Mass of person on top of floating icecube.

[m0286]

Hello!
I am working onmy physics assignment. and i am having trouble with one question...
it says:
If you are standing atop of a large ice cube of pure ice in a freshwater river that is rapidly moving out to see. In the river the ice cube is just submerged, Given that each side of the cube is 1m, what is your mass? As you move into sea will your shoes get wet? What distance do you rise or fall above the surface of the seawater? In either case, estimate how much more or less mass is requred so that the ice cube is just submerged and your shoes are preserved. (Neglect any melting)
What I got:
the volume of the ice v=1*1*1=1m^3
the density of ice is D=920kg/m^3
the density of river is 970kg/m^3
density of seawater = 1030kg/m^3

i believe i have found the mass of the ice cube =
density of ice*volume of ice= 920kg...
But whut do i need to do to find the mass of me on the ice?

If you can't tell me that, can you atleast tell me if its possible or is there some information missing from the question? I am soo confused!
Once I can figure out the mass I am sure I can figure out the rest! THANKS!

 

[mathlete]

Your mass won't change at all.

 

[Doc Al]

i believe i have found the mass of the ice cube =
density of ice*volume of ice= 920kg...
But whut do i need to do to find the mass of me on the ice?

Consider the forces acting on the ice cube. What are they? (Hint: I count three forces acting on the ice.)

Once you've identified the forces, realize that the cube is in equilibrium.

 

[topsquark]

And that the density of sea water is not the same as that of freshwater.

-Dan