# Why would a suddenly applied force have more impact than the same force being applied over time?

I'm a fan of computer keyboards, and I was thinking about the durability of electrical contacts that lay below the keys on certain kinds of keyboards, such as this one.

I tried to wonder how many keystrokes that a PCB could take before wearing down, but then it occurred to me that a normal keystroke, as in, pressing down on the key until the key buckles and taps down on the PCB, seems like it'd do a lot more wear than starting with the key down on the PCB, and then pressing down.

Intuitively, I know that doing the "up-down" keystroke has more impact on the PCB, but since you could exert more force with the "down-down" press, with the key starting directly on the PCB, I can't figure out how the first one is supposed to have more impact.

I figured that the "down-down" press would transfer the force through the key, through the PCB, through the keyboard base, through the table into the ground, but I figured that the "up-down" press does the same thing. What's responsible for the latter having more of an impact on the PCB, assuming that the resulting force of the accelerating pressing finger is the same as the non-accelerating pressing finger over time?

Thanks!

• In many cases, it's the "jerk" $d^3x/dt^2$ that is a better measure of motion than the acceleration $d^2x/dt^2$. For example, a perfectly steady acceleration can't create vibration, and often vibrations are damaging.
– user4552
Mar 7, 2018 at 6:06
• I remembered an old nerd joke; how do you call $d^{4}x/dt^{4}$? Elections! Mar 7, 2018 at 7:28

Here is what happens when you press a key:

• Your fingers cause the key to accelerate downwards.
• If a balance is achieved between the force you apply and the force of the spring keeping the keys in the 'up' position, constant velocity downwards results.
• On contact with the PCB, rapid deceleration occurs, causing a force according to Newton's Second Law of motion:

$F = ma$

In the figure below, the dotted line is point where rapid deceleration ($\Delta v/\Delta t$) starts.

In the case where there is a keyboard-PCB gap, during travel from 'up' to 'down' position, the force does work, quantified by force multiplied by distance in the direction of the force ($W = Fd$). This work increases the speed of the key. At contact with the PCB, the total force on the PCB is: the force due to the deceleration of this moving key ($F_a = ma$), plus the force that your finger is still applying to the key ($F_s$, 'static' force):

$F_{tot} = F_s + ma$

In the case where there is no keyboard-PCB gap, since the key moves very little, the direct force from your finger ($F_s$) is virtually all the force applied to the PCB, since deceleration is small.

$F_{tot} = F_s$

This question has some relation to the mass dropped on scale question.