Question

Rainwater collects behind the concrete retaining wall shown here: https://dl.dropbox.com/u/63664351/MATS2005/Forces%20on%20submerged%20surface.PNG
If the water saturated soil (specific gravity 2.2) acts as a fluid, determine the force and center of pressure on a 1-m width of the wall.

Answer 1

From here:
http://books.google.co.uk/books?id=wQCMC3ryYzkC&pg=PA394&lpg=PA394&dq=equation+for+force+behind+a+dam&source=bl&ots=SL_b8Z6K9y&sig=phVrDEYfm68Vo45u7M5YhQ9Epo0&hl=en&sa=X&ei=aTbXUKX-CcjM0AWv64CQDQ&sqi=2&ved=0CDMQ6AEwAQ#v=onepage&q=equation%20for%20force%20behind%20a%20dam&f=false
I think width here means the horizontal line (though in real life I usually call this the length and call the thickness width).

Attempted solution:
https://dl.dropbox.com/u/63664351/MATS2005/Forces%20on%20a%20dam.PNG

Problem:
Since I don't know the height of the force, I use arbitrary y and integrate.
Still, I am confused. Where is this force applied? I don't know the location (height).
If the height were given, it would lead to different result, won't it?

Answer 2

Or what I am doing are merely finding the average Force? How so?

Answer 3

Hang on. For the pressure difference in the region in contact with soil,
shouldn't the limit of integration be from y=1 to y=4 ?

Answer 4

Am I too late?

Break the problem up into smaller pieces. Treat each layer individually to find the force that layer produces within the layer. Find the force due to the water on the soil layer (remember the pressure of the water layer on the soil layer is transmitted uniformally throughout the soil layer and is equal to the pressure at the bottom of the water layer. These three forces are summed for the net force. Now were does this net force appear to be acting. Use the concept of a moment of force ie. force *distance . Find the moment of each force you found above at a common point. this should be equal to the net force *distance to that same point\\[eq 1 ....F _{net}\times d _{eff}=F _{water}\times d _{1}+F _{soil}\times d _{soil}F _{water-on-soil }\times d _{3}\] To find the effect point of action of a column of a fluid of height H integrate the force at depth times the distance l to the bottom of the column ie\[l*dF=l*P(l)*dA=l*P(l)*Width*dl\]
apply this result to the above situation to get the moment arms, the d's in eq 1.