Force at a particular point on a submerged vertical surface (other than Ycp)

In summary: It doesn't mean the weight is a point load that only exists at the center of gravity; the weight acts on every part of the beam. Similarly, the hydrostatic force acting on a submerged body is considered to act through the center of pressure, even though the force acts on every point of the body. Does that make sense? In summary, the conversation is about determining the forces at points A, B, and C on a 1-m wide by 2-m gate in the vertical wall of a reservoir when the water surface is 700 mm above the top of the gate. The resultant force and the center of pressure are calculated using equations, but the forces at the specific points are still unknown. The solution is
  • #1
lechatelier
7
0

Homework Statement



https://www.dropbox.com/s/mw1ol8tnrdf0ajb/fm%20prob.JPG?dl=0 [/B]


A 1-m wide by 2-m gate in the vertical wall of a reservoir functions like a door. It is hinged at A and B and latched at C. Determine the forces at A, B, and C when the water surface is 700 mm above the top of the gate.

Hinge A is 0.3 m below the top-left corner of the door
Hinge B is 0.3m above the bottom-left corner of the door
Latch C is in the middle of the right edge of the door.

(Answers:
Force at A 3.57kN
Force at B 13.10kN
Force at C 16.68kN)2. Homework Equations

F = (rho)(g)(hcg)A gives the resultant force on the full door

Ycp = Ycg + (Icg/(Ycg. A)) gives the center of pressure3. The Attempt at a Solution

The resultant force was calculated by the above equations but I am struggling to find forces at the particular points A, B and C

Any help will be greatly appreciated

Thanks
 
Last edited:
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  • #2
The door doesn't move, it doesn't rotate around a horizontal axis and it doesn't rotate around a vertical axis. That gives you three balances to calculate the desired F magnitudes.
 
  • #3
Thanks for the reply. But I still have no idea how to go about calculating the forces at A,B and C
I easily calculated the total force on the full door as 33.354 kN and this is acting at Ycp = 1.896m from the surface.
I don't see how to use "sum of moments = 0" since there are three unknowns.
Thanks for your help
 
  • #4
Consider two axes for the moment calculation: the one where the hinges are located, and the one perpendicular to it and passing through the latch. The resultant moment with regard to either axis must be zero.
 
  • #5
Hi voko Thanks for the help. I was able to calculate Fc (the latch) as you mentioned by taking moments about the hinges (the top view) so that only the Fc and the resultant are considered. This gave me the correct answer for force at C = 16.68kN.

But I am still not able to picture how to take moments about the axis that runs through the latch. I am assuming I look at the side-view of the door. But still there are two unknowns then (Force A and Force B) And the resultant force is acting in the middle of the door and it's in a different plane.
Thanks again for your help

I have the picture of the problem here

https://www.dropbox.com/s/mw1ol8tnrdf0ajb/fm prob.JPG?dl=0
 
  • #6
By considering the moments about the latch axis, you get one equation relating force A and force B. Another equation relating these two unknowns is obtained by equating the total resultant force to zero. Two equations, two unknowns; solve.
 
  • #7
Have been trying to solve this but no luck. The issue that I'm having is that (from what I understand, please correct me if my understanding is wrong) that if there is a force at a particular point on a submerged body, let's say it acts a distance of Ycp. We know that from the surface it is at a distance of Ycp BUT this force acts at all points on that submerged body that are at a distance of Ycp. So how to take a moment arm when we don't have one particular point?
 
  • #8
lechatelier said:
Have been trying to solve this but no luck. The issue that I'm having is that (from what I understand, please correct me if my understanding is wrong) that if there is a force at a particular point on a submerged body, let's say it acts a distance of Ycp. We know that from the surface it is at a distance of Ycp BUT this force acts at all points on that submerged body that are at a distance of Ycp. So how to take a moment arm when we don't have one particular point?
It is analogous to how we consider the force of gravity on an extended body to be working through its center of gravity. When you analyze a beam, the force vector for the weight is placed at its center of gravity, right?
 

1. What is the formula for calculating force at a particular point on a submerged vertical surface?

The formula for calculating force at a particular point on a submerged vertical surface is F = ρghA, where F is the force, ρ is the density of the fluid, g is the acceleration due to gravity, h is the height of the fluid column above the point, and A is the area of the surface.

2. How does the depth of the fluid affect the force at a particular point on a submerged vertical surface?

The force at a particular point on a submerged vertical surface is directly proportional to the depth of the fluid. This means that as the depth of the fluid increases, the force will also increase.

3. Does the shape of the submerged surface affect the force at a particular point?

Yes, the shape of the submerged surface does affect the force at a particular point. The shape of the surface will determine the area (A) in the formula for calculating force. A larger surface area will result in a larger force.

4. How does the density of the fluid affect the force at a particular point on a submerged vertical surface?

The density of the fluid has a direct impact on the force at a particular point on a submerged vertical surface. The higher the density of the fluid, the greater the force will be. This is because denser fluids have more mass per unit volume, resulting in a greater force.

5. Can the force at a particular point on a submerged vertical surface be negative?

No, the force at a particular point on a submerged vertical surface cannot be negative. Force is a vector quantity and can only have positive values. If the force is acting in the opposite direction, it would be considered a negative force, but the magnitude will still be positive.

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