# Impact force under water

1. Jan 7, 2010

### jayzedkay

i'm trying to calculate the impact force of a slide hammer under water. i can calculate it in air using (mv^2/2d), from work energy principle info i got off the net. and i have formulas for calculating drag; (0.5CpAv^2), also off the net, which i believe i have to account for? but no idea how to put them together really, differential?. i'm trying to do this to appropriately size a mems accelloremeter to measure force impact of a sediment corer. i thought if i work it out for air, it'd be less in water? but if i can work it out properly it'd be good. any help greatly appreciated. thanks in advance.
j.

mmm, looked into this abit more, not sure if i was going the right way about it? ideally want the answer in 'g' force. i don't think i could/can relate 'g' (accel force due to gravity) to force (newtons) on impact? or can i? it's all abit confusing

example:if a weight drops 2m the velocity prior to impact i get is sqrt(2gh) = 6.26m/s^2. if it impacts and comes to rest over 0.1m, the deacceleration is v^2/2d = 196m/s^2. converting this to g-force i divide by 9.8m/s^2? this gives me 20g in this example.

but i still have the problem of it all happening under water and the drag factor on the acceleration and deacceleration etc?

am i making sense?

Last edited: Jan 7, 2010
2. Jan 7, 2010

### mgb_phys

Thats only true for high speed-turbulent drag (what we call high Reynolds Number) at low speed in water the drag is given by stoke's law (http://en.wikipedia.org/wiki/Stokes'_law)

Correct

Since the drag depends on the speed, and the speed depends on the drag then yes you need calculus.
Or you can do it numerically with a spreadsheet.

Divide it up into small increments of time.
For each slot work out the acceleration (due to falling) and the drag for this speed.
Then from the acceleration work out what the speed will be for the next time increment and so on.

3. Jan 7, 2010

### jayzedkay

thanks for the stokes pointer. stokes refers to a sphere, my body is a cylinder, but not concerned with that for now?

got this info and related it to my problem; free body diagram shows that the body has acting on it a buoyancy force Fb, which is the displayced fluid weight combined with fluid density and a drag force Fd. these are equal to the gravitational attraction.

so, Fb + Fd = mg

expanding on Fb for my cylinder i get

Fb = PIr^2hpg, where PIr^2h is vol of cyl. p is fluid density and g is gravity acceleration.

Fd i get from stokes, Fd = 6PIuRV

combining i get PIr^2hpg + 6PIuRV = mg

i can now solve for V.

is this 'v', velocity of the body in free fall, in the fluid?

can i then use this velocity, which considers the fluid. back in the previous equations i used to calculate the forces on impact?