I typically try to avoid all those derived so-called simplifications for the force and line of action of a force on a submerged flat surface. Instead I teach the students that the force is the integral of the pressure over the area (with the pressure varying with depth) and that the line of action of the force can be calculated by taking moments about some point on the surface. This is how many text books introduce the topic but they then revert to the force being the pressure at the center of area times the area and use that horrible equation for the line of action that uses the second moment of area. I really dislike this equation as the geometry of applying the equation, particularly for a fully submerged angled surface, is very complex. I prefer to teach my students how to do the integrals. However, they also find this challenging so I use a simple visualization demonstration using piles of textbooks.
All you need is a large number of mechanics textbooks (approximately 30), a yard stick, and a large flat surface such as a desk (I use the trolley that has the OHP on it in our class rooms). The topic of the books is obviously irrelevant. I am also typically not a big fan of textbooks. I find that the students do not use them the way they were written to be used so they can cause as many problems as they solve. I therefore use this demonstration to joke about having found a good use for all the textbooks in my office.
I start off by reviewing distributed loads from statics. I make 4 piles of books all of the same height with all the piles touching each other. This is a simple uniformly distributed load with the total load being equal to the weight of all the books.
I then adjust the book piles so that the first pile has one book, the second has two books, the third has three and the forth pile has four. I then lay the yard stick across the top of the piles so that the stick represents a distributed load that increases linearly with distance along the table (or along the floor as in the pictures in this post).
I then say that if you wanted to approximate the total load you would add up the load due to each pile.
F=Σ pile height x (weight/unit height of the books)
The approximation would improve if we used more piles of narrower books with the actual load being given by the sum of an infinite number of infinitesimally thin book piles. You can write this on the board as
F=limit (pile width →0) Σ pile height x (weight/unit height of the books ) (=∫fdx)
where f is the distributed load function.
To calculate the distance from some point (denoted by a) to the point of action of the force we calculate the moment about a (Ma) and divide that by the total load F. The moment can be calculated by adding up the moments due to the individual piles of books. Again the estimate is improved by using a large number of thin piles. For a load that increases linearly from zero, the force acts 2/3rds of the way along the distributed load. This often leads to confusion as this only applies to hydrostatic forces on a submerged surface if the surface is rectangular, and the top of the surface is at the free surface.
To illustrate this I ask the class to consider the hydrostatic pressure force on a non-rectangular flat surface. Without loss of generality, I consider a triangular surface that increases in width with depth. In this case I start with one book, next to that I build two piles of two books, then three piles of three books, and four piles of four books. The layout is shown in the images below. Again, I lay the yard stick over the pile showing the linear increase in height with distance from the first pile.
This time, because each pile of books has a different width, the weight of the pile is related to the height of the pile and the transverse width of the pile, that is, the width of the surface. The force is given by
F=Σ pile height x (weight/unit height of the books) x # piles across
This is analogous to the force on the surface being given by
F=∫P da=∫ P(L) W(L) dL
where P(L) is the pressure as a function of distance, L, along the surface and W(L) is the width of the surface at a distance L along the surface.
The point of action can be calculated in the same manner as for the earlier distributed load. Simply take moments about the top of the surface (L=0) by summing the moment due to each pile of books
ML=0= Σ L x pile height x weight/unit height of the books x # piles across.
This is directly analogous to the integral form for the moment given by.
ML=0=∫ L P(L) W(L) dL.
The distance to the point of action is D= ML=0/F. In general this is not 2/3rds of the way along the distributed load.
You can also use this demonstration to discuss numerical approximation of integrals. The nature of the integral does not matter; simply draw the analogy between the finite thickness slices (book piles) and the limit of infinitesimally thin book piles (integration).
Don’t forget to follow @nbkaye on twitter for updates to this blog. If you have a demonstration that you use in class that you would like to share on this blog please email me (firstname.lastname@example.org). I also welcome comments (through the comments section or via email) on improving the demonstrations.
Thanks to my father-in-law for lending me his books and floor for the photos.