G Force upon falling object on impact

Click For Summary
To determine the G force experienced by a measurement device upon impact, it's essential to know the deceleration during the stop, as this is where the majority of G forces occur. While the device experiences 0 Gs during free fall, it feels 1 G when resting on the floor. Accurate calculations of G forces are challenging without precise data on how quickly the object stops upon impact. Attaching accelerometers to the device can provide the necessary measurements for a more accurate assessment. Understanding these dynamics is crucial for establishing the calibration threshold for the device.
kiwisoundguy
Messages
1
Reaction score
0
Hi Guys,

I have been asked to find out at what point a measurement device go out of calibration when dropped and express that in a g as in G force.

Its been 15 years since I studied physics (for 3 years), I do rememeber a bit but I need some help..

I have the size and weight (mass) of the device
It is dropped from a set height to a concrete floor. Assume no absortion by the floor.
After dropping we check to see if it still measures properly and if it does, increase the height until the readins is incorrect. At this point we will say the unit can expreience 'x' G force before calibration is required again.

mass=0.59kg
Height = 1m (for example)
Gravity = 9.8 (unless something has changed since I was at school)

What other parameters do I need to know and what formula do I need to use.

thanks in advnance..

Cheers, Ben
 
Physics news on Phys.org
You need to know how fast the object stops...that's basically the only information you need.

In other words, during the entire fall, the object is essentially weightless (feels 0 g's), while it is on the floor, it experiences 1g of force acting on it; but most of the g's will come from when the object is stopped by the floor. Without knowing how fast the object stops, we can't tell the g's.
 
Sine you're actually testing, it would be best to attach accelerometers and measure the acceleration. It is extremely difficult to calculate accurately.
 

Similar threads

Replies
4
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 26 ·
Replies
26
Views
9K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 19 ·
Replies
19
Views
7K
Replies
6
Views
3K