring-mass system) and your net force acting on the system as a whole. The balanced spring force can determine the distance between the masses, and the additional net force that isn't balanced would determine how the whole system accelerates. $\endgroup$
    – JMac
    Jul 19, 2018 at 10:47
  • $\begingroup$ Before equilibrium if the spring is accelerating then how come it is compressing too...like you said the additional net force that isn't balanced would determine how the whole body accelerate.... I am sorry for asking this silly doubt but I am not getting the big picture like what is the cause of the force acting on the spring which is making it compress instead of moving like you said it should if pushed by a single force $\endgroup$
    – Abhinav
    Jul 19, 2018 at 11:39
  • $\begingroup$ It’s hard to tell from your picture and description IMO, but it looks like m2 is attached to the wall at first, and there’s tension being held in the spring from the force compressing it. Once the wall is removed, there’s nothing holding the compression anymore, and the spring extends (and starts accelerating due to applied forces). $\endgroup$
    – JMac
    Jul 19, 2018 at 11:56
1
$\begingroup$

In the beginning the free body diagrams for each of the components in your question are as shown below.

enter image description here

All you need to do is now remove the force on mass $2$ due to the wall $F_{\rm 2w}$ to see which forces are left.

Another way of looking at the system is to imagine that $m_1 \gg m_2$ and so when the force due to the wall was removed mass $1$ would have a very small acceleration, ie would hardly move at all, and mass $2$ would move much more than mass $1$ because the compressed spring is trying to expand.
So the spring must be exerting a force on mass $2$.

Improve this answer
$\endgroup$
1
  • $\begingroup$ At any finite instant after wall is removed would. F s1 and F s2 would be equal........would this mean the spring will remain at rest but would be accelerating m2 and the magnitudes of F s1 and Fs2 would change but remain equal How can I calculate this new magnitude if this is true $\endgroup$
    – Abhinav
    Jul 19, 2018 at 23:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.