Calculating Water Force and Moment in a Fresh Water Storage Tank

[AutumnBeds]

1. Homework Statement

A fresh water storage tank in a factory is shown in the figure above. Its top side is denoted as C, the narrow side as B and the long side as A. In order to allow regular maintenance and cleaning side A is hinged at its bottom edge and secured using a heavy-duty clasp at its top edge. L = 1.25m, H = 0.6m & W = 0.5m

Side a has a length of 1.25m and a height of 0.6m

(Assume the density of fresh water is 1000 kg/m3 and the acceleration due to gravity is 9.81 m/s2)

a) Draw a side on view of side A (looking along arrow D) and show the size of the two forces acting on the tank side and how high they are located from the bottom of the tank

Homework Equations

Resultant Force = Equilibrium Force in opposite direction

The Attempt at a Solution

I've calculated the water thrust force to be 2207.25 Newtons and it's height to be 0.4m meters from the top of the box.

I am correct in assuming that water thrust force is the resultant force? and the second force being the equilibrium force?

 

[haruspex]

the two forces acting on the tank side

Including the water, I count three.

it's height to be 0.4m meters

Check that.

water thrust force is the resultant force?

A "resultant" force is the sum of a chosen set of forces. The sum of all the forces, in a static system, is zero. If you are choosing a subset, it depends which subset you choose.

 

[Chestermiller]

Let's see your free body diagram.

 

[Chestermiller]

I don't confirm your water total force magnitude, but I do confirm the location of its moment arm as 0.4 m from the top (0.2 m from the hinge).

 

[AutumnBeds]

I don't confirm your water total force magnitude, but I do confirm the location of its moment arm as 0.4 m from the top (0.2 m from the hinge).

I've redone the workings, and hope this is somewhere along the lines of being correct.

I've based my workings on the following - http://physics.usask.ca/~chang/course/ep324/lecture/lecture8.pdf

 

[Chestermiller]

You have drawn a front view of the hinged door, not a side view (as requested in the problem statement). Please draw a side view of the hinged door, showing the force exerted by the hinge, the force exerted by the clasp, and the force exerted by the water on the door. Your calculation of F1 is correct, except that you forgot to multiply by g. I don't understand why you calculated F2, since that is not part of the problem (and I don't understand your derivation of F2). After you correct your calculation of F1, please calculate the moment of the force of the water about the hinge (please express it algebraically first).

 

[AutumnBeds]

You have drawn a front view of the hinged door, not a side view (as requested in the problem statement). Please draw a side view of the hinged door, showing the force exerted by the hinge, the force exerted by the clasp, and the force exerted by the water on the door. Your calculation of F1 is correct, except that you forgot to multiply by g. I don't understand why you calculated F2, since that is not part of the problem (and I don't understand your derivation of F2). After you correct your calculation of F1, please calculate the moment of the force of the water about the hinge (please express it algebraically first).

Hi Chester,

Thanks for your input.

However, I am slightly confused. I had to ask the tutor who issued the work for clarity on the question;

Is this the forces acting on the wall and the base of the tank? YES

The way I read the question is that it is two forces acting on one side? YES - Water thrust on one side and then a force at each of the clasp and hinge on the other.

If I was to tale F2 again, and use it as the force acting against the centre of pressure on the door, would this be correct? and treat F3 as the force against the clasp (required force to keep door shut)?

Thanks,

 

[Chestermiller]

Hi Chester,

Thanks for your input.

However, I am slightly confused. I had to ask the tutor who issued the work for clarity on the question;

Is this the forces acting on the wall and the base of the tank? YES

The way I read the question is that it is two forces acting on one side? YES - Water thrust on one side and then a force at each of the clasp and hinge on the other.

If I was to tale F2 again, and use it as the force acting against the centre of pressure on the door, would this be correct? and treat F3 as the force against the clasp (required force to keep door shut)?

Thanks,

His 2nd answer is correct. His first answer doesn't seem correct. You should only be looking at the forces acting on the door. I'm still waiting to see your correctly drawn diagram of the side view of the door, showing the forces of the water, clasp, and hinge.

 

[AutumnBeds]

His 2nd answer is correct. His first answer doesn't seem correct. You should only be looking at the forces acting on the door. I'm still waiting to see your correctly drawn diagram of the side view of the door, showing the forces of the water, clasp, and hinge.

Hi Chester,

I believe that he is making reference to the final question which is to calculate the force required to keep the clasp shut;
a) Calculate the required resisting force at the clasp to keep the panel shut.

Amended work

 

[Chestermiller]

This is the figure I had in mind:


At elevation z above the hinge, the water pressure is $$p=\rho g (h-z)$$
The differential pressure force on the section of door between z and z + dz is
$$dF=pwdz=\rho g w (h-z) dz$$where w is the width of the door (1.25 m).
The differential moment of pressure force (about the hinge) on the section of the door between z and z + dz is
$$dM=pwzdz=\rho g w(h-z)zdz$$
Please integrate these equations to get the total force F and the total moment M.

 

[AutumnBeds]

Hi Chester,

Thanks for the info. To be totally honest this all pretty much new to me and will take me a day to comprehend the above. I understand the basic formula, but when it's come to algebraic terms and integration I'm lost.

I've looked over my notes again and the only I can resolve which may be totally wrong is to calculate the water thrust on the opposite of the hinge and clasp and solve the resisting force on the clasp with F1D1=F2D2. F1 being the water thrust force and D1 the overturning moment from the hinge, D2 being the over all height of the door. The remaining force will on the hinge F3, and will be F1 - F2.

F2, Fclasp
F3, hinge

This may be the total wrong approach.

Your help so far is greatly appreciated.

 

[Chestermiller]

Hi Chester,

Thanks for the info. To be totally honest this all pretty much new to me and will take me a day to comprehend the above. I understand the basic formula, but when it's come to algebraic terms and integration I'm lost.

I've looked over my notes again and the only I can resolve which may be totally wrong is to calculate the water thrust on the opposite of the hinge and clasp and solve the resisting force on the clasp with F1D1=F2D2. F1 being the water thrust force and D1 the overturning moment from the hinge, D2 being the over all height of the door. The remaining force will on the hinge F3, and will be F1 - F2.

F2, Fclasp
F3, hinge

This may be the total wrong approach.

Your help so far is greatly appreciated.

This is actually very close to the way to do it.

If you integrate the water pressure distribution to get the water force F1, you get $$F_1=\rho g w\frac{h^2}{2}$$If you integrate the water pressure distribution to get the water moment M about the hinge, you get $$M=\rho g w \frac{h^3}{6}$$The distance you call D1 is the moment divided by the force: $$D_1=\frac{M}{F_1}=\frac{h}{3}$$Or, $$F_1D_1=M=\rho g w \frac{h^3}{6}$$So, now you have $$F_2D_2=\rho g w \frac{h^3}{6}$$
This gives you what you need to complete the solution (using the equations that you derived).

Chet

 

[AutumnBeds]

This is actually very close to the way to do it.

If you integrate the water pressure distribution to get the water force F1, you get $$F_1=\rho g w\frac{h^2}{2}$$If you integrate the water pressure distribution to get the water moment M about the hinge, you get $$M=\rho g w \frac{h^3}{6}$$The distance you call D1 is the moment divided by the force: $$D_1=\frac{M}{F_1}=\frac{h}{3}$$Or, $$F_1D_1=M=\rho g w \frac{h^3}{6}$$So, now you have $$F_2D_2=\rho g w \frac{h^3}{6}$$
This gives you what you need to complete the solution (using the equations that you derived).

Chet

Chet,

Both solutions return the same figures. I am confident I have submitted the work correct.

Thank you kindly for your assistance. It has been greatly appreciated.