# Homework Help: Specific Gravity in Fluids

1. Jan 17, 2005

### AoshiShinomori

This one got me.
Any help?
Regardless of if its being homework, respond if you can.

A life preserver with a volume of .05m^3 will support a 80kg person with a specific gravity of .92, in seawater, specific gravity of 1.03, the person is 20% above water(volume-wise) and the preserver is fully submerged, find the density of the life preserver.

Okay, well so far, I took all of the specific gravities and changed them to weight densities. Therefore, I had seawater as 10094 N/m^3 (not working in lb/ft^3), the human as 9016, and of course, pure water is9800. I took the 9800 and multiplied through all the specifics to get there weight densities. Some formulae I've pulled together include:

9800*specific gravity=weight density
Pressure=weighdensity*height (irrelevant)
Force Buoyancy=Force Weight - Force Apparent (where Fb is weight of fluid displaced, naturally)
Specific gravity*1000=density of certain solution
Density=mass/volume(...)
Force Buoyancy=(Volume of object)(weight density of fluid)

Anyways, given the volume, we need to find the mass of the preserver. Since the preserver is just submerged, the 80kg person, plus the mass of the raft caused the force weight to equal the force buoyancy (may be off).

Using the latter of the equations, I took the weight density of the seawater, which was 9800*1.03 and multiplied that by the volume of the buoyant preserver, which, I used to equal force weight plus the mass of the raft(which I'm guessing is force b....it should be anyway). Solved for Mass.......and plugged into mass/volume=density. Is that right? Let's just say that was the bonus, and he didn't put a correct answer down...

In numerical format.....(9800)(1.03)(.05)=(80*9.8 +/-(?) x(mass of raft)*9.8)

Last edited: Jan 17, 2005
2. Jan 17, 2005

### dextercioby

Cute problem,any ideas...??

Daniel.