I need help.
An open tank contains a layer of oil floating on top of a layer of water (of density 1000 kg/m^3) that is 3.0 m thick, as shown. What must be the thickness of the oil layer if the gauge pressure at the bottom of the tank is to be 6.4×10^4 Pa? The density of the oil is 820 kg/m^3.

Question

I need help.
An open tank contains a layer of oil floating on top of a layer of water (of density 1000 kg/m^3) that is 3.0 m thick, as shown. What must be the thickness of the oil layer if the gauge pressure at the bottom of the tank is to be 6.4×10^4 Pa? The density of the oil is 820 kg/m^3.

anonymous · Answer

The gauge pressure at the bottom of the tank is simply equal to the sum of the pressure exerted by the oil and water so;

$$P =(\rho _{water}*g*h _{_{water}}) + (\rho _{oil}*g*h _{oil})$$
where g =9.81 m/s^2
then you can substitute and solve it easily

anonymous · Answer

You have also to add the pressure exerted by the air because its an open tank. Sorry i forgot it

anonymous · Answer

This is the information  I typed up 
6.4x10^4 = (1000 x 9.8 x 3) + (820 x9.8 x h)
H= 63996 
Which did not make sense why it's so thick
How do I add the pressure exerted by air
Thank you

anonymous · Answer

Ops! I think adding the air pressure is wrong because they gave you already the gauge pressure not the absolute.
If you substitute you must get h equal to 4.3m
$$h= [(6.4*10^4)-(1000 * 9.8 * 3)]/(820 * 9.8) = 4.3 m$$

anonymous · Answer

Great, thank you