Hydrostatics question - angled triangular gate?

[Sunil Simha]

So can we conclude that our answer is right? We do have quite logical reasoning.

Well, in the midst of all this discussion, we have completely ignored theBEAST's original question

By the way theBEAST, I think you are getting a wrong answer because ρgh_bottomA is actually the force on the water beneath the gate due t hydrostatic pressure. The vertical component of force on the gate is actually that force minus the weight of the water.( See your own diagram at post #3; the vertical component of force on the water is simply mg-F_u)

 

[TSny]

It looks to me that the textbook solution is correct as far as shown. But, if the textbook is claiming that the force $F_y$ is the same as the vertical force on the cylinder, then, yes, the textbook is wrong. The vertical force $F_V$ on the cylinder would be $F_V = F_y - W$ where $W$ is the weight of the water below the cylinder as shown in the figure in the textbook. I'm wondering if the solution in the textbook continues onto another page that was not posted.

 

[Sunil Simha]

Yes that sounds right. Sorry for the last second edit

 

[TSny]

So I decided to try the other method I mentioned where you take the magnitude of the Fhorizontal and Fvertical. However, my answer is off by around 3000N (the answer key gives 39000N). Do you guys know what is wrong?

I think you're off by a factor of 2 in finding the weight of the water. In particular, look at how you found the area of the base of the region of water.