Verifying information on Archimedes

[fessia175]

Homework Statement

This is the information and calculations I have been given, but am not sure it is correct. Please verify.
An object in the form of a cube with sides of 50 cm is immersed in water. Determine the height of immersed object, knowing that the density of water is 1000 kg/m^3 and the density of the object is 800 kg/m^3.

Homework Equations

V = lxbxh
Specific weight = density x gravity
Weight of Object = Specific weight x Volume

The Attempt at a Solution

/ = devided
Volume object = 0.5^3 = 0.125 m^3
γ Object = 800 x 9.81 = 7848 N/m^3
γ Liquid = 1000 x 9.81 = 9810 N/m^3
Weight of Object = 7848 N/m^3 x 0.125 m^3 = 981 N
Ab = Area of base

If the object floats Weight of Object = Upward thrust of Archimedes

981/ 9810 (0.5) = 1/5 = 0.2 m
Why is the 9810 multiplied by 0.5 and where does the 0.5 come from?
The calculation given for height immersed is : upward thrust / γ liquid x Ab (which in this case should be .25 m^2.

I can only continue figuring the rest of this out once the above query has been answered.

Thanking in advance
MOM ( trying to) helping with homework

 

[LawrenceC]

"981/ 9810 (0.5) = 1/5 = 0.2 m"

The above makes no sense to me. Newtons are being divided by N/m^3 implying the unit of 0.5 is m^2 in order for the result to have the unit of m.

A straightforward way to look at this problem (because it has straight vertical sides and bottom is planar and horizontal) is to determine what pressure multiplied by the area of the bottom gives a force that equals the weight of the object. The pressure is fluid density multiplied by depth.