How does a fluid force behave at different angles?

Question

I have a mock exam problem that deals with fluid forces acting at different angles. Here are the scans: Based on the description of the systems, I know that in (a) and (c), F is going to equal the sum of the weight of the piston and the force exerted by the water on the surface of the piston. In (b), the weight of the piston is negligible so F is equal to just the force exerted by the water on the piston.

This is what I got for F in (a):

$$ F = mg + \rho_{water}XY(H-d)g $$

The solution to (b) is essentially the same, except for the weight of the piston. My question is about (c)... I think it's a trick question. The fluid force acts at a 45 degree angle to the surface, so the y-component will be smaller than in (a). However, there is more surface area.

To me it seems intuitive that the y-component of fluid force will be weaker in (c) than it is in (a), but in my rough calculations they wind up being essentially the same. So how does it work? Are they actually the same?

Thank you!

Answer 1

The solution to part (B) is not same as part (a) because the fluid pressure along YZ plane would vary as $$P=P_o+\rho(gh)$$

Thus

$force_{yz}=\int_{y}^{z}(P_o+\rho(gy))Zdy$

$$force_{yz}=P_o(Z)(Z-Y)+\frac{\rho(g)(Z)(Z^2-Y^2)}{2}$$ Therefore force on YZ plane would have to found by integration.

On similar grounds, force in part(c) would have to found by integration keeping in mind the fact that only Y component of force is taken into account(i.e $cos45$).