Resultant Force Due to Hydrostatic Pressure

[MJay82]

I've got a question concerning the resultant force due to hydrostatic pressure. I understand how to calculate an object with uniform dimensions as depth (and pressure) increase. But I got thrown a serious curve ball on a Statics test.

The situation was such:

A downward facing isosceles triangular gate with base b and height a is hinged on the top at point O, which is a distance h below the surface of water. Calculate the force exerted on the back side of the gate to keep it closed.

I feel confident in finding the point that the resultant pressure force is applied at, but I got really confused because surface area of the gate is decreasing with depth while pressure is increasing with depth.

I see this equation here,
dp = \gammadh
Can I use this to find the magnitude of the resultant pressure force? I've got a final tomorrow and I feel that if only I could understand this one concept, I could work out the rest of my problems fairly easily. Thanks for any help.

 

[tiny-tim]

Hi MJay82!

Standard method: slice the gate into horizontal slices from depth h to h+dh, find the force on each slice, and integrate.

(and there's a hinge, so this is a rotational problem, so in this case you'd need the moment of the force on each slice, about the hinge)

 

[MJay82]

Ok - that's what I thought would be going on, thanks for the speedy reply!
I should have mentioned that I did understand the moment part. Using the scalar moment calculation (force x perpendicular distance), finding the force necessary to keep the gate closed is a cinch.

Let's see if I can maybe more clearly articulate what is baffling me, because although your answer let's me know that I'm at least thinking correctly, I can't conceptualize the changes.

You've got a pressure which is varying with depth - increasing as depth increases. And you've got an area which is also varying with depth, but is decreasing as depth increases.

I'm not familiar with double integration yet - is this a problem that would call for such, or since area and pressure are both varying with respect to depth, is it just a matter of relating both area and pressure in terms of one variable?

 

[tiny-tim]

I'm not familiar with double integration yet - is this a problem that would call for such, or since area and pressure are both varying with respect to depth, is it just a matter of relating both area and pressure in terms of one variable?

Forget about variables, just ask yourself how many d's are there?

In this case, there's only dh …

so you'll be summing over all the slices, which becomes a single integral ∫ … dh.

Just insert the moment of the force on the whole slice.

(If the pressure was also varying horizontally, then you'd need to slice-and-dice, and you'd have to sum over little squares of height dh and width dx … two d's, so a double integral. )