What is the compression force of the spring being applied to the block?

[RohansK]

When thre is a spring attatched to a certain body (like take the example of the suspensions/shock absorbers of an automobile) and a force is applied through that spring.

The automobile suspensions can be the best eamle for this discussion. We say that the force generated (which we experience as the jerk/jolt/shock upwards) due to the vehicle falling in any ditch/pit (this is supposedly the reaction force to the action generated by hte weight of the vehicle sort of falling down the ditch/pit).

The suspension is attatched with the springs because of the following phenomena:

The upwards jerk/force which is generated is first absorbed by the springs and due to the spring stiffness 'k' some part of this jerk/shock force is consumed to overcome the stiffness of the spring and the spring is compressed in due course.

This which results in a much lesser amount of force out of the total shock to be transmitted further upwards to the passengers of the vehicle. And so the shock transmitted to/experienced by the passengers is much lesser and bearable.

The POINT HERE IS : that say a total force (shock) of 100N was applied to the springs initially, then say a certain amount of it say 60N was utilised in compressing the springs (let us suppose the value of 'k' and avvailable distance for compression 'x' be such that force required by the spring is 60N) then naturally only the remaining part of 40N will be transmitted further upwards to the body (the vehicle) attatched to the springs. This is what the principle of the suspensions suggests.



Now consider a similar scenario where there is a block of mass 'm' in the positive X-Y plane with a spring atatched to its right hand side face. And the mass can be moved towards the left by applying a force to it from the right side through the spring (block/spring/<-- force)

Now suppose we apply a force say 10n to it through the spring and the values of 'k' and the length of spring was such taht the total compression of the spring would require only 4N force.

Then what I have been told is that the block of mass 'm' (to which the springs are attatched) will have only 10-4 = 6N froce to move itself towards the left.

And thus it will have only a fraction of the total force applied (from the right ) to cause its motion (related to F = m.a)

And this absolutely complies with the spring - force principle/phenomennon used in the example of the working of automobile suspensions.

Is it true for the block-spring case mentioned above.


Secondly, if the block had a wall on its left as a support, then if the force of 10N be applied to the block through the springs then the REACTION at the support wall was given as FULL 10N. HOW?

I couldn't understand this clearly at all.

Say in the earlier case where the block could move, we said that the force available for moving the block was 10 - 4 = 6N , so now as the same method of applying the force is used here also, then the same amount of force must cause the reaction to the block i.e. 6N only. WHY is it the full 10N? if a part of that 10N was utilised in overcoming the spring stiffness ( the 4N mentioned earlier).

I know this sounds a bit basic but these are the small things which we do not think of in depth and run after the bigger concepts.

Can somebody please explain this in detail

 

[jack action]

You are wrong: The spring doesn't absorb the force, it absorbs the motion.

No matter what force you have at one end of a spring, you have the same one at the other end.

As for the car suspension, let's analyze it:

The spring is subjected to a certain wheel force F caused by a bump of height dx. That force goes through the spring to reach the mass m of the car. The car is then subjected to an acceleration a = F / m = (K / m) * dx.

Now, if K is very large or m is very small, then a will be very large. This means that the car will move very easily (upward). Hence, the spring will decompress quickly and the car motion will follow the road bump.

But if K is very small or m is very large, then a will be close to zero. This means that the car will barely move. Hence, if the car doesn't move, the spring will compress a distance equivalent to the bump height. And if the car goes fast enough, it will have time to go over the bump, thus decompressing the spring and removing the force before the car starts moving upward noticeably.

Note that if the car goes slowly enough over the bump, no matter how soft is the spring or how heavy the car is, the car will follow the road shape and the spring will never compress. So the goal of a suspension is to absorb the wheel acceleration (not the force) long enough such that the car have the time to go over the disturbance before it reaches the passengers (high acceleration are what causes discomfort and shocks).

 

[RohansK]

The theory regarding How a Car experiences a bump does not appeal to me much.

The statement that

Now, if K is very large or m is very small, then a will be very large. This means that the car will move very easily (upward). Hence, the spring will decompress quickly and the car motion will follow the road bump.

does not convince me much.

Anyway, my question is regarding the transmission of force through the spring and reducing the shock/jerk experienced by the passengers (just check if my understanding of the basic suspensions os correct)

And the real question is: Can we apply the same logic to a block attatched with spring and made to move forward by applying the force to the spring end, as I have mentioned in the original question.

And the difference between this example and the case where the block has a support behind it. (Read my original Question please).

Please give in your replies and explain this.

 

[jack action]

Anyway, my question is regarding the transmission of force through the spring and reducing the shock/jerk experienced by the passengers (just check if my understanding of the basic suspensions os correct)

Again, force "passes" through a spring. It is not "filtered". However, displacement is. And from displacement you obtain velocity, from velocity you obtain acceleration and from acceleration you obtain jerk (the 3rd derivative of displacement wrt time). Motion is what's felt by the passengers, that's what create discomfort.

And the real question is: Can we apply the same logic to a block attatched with spring and made to move forward by applying the force to the spring end, as I have mentioned in the original question.

And the difference between this example and the case where the block has a support behind it. (Read my original Question please).

Please give in your replies and explain this.

Whether the motion is vertical or horizontal, the same logic applies; but not the logic mentioned in the OP.

The force that compresses the spring doesn't "go away". To be compressed, a spring needs to have two forces applied to itself: One at each end. Otherwise, the spring will not compress, it will move.

Just try it yourself. Take a spring, hold one end with each hand and compress it. Are you applying force with only one hand? If so, which one? The fact is, there will be an equal and opposite force in each of your hand. Second experiment: Put the spring on a table (with a low friction coefficient) and push it very slowly with one hand. Does it compress? No, it will just start to move in the direction you push. If you push it fast enough, giving it a good acceleration, you will see it compress and that will be because of the effect of the spring mass itself resisting your force (just like the car suspension does with the mass of the car).