What would put a harddisk drive (HDD) under 350G's of force? I always see the label and it says 350G's withstandable. What would put this over 350G's? Is it even possible to hit 350Gs of force to a hard drive?
 A: You are not the first person to ask this question.
https://superuser.com/questions/925826/what-would-put-a-hdd-under-350gs-of-force claims that 350 g of force is slightly more than a soccer player kicking a football. What this means is that you basically can kick your case, and it shouldn't brick your hard drive. It might cause other issues though, so don't go around kicking your computer case.
And I just noticed that you also posted the linked question. So technically, you are the first person to ask this question on the SE network.
A: 
Is it even possible to hit 350Gs of force to a hard drive?

Sure is. Drop it on the floor.
You are thinking about sustained forces. 350g sustained won't happen even in rocket launches. But momentary forces can easily peak at this level.
Note that the G limit on the drive is for when it's not running. No spinning drive will like 350g, except maybe in particular directions that will never happen in reality.

If you drop your hard drive from $1~\text{m}$ it will hit the floor at around:
$$
\sqrt{2\times 1~\text{m}\times 1~g}\approx 4.4~\text{m}\cdot\text{s}^{-1}
$$
At exactly $350~g$ it would come to a stop in:
$$
\frac{\left(4.4~\text{m}\cdot\text{s}^{-1}\right)^2}{2\times 350~g}=2.8~\text{mm}
$$
(Note that due to the way the math works out, the stopping distance when dropping from a height $h$ is just $hg/a$).
Since the actual impact will probably be a varying acceleration and the rigid case of the hard drive will probably deform less than $3~\text{mm}$, the actual peak acceleration can easily exceed $350~g$.
A: Here's an application where an ability to withstand high shock is important.
Explosions. In the mid 1980s I did work for a mining company's research laboratory (BHP Research, now defunct like all Australian corporate research).
We would lower data-logging computers into boreholes to set up a grid of dataloggers, then detonate a charge of known energy at a known location with the data loggers running. We would then recover the now utterly destroyed computers: their circuit boards shredded into resin and glass fibres. But their log of seismic data was perfectly recoverable from the HDD and we could seismically map the area for minerals prospecting. We would have destroyed thousands of computers in this way. But I don't believe there was a single unreadable HDD, even those within meters of the blast.
At the time, this was actually cheaper than running communications links of the required datarate to all the points in the grid.
As Paul's Answer Points Out, the specification is likely for a parked HDD. Our HDDs were running, but of considerably less data density than HDDs today (they were 160MB drives), which probably explained their ability to withstand such shock whilst working.
A: A g is a unit of acceleration. I am going to assume the force meant here is 350 times the earth weight of the hard drive. Assuming the drive is 625 grams, that works out to a force of about 450 lbs. This is easy to achieve: just hit it with a hammer.
Edit: as many have pointed out, the rating is given in units of acceleration for good reason. An accelerating reference frame is a better way to picture the problem, so you don't need dynamics here, just kinematics.
