# Quick static fluids question

1. Dec 8, 2007

### islanderfan

1. The problem statement, all variables and given/known data

A small sphere 0.75 times as dense as water is dropped from a height of 11 m above the surface of a smooth lake. Determine the maximum depth to which the sphere will sink. Neglect any energy transferred to the water during impact and sinking.

for clarity's sake, i'm letting v = volume, and v = velocity

2. Relevant equations

vf^2 = Vi^2 + 2ad, Fb = pgv, K = 1/2mv^2, Work(non conservative) = change in mechanical energy

3. The attempt at a solution

the first thing i did was basic kinematics to get a velocity of 14.68 m/s upon hitting the water.
next i set Fb equal to nonconservative work and i got pgv(d) = Kf -Ki...
simplified down i got it to pvg(d) = 1/2mv^2
= pvg(d) = 1/2pvv^2

at this point i cancled out the volumes and got a final equation of
1000*9.8*(d) = .5*750*(14.68^2)

this got me a final answer of 8.25 meters which seems to be too large, and in fact it was wrong.
can anyone find where i'm going wrong?

2. Dec 8, 2007

### Staff: Mentor

Gravity still acts on the object when it's under water.

3. Dec 8, 2007

### islanderfan

oh i see, but how do i take that into account? the buoyant force still is non conservative right?

4. Dec 8, 2007

### Staff: Mentor

Yes, buoyant force is the only non-conservative force acting (in this simplified problem).

You want the super-duper easy way to solve this? Measured from the final position of the object (distance d under the water), what's the total mechanical energy of the object just before it's dropped? After it reaches the final position? So what must the non-conservative work equal?

5. Dec 8, 2007

### islanderfan

or can i make an Fnet equation

like, mg - Fb = ma

p(ball)vg - p(water)vg = p(ball)*v*a

volumes cancle, so => 750(9.8) -1000(9.8) = 750*a
this yielding an acceleration of 3.267 m/s^2

would that be a correct approach, or is what i just laid out above incorrect?

6. Dec 8, 2007

### Staff: Mentor

Nothing wrong with that approach. Go for it.

7. Dec 8, 2007

### islanderfan

ok so since theres potential at the top and nothing(?) at the end, can i say

p(water)vg(d) = p(ball)vgh

1000(d) = 750(11)

d = .12 meters????

i really appreciate the help

8. Dec 8, 2007

### Staff: Mentor

That "h" has to be the total change in height, not just the height above the water surface.

9. Dec 8, 2007

### islanderfan

ok that makes sense, but won't that leave me with two unknowns?

10. Dec 8, 2007

### islanderfan

or can i make h = 11+d?

but when i did that i got a final d of 33 meters. this seems way too large to me

11. Dec 8, 2007

### Staff: Mentor

Looks OK to me. Solve it the other way and compare.

12. Dec 8, 2007

### islanderfan

i just don't understand how it can sink that far...

btw thanks for all the help, i really understand how to solve this type of question now, i appreciate it, thanks!

13. Dec 8, 2007

### Staff: Mentor

That's because you have some common sense.
The problem is the bogus assumption: "Neglect any energy transferred to the water during impact and sinking." That's a huge effect to ignore!

14. Dec 15, 2007

### craanberry

im still confused, my height above water was 12 instead of 11, and i tried what he did, with 1000(d)=750(12+d) because h=12+d..... i got my h to be 48 and d to be 36... i tried both those answers and neither worked. his was the correct answer and im pretty sure what i did is what he did... how did he get 33 for his final answer?

nvmmmm i used the wrong coefficients!

Last edited: Dec 15, 2007