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fessia175
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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