Fluid mechanics civil engineering

[Mikee Carlos]

As shown in the attached image, a tank has an inclined wall at an angle of 450 to the horizontal. On this wall, there is a 1m square door that is hinged at A and has a simple latch at B. The distance between A and the free surface of sea water measured along the inclined wall is 2 m. (Specific gravity of sea water sg = 1.1). Calculate the force F on the door due to the oil
MY WORKING OUT could anyone pls help check if i am right :) P=F/A SO F=PA than F=densitygravitysin(angle)Areaheight now in order to find the height cause it's inclined i did sin 90/2=sin45/x cross multiply to get x-1.414 back to the formula 1100*9.81*sin45*1m^2*1.414 and i got 10.789kN as my force

 

[BvU]

Hello Mikee,

I find the problem sattement confusing: it asks for

the force F on the door due to the oil

but it only gives the specific gravity of sea water. Why ? Where is the sea water ?
Your calculation also seems unlikely to me: you use 1.41 m (the depth of point A) as the depth for every point on the door ?