Wader, not entiiirely correct....
So we have a crash, and the tristar decelerates at 3g consistently over the space of a second or 2 (for our hypothetical scenario!) , in a direction that causes the luggage to want to leave said overhead locker through it's weakest point (the door latch one might reasonably assume).
So our 25lb luggage hits the locker door at a velocity based on it's acceleration (3g - ie 3x9.81ms-2, so about 29.5 meters per second per second) and the distance between it and the locker door. This is the "initial impact" may or may not break the door, depending on it's designed strength. Then, if we assume the luggage is now sitting on the door, it will exert a force of 3x its weight at 1g, so your "75lb" example.
Which is the bigger of the two forces will depend on how tightly your overhead bin is packed (i.e. how much distance the luggage has to accelerate over at 3g) and how "squidgy" your luggage is (i.e. it's compressibility. This relates to the distance over which it will decelerate, thus affecting the force needed to slow it down to rest relative to the bin door using F=ma)
Two examples of the above:
A 25lb small hard briefcase resting at the back of the overhead bin by itself at the moment of crash: Leaps forwards through the locker at impacts with a forward corner of the door. Say the distance of travel is 1m. Using Vsqd = Usqd + 2aS (i.e (End Velocity) squared is equal to the (initial velocity) squared + twice the acceleration times the distance traveled)
So, Vsqd = 0 + 2x29.5x1
Vsqd = 59
V=7.7 meters per second(1dp)
So the briefcase hits the door at 7.7ms-1. It then decelerates to "rest" (relative to the door, which is itself slowing down at the same time, but essentially starts leaning on the door and being decelerated by it). If we say the briefcase is metal, and the corner dents inwards by approximately 1cm we can use vsqd = usqd + 2as again to calculate the deceleration.
End velocity, V = 0 (relative to the door, our frame of reference for this entire calculation)
Initial velocity U = 7.7ms-1 (see above)
Distance travelled, S = 1cm, or 0.01m (dent in corner of case)
Deceleration, our unknown = a
So
0 = 59 + 0.02a
-59 = 0.02a
a= -2950ms-2
or 300g!
So this would put 100 times the force on your luggage bin door than just a simple bag already resting on it weighing the same. The morale is: Dont allow your "load" to shift in flight, as all good loadies will know!
(I suspect there's bound to be an error in there somewhere, maths was never my strong point! So flame away... But it shows a point)