How Do You Calculate Thrust on a Circular Trap Door Immersed in Oil?

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To calculate the thrust on a circular trap door immersed in oil, the relative density of the oil is 0.7, and the fuel level is 1.8m above the door. The initial calculation using P=pgh resulted in 24.7212 kN, while using the area formula gave 62.7918 kN. The correct thrust value is 47.2 kN, which can be derived by recognizing the relationship between the calculated values. A hint suggests that the ratio of the correct answer to the initial area calculation is 3/4. Understanding this relationship helped clarify the calculation process.
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A fuel tank contains oil of a relative density of 0.7 on one side there is a circular trap door of 1.8m in diameter. The fuel level is 1.8m above the top edge of the door calculate the thrust on the door.
 
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Hi Twowaypower! Welcome to PF! :wink:

Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
 
I tried using
P=pgh. Which works out as 700*9.81*3.6= 24.7212kN
Then thought many had to put that int P=FA. Which as area is Pi r^2 area works out as 2.54m^2
That gave me 62.7918kN.
I have used 1/2pgbh^2 to find out thrust on submerge rectangular objects but couldn't make that one work either. The answers in my book is 47.2 kN but I just can't get it.
:/
 
Hi Twowaypower! :smile:

Hint: 47.2/62.7918 = 3/4 :wink:​
 
I was scratching my head for so long and it was so obvious. Thank you. Really need that.
 
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