# Homework Help: Hydrostatics on Plane surface

1. Nov 27, 2015

### RAP1234

1. The problem statement, all variables and given/known data
A 3 m long gate of weight 4 kN per unit width is hinged at O and sits at an angle θ as a function of water height h above the hinge. (A) Using a y axis measured up from the hinge, derive a general relation between h and θ, with all other variables evaluated in the relation. (ANS: h ∼ C(cosθ sin2 θ) ^1/3 , where you must find C.) (B) Using your result, what h will cause the gate to be inclined at θ = 45◦ ? (ANS: h(45◦ ) ∼ 1.2m)https://www.physicsforums.com/attachments/asd-png.92500/ [Broken]

2. Relevant equations

3. The attempt at a solution
My answer is not of the form h ∼ C(cosθ sin2 θ) ^1/3

I am obligated to use the coordinate system with y increasing from the hinge, up along the gate.
I chose z positive direction going straight vertically upwards from the hinge

dF= PdA eq(1)

dM=ydF eq(2)

dP/dz = rho*g eq(3)

Integrating eq(3)

P = rho*g*z+C

but at z = h , P = 0 gage
--> C =rho*g*h

--> P = rho*g*z- rho*g*h\

replacing this into eq(1)

dF = (rho*g*z- rho*g*h) *w*dy

needing z as a function of y, I find that z = ysin(theta)

--> dF = (rho*g*ysin(theta)- rho*g*h) *w*dy

replacing this into eq(2) I find that

dM = (rho*g*y^2*sin(theta)- rho*g*h*y) *w*dy

Integrating the right side with bounds from y=0 to y= 3 I find that

M = w*(9*rho*g*sin(theta)-4.5*rho*g*h) eq(4)

which should be the moment due to the fluid

I thought that since the moment due to the fluid should equal the moment due to the weight of the gate acting n the centroid I would have something like

Mfluid =M weight of gate

Since I am given M due to weight of gate is 4kN per unit width, dividing eq(4) by w

(9*rho*g*sin(theta)-4.5*rho*g*h) = 4,000*cos(theta)*(L/2)

but given in the problem statement is L = 1m

so I replaced in the appropriate values for the density of water rho = 10^3, g = 9.8, L = 1, and solved for h in terms of theta.

But my answer is not of the proper form that problem suggested h ∼ C(cosθ sin2 θ) ^1/3

Please help. I am truly not knowing what I am doing wrong here and don't have anyone I can ask to verify my method.
apologize. I didn't realize the image would be so small when uploaded.
I am obligated to use the coordinate system with y increasing from the hinge, up along the gate.
I chose z positive direction going straight vertically upwards from the hinge

dF= PdA eq(1)

dM=ydF eq(2)

dP/dz = rho*g eq(3)

Integrating eq(3)

P = rho*g*z+C

but at z = h , P = 0 gage
--> C =rho*g*h

--> P = rho*g*z- rho*g*h\

replacing this into eq(1)

dF = (rho*g*z- rho*g*h) *w*dy

needing z as a function of y, I find that z = ysin(theta)

--> dF = (rho*g*ysin(theta)- rho*g*h) *w*dy

replacing this into eq(2) I find that

dM = (rho*g*y^2*sin(theta)- rho*g*h*y) *w*dy

Integrating the right side with bounds from y=0 to y= 3 I find that

M = w*(9*rho*g*sin(theta)-4.5*rho*g*h) eq(4)

which should be the moment due to the fluid

I thought that since the moment due to the fluid should equal the moment due to the weight of the gate acting n the centroid I would have something like

Mfluid =M weight of gate

Since I am given M due to weight of gate is 4kN per unit width, dividing eq(4) by w

(9*rho*g*sin(theta)-4.5*rho*g*h) = 4,000*cos(theta)*(L/2)

but given in the problem statement is L = 1m

so I replaced in the appropriate values for the density of water rho = 10^3, g = 9.8, L = 1, and solved for h in terms of theta.

But my answer is not of the proper form that problem suggested h ∼ C(cosθ sin2 θ) ^1/3

Please help. I am truly not knowing what I am doing wrong here and don't have anyone I can ask to verify my method.

Last edited by a moderator: May 7, 2017
2. Nov 28, 2015

### haruspex

Yes, if you will be giving g a negative value.
That extends beyond the region in contact with the water.
Looks like you forgot a step in integrating y2.

3. Nov 29, 2015

### RAP1234

I integrated and evaluated the bounds of the definite integral from y = 0 to y = 3

BUT as you said before the water does not go to the height of the gate.I will fix this and get back to you.

4. Nov 29, 2015

### RAP1234

I am getting an answer of the proper form, but the constant C in front of it I am not sure of. Will the moment due to the weight of the gate have a change in the length of the lever arm since it is partially submerged?

5. Nov 29, 2015

### SteamKing

Staff Emeritus
Yes. The net moment about the hinge should be zero for the gate to remain in equilibrium.

6. Nov 29, 2015

### RAP1234

so we would still take 3/2 to be the length of the lever arm, as gravity still acts at the centroid

7. Nov 29, 2015

### SteamKing

Staff Emeritus
The weight of the gate is fixed, and so is its lever arm about the hinge. The same cannot be said of the hydrostatic force acting on the gate, nor its lever arm reference to the hinge. Those are functions of the water depth h.

8. Nov 29, 2015

### RAP1234

Thank you. I did fix the integration with the moment portion for the hydrostatic force from 0 t y = h/sin(theta)

9. Nov 29, 2015

### haruspex

It depends what you mean. The length 3/2m is not orthogonal to gravity.

10. Nov 29, 2015

### RAP1234

I posted pictures of my new working below. I do appreciate the help

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11. Nov 29, 2015

### haruspex

That all looks pretty good, except that you went wrong early on.
I gather you are taking g as having a positive value, so switched the sign on dP/dz to minus. But that should change the sign on the constant of integration too. Look at your expression for P(z) and consider its value when z=h.

12. Nov 29, 2015

### RAP1234

Thanks for the response. I will have to fix that as P should be zero when z = h

Update: Wonderful . After fixing it properly, everything seems to be within range. Thanks again for your help :)

Last edited: Nov 29, 2015