Finding the magnitude of the hydrostatic force per unit of circumference

[tomtomtom1]

Homework Statement

Finding magnitude of hydrostatic force per unit of circumference

Relevant Equations

Force = Density * Gravity * Distance from Surface to centriod of tank * Area
Density = 1000
Gravity = 9.8
Distance from surface to centriod of tank = 4.25
Area = 28.274

Hello all

I am trying to find the hydrostatic force of a cylindrical tank of water. I was hoping someone could help point me in the right direction.

I have a cylindrical water tank. The dimensions are as follows:-

I need to find the magnitude of the hydro static force per unit of circumference.

I know the answer which is 354000 i just don't know how to get to it.

I currently do the following:-

Force = Density * Gravity * Distance from Surface to centriod of tank * Area
Density = 1000
Gravity = 9.8
Distance from surface to centriod of tank = 4.25
Area = 28.274

Force = 1000 * 9.8 * 4.25 * 28.274 = 1177612 Newtons

Circumference = 2 * Pie * Radius = 18.8496

I don't really know where to from here?

I thought i could just divide the Force by the Circumference but i have been told that this is incorrect.

Its the "per unit of circumference" that's really throwing me.

Can anyone help?

Thank you.

 

[Chestermiller]

The pressure (force per unit area) is increasing with the distance z beneath the surface according to the equation $$p=\rho g z$$ So the pressure force per unit area on the wall is lower near the surface than near the bottom of the tank. So there isn't just one value of p that you have to work with; you need to take into account how p is varying with depth. The differential area on the circumference of the tank wall is given by $dA=Rd\theta dz$. What is the pressure force acting on this differential element of area?

 

[TSny]

Homework Equations:: Force = Density * Gravity * Distance from Surface to centriod of tank * Area
Density = 1000
Gravity = 9.8
Distance from surface to centriod of tank = 4.25
Area = 28.274
.
Force = 1000 * 9.8 * 4.25 * 28.274 = 1177612 Newtons

Your work looks OK, except for the value of the lateral area of the cylinder. (Looks like you calculated the area of the bottom.) Recall that the lateral area is given by the circumference times the height. (You don't need to evaluate the circumference since you are going to divide F by the circumference to get the force per unit circumference.)

 

[Chestermiller]

Your work looks OK, except for the value of the lateral area of the cylinder. (Looks like you calculated the area of the bottom.) Recall that the lateral area is given by the circumference times the height. (You don't need to evaluate the circumference since you are going to divide F by the circumference to get the force per unit circumference.)

It shouldn't be the distance from the centroid. It should be half the depth, 1.75 m.

 

[TSny]

The only way I could get the stated answer was to assume that the question is asking for the hydrostatic pressure on just the vertical side of the tank per unit circumference. Half the depth (4.25 m) would then be the same as the depth of the centroid of the vertical side of the tank. This is the value that @tomtomtom1 used. He referred to it as the depth of the "centroid of the tank" rather than the depth of the "centroid of the vertical side of the tank".

 

[Chestermiller]

The only way I could get the stated answer was to assume that the question is asking for the hydrostatic pressure on just the vertical side of the tank per unit circumference. Half the depth (4.25 m) would then be the same as the depth of the centroid of the vertical side of the tank. This is the value that @tomtomtom1 used. He referred to it as the depth of the "centroid of the tank" rather than the depth of the "centroid of the vertical side of the tank".

I misread the depth as 3.5 rather than 8.5. My eyes are not what they used to be.