I am unable to understand how springs and shock absorbers absorb impact. So it would be great if someone could explain how impact is absorbed; and if it could be used to protect someone's leg from beaking when falling from a height. Thanks in advance.
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1$\begingroup$ impact is high force but limited energy. when sufficient energy is introduced to a material, it breaks bonds. For an initial crack, rigid materials don't need much energy but do need a fairly high force per area (which a large force like an impact with a rigid mass can provide). By putting a less rigid material in, it converts impact forces to deformation (aka movement aka acceleration). This way the forces on the rigid entity are reduced because the same energy is dissipated over a longer time (and distance) $\endgroup$– AbelCommented Feb 6, 2022 at 12:13
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$\begingroup$ I think you can find excessive information on various websites on the internet about this. $\endgroup$– Rameez Ul HaqCommented Feb 6, 2022 at 13:44
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4 Answers
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During impact what usually happens is that a moving object with mass $m$ with an initial velocity $v_0$ stops (or rebounces) in a very short time $\Delta t$. Assuming a body stops during impact this means that there is a change of momentum which is equal to :
$$\Delta mv = - m \cdot v_0 $$
According to the impulse momentum theorem the average force applied to the object can be estimated by:
$$F_{av} = \frac{\Delta mv}{\Delta t}$$
I.e. for the same change in momentum if the duration is greater then the forces are greater.
So when for example a vase drop from a height to the floor (which is assumed rigid) the duration of the impact is very small, so the forces are high, and the vase brakes.
On the other hand if the vase fall onto a very thick pillow or mattress, the $\Delta t$ is greater (it takes loooonger to stop), and as a result the forces are lower (and sometimes you might get away without braking the vase).
So Spring and shock absorbers do just that, they increase the time it takes to slow down to stop a moving object thus reducing the forces.
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$\begingroup$ Which physical phenomena explains that it is necessary for the object to stop or rebound after striking ground, for example. $\endgroup$ Commented Feb 7, 2022 at 16:30
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Your question has two parts, springs and shock absorbers.
we are concerned with the effect of these in relation to impact, otherwise, there are other situations that we don't get into.
Impact in this scope means a collision with fast speed which is likely going to cause undesired damage.
In an impact, two factors come into play the speed of the moving object and the short time duration it comes to a sudden stop. It is called impulse,
$ \ J = F \cdot \Delta t =\Delta P=m\Delta v $
- v is speed
- m is mass
- J is impulse
- P is momentum, mass times speed
It grows as the speed grows or the time decreases.
A spring can receive the impact, resist it, and slow it down increasing the time the collision takes to stop, therefore making it less intense.
The same idea is applied in airbags, they deploy when they sense a crash, and they offer just a bit of extra time for our head before it hits the dashboard, the same idea as a spring, softening the impact.
The shock absorber in cars has the job of reducing the bounciness of the suspension and springs.
When spring is loaded during an impact it has potential energy and will bounce back, a shock absorber will dampen the bouncing beck, offering a smoth ride.
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$\begingroup$ Thank you very much $\endgroup$ Commented Feb 25, 2022 at 12:26
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The answer by NMech addresses one of the aspets of this matter, here is another aspect, based on theories of work and energy. .
The energy of a moving object crash into another object is $E = \dfrac{m(v_f^2-v_o^2)}{2}$. For a complete stop, $v_f = 0$, thus $E = -\dfrac{mv_o^2}{2}$.
Deformation in the direction of motion will result, that yields energy of, $E = F\Delta$ (on a non-yielding object), or $E = \dfrac {k\Delta^2}{2}$ (on a spring). At the time of the complete stop, the energy is absorbed, that is
$E = F\Delta - \dfrac{mv_o^2}{2} = 0$
Rearrange the terms, the force at the impact is
$F = \dfrac{mv_o^2}{2\Delta}$
From the equations above, we conclude that the crash introduces energy, and the energy is consumed/absrbed by the deformation/displacement of the objects. The smaller the deformation/displacement, the greater the impact force.
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$\begingroup$ Thank you very much $\endgroup$ Commented Feb 25, 2022 at 12:25
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The classic case of this are the springs and shock absorbers as part of a road vehicle suspension system
The simple answer is that the spring coupled with some sort of lever suspension an object that is subject to the shock is simply a device to allow the shock seen as an impulse like a hammer blow to be absorbed with a modest amount of movement.
The critical part of the mechanism us the damper or shock absorber which is a mechanical device that dissipates energy as the shock is absorbed by the movement. This is often a hydraulic device with a piston that forces a fluid through a narrow nozzle to make the liquid very turbulent this causes the liquid to heat up and this dissipate the energy of the shock.
Without the shock absorber part the spring would just bounce back snd push the infuse energy into a series of oscillations
The clever bit is to know the range of impulsive and to oscillatory shocks that need to be absorbed the range of load on the suspension and the amount of movement that is needed and ro get them matched correctly together to produce the desired results.
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$\begingroup$ Thank you very much $\endgroup$ Commented Feb 25, 2022 at 12:25