1
$\begingroup$

I have a mock exam problem that deals with fluid forces acting at different angles. Here are the scans: enter image description here enter image description here 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!

$\endgroup$
  • $\begingroup$ Pressure always acts perpendicular to a surface. You need to make use of this in problem c with the slanted surface. $\endgroup$ – Chet Miller Apr 15 '16 at 20:55
0
$\begingroup$

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 $cos45$).

$\endgroup$
  • $\begingroup$ Thanks for the thorough response! When I attempted (b) I was using the pressure at (H - (Y / 2). So integration is required for (b) and (c), but in (c) only the Y-component of the force will affect F. Got it. Thanks again! $\endgroup$ – ErikW89 Apr 18 '16 at 19:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.