# Understanding Force / Pressure Equation

• Engineering

## Homework Statement:

Understanding the Force / Pressure Equation along a wall and across a surface area

## Homework Equations:

p = density * gravity * Distance from the water surface.
Hello all

I am trying to understand the pressure equation.

I have the following question:-

I understand this, to me what this says to me is the total force acting on the wall as a result of the water of 12m is 706320 Pascals.

This is where I get a little confused, I have the following question:-

Now I am asked to calculate the total force on a surface area of a wall and because of this I have been told to use a difference equation which involves w i.e. the width.

Am I correct in thinking that the reason why the equations are different is because the first question is only asking me to find the total force in 1 dimension whereas the second question I am asked to find the force across an area which is in 2 dimensions

Is this correct?

Thank you.

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BvU
Homework Helper
the total force acting on the wall as a result of the water of 12m is 706320 Pascals.
Can't be: check the dimensions. Perhaps somewhere it is said 'per meter', or 'the width is 1 meter' ?

If not, then the first picture is simply wrong. Second is correct.

jack action
Gold Member
The dimension pascal is wrong in both pictures. In the first one, it should be newton per meter (or the less meaningful pascal-meter) and in the second one, it should be newton (which is the unit of force).

Can't be: check the dimensions. Perhaps somewhere it is said 'per meter', or 'the width is 1 meter' ?

If not, then the first picture is simply wrong. Second is correct.
Hi Jack

The equation for question one states per unit length but I am not sure what them means?

As for the units, I go that wrong I did the calculation again to get the following:-
1000 * 9.81 * (12^2)/2 = 706320
kg * m * m^2 / m^3 * s^2 = kg/s^2
706320 kg/s^2 - not sure what these units are?

Can't be: check the dimensions. Perhaps somewhere it is said 'per meter', or 'the width is 1 meter' ?

If not, then the first picture is simply wrong. Second is correct.
BvU

The equation for question one states per unit length but I am not sure what them means?

BvU
Homework Helper
kg * m * m^2 / m^3 * s^2 = kg/s^2
You mean kg * m * m^2 / (m^3 * s^2) ...
706320 kg/s^2 - not sure what these units are?
kg/s^2 which can also be written as (kg m/s2)/m or: Newton per meter (see also: #3)
(implict advice: use the sub/superscript buttons)​
So you (they) found the force per unit length (or rather: per unit width)

Second is correct
was put right by @jack action : Not Pa but N

The equation for question one states per unit length but I am not sure what them means?
That is right. Since BvU answered this, I will provide an example of how this is used in an effort to clarify for you.

The force you calculated is Newtons per meter. This means Newtons per unit width. Since we are not given the width, your solution is as much as can be given. It is used like this. If the wall is one meter wide, the force on the wall is:

$7036320 \left[\frac{kg}{s^2}\right] 1\left[m\right] = 7036320 \left[\frac{kg m}{s^2}\right] = 7036320 \left[N\right]$

If the wall is 2 meters wide, then the force on the wall is:

$7036320 \left[\frac{kg}{s^2}\right] 2\left[m\right] = 1412640 \left[\frac{kg m}{s^2}\right] = 1412640 \left[N\right]$

HTH

Mfig

Thank you for the worked example.

What you have said makes sense because you have assumed the wall to be one meter wide - that along has cleared things up.

However, looking at the screen shot below:-

The equation I was trying to understand was the one circled in Red - as you can see that there is no mention of the wall width just the comment that states "..per unit length".

This is the confusing part.

Mfig

Thank you for the worked example.

What you have said makes sense because you have assumed the wall to be one meter wide - that along has cleared things up.

However, looking at the screen shot below:-

View attachment 253356

The equation I was trying to understand was the one circled in Red - as you can see that there is no mention of the wall width just the comment that states "..per unit length".

This is the confusing part.

Right. They leave that as an unknown. So you are to find the force per unit width. That is what you did! That is why the units look like they do, and why we can find the total force if we ever find out the width (as my examples showed). They are just saying length instead of width, but it is the same thing as far as the physics goes. Perhaps they mean the area of the wall is length*height whereas I am saying the area is length*width or height*width. That part is just English, not physics.

Mfig

It has started to hit home.

Stepping back for one moment; I can therefore interpret this equation as saying as saying:-

For every 1m of wall width with a height of Y, the total force acting on the wall is F.

Looking at the equation like this along with your worked example makes sense to me now.

Right. They leave that as an unknown. So you are to find the force per unit width. That is what you did! That is why the units look like they do, and why we can find the total force if we ever find out the width (as my examples showed). They are just saying length instead of width, but it is the same thing as far as the physics goes. Perhaps they mean the area of the wall is length*height whereas I am saying the area is length*width or height*width. That part is just English, not physics.
Hi mfig

My apologies for bring this thread back to life but something has come up and I was hoping you clarify , as I did not want to create a new post.

I thought that to calculate the total force acting on a wall caused by a fluid in this case water is:-

Where:-
p = density
g = gravity
w = width of wall
Y = height of the water on the wall

However I have since been given the following equation which is meant to tell me the force acting on a submerged object:-

Where
F = Force
p = density
g = gravity
yc = Distance from water surface to the centroid of the submerged object

I am really struggling to understand the difference between the two equations and I was hoping you could or anyone else could shed some light.

Thanks .

Look carefully at your first equation. It applies to a rectangular wall with area A=wy.

$F = \rho g w \frac{y^2}{2} = \rho g w y \frac{y}{2} = \rho g A \frac{y}{2} = \rho g A y_c$

They are the same equation, it is just that the depth of the centroid of a rectangular wall with one edge at the surface is given by half the wall height.

Look carefully at your first equation. It applies to a rectangular wall with area A=wy.

$F = \rho g w \frac{y^2}{2} = \rho g w y \frac{y}{2} = \rho g A \frac{y}{2} = \rho g A y_c$

They are the same equation, it is just that the depth of the centroid of a rectangular wall with one edge at the surface is given by half the wall height.

mfig

= p*g*w*(y^2/2)
= p*g*w*y*y*1/2
= Since w*y the same as area I can state:-
= p*g*Area*y*1/2
= Since y*1/2 is the same as y/2 I can state:-
= p*g*Area*y/2
= Since y/2 = the distance from the centroid it is the same as yc giving me:-
Force = p*g*Area*yc.

You are BRILLANT BRILLANT BRILLANT BRILLANT BRILLANT, thank you.

Look carefully at your first equation. It applies to a rectangular wall with area A=wy.

$F = \rho g w \frac{y^2}{2} = \rho g w y \frac{y}{2} = \rho g A \frac{y}{2} = \rho g A y_c$

They are the same equation, it is just that the depth of the centroid of a rectangular wall with one edge at the surface is given by half the wall height.

Hi mfig

I wanted to consolidate my understanding what you have showed me.

If I had a triangle submerged underwater and I wanted to find the force acting on one side then would the following be correct:-

I guess I wanted to confirm that Yc is always the distance from the water surface to the objects centroid, in this case my object is a right triangle, to find the centroid of a right triangle I would divide the height by 3 and the breath by 3.

If my submerged object was a semi circle then to find the centroid in the Y dimension it would be 4r/3Pie, I would the force on a semi circle would be 1000 * 9.81 * (14-4r/3Pie).

Would this be correct?

Thanks