Fluid Mechanics, Hydrostatic force on submerged plane.

Question

A vertical rectangular gate, 3m wide and 2m high is hinged at H.
On the left side of the gate, sea water lies up to a height 1.2m above the top of the gate.
Whereas on the right side of the gate, water is available up to a height touching the top of the gate.

Determine the horizontal force at the top of the gate, which is required to open it, FA

[Given: Pwater = 1000 kg/m3 & Psea water = 1050 kg/m3]

DIAGRAM HERE: https://i499.photobucket.com/albums/rr354/sdsdasdsdsdsdsd/Flu_zps1rhhtspz.jpg


I'm definitely stumped. My answers don't add up.
Thank you for your time.

Answer 1

Haizal,

Here is brief outline of how to answer this question. 'g' is acc. due to gravity.

 

1) Calculate the pressure on the gate at any depth due to the sea water (total column height 3.2 m) as a function of the depth itself. 1050g(3.2-y) where y is measured from the Point H.

 

2) Calculate the pressure on the gate at any depth due to water (total column height 2 m) as a function of the depth itself. 1000g(2-y) where y is measured from the Point H.

 

3) The difference in pressure results in a net force which acting about H produces a torque.

 

4) Consider an incremental area dy high by 3 m wide and thus the net force is [1050g(3.2-y)- 1000g(2-y)]3.dy

 

5) this force is acting at a distance y from H, thus the Torque produced is  { [1050g(3.2-y)- 1000g(2-y)]3.dy}.y

 

6) Total net torque is found by integrating this expression over the interval y goes from 0 to 2.

 

7) This torque ( turns out to be 7760g is to be countered by F_A x 2, thus F_A = 3880g N