I just gave my second “calculus two” exam; the final two problems involved the following:
Suppose a trough with semi-circular ends is filled with water (say, length = 40 feet, radius = 5 feet).
1. How much work does one do in pumping all of the water to the top of the tank and out? (work against gravity only)
2. How much hydrostatic force is there against one of the semi-circular end plates?
Assuming water is 62.5 pounds per gallon (I gave that to them):
1. Work = ; of course, is the distance a molecule of water is lifted and is the cross sectional volume of water.
2. Force = ; of coure, is the depth of the water (hence is the pressure at that depth) and is the area at depth that the pressure is applied to.
The student can easily notice that the two answers differ by a factor of 40, which is the length of the trough.
So, what is the lesson here? Well, for one, I always envisioned a wall of a tank holding back a long mass of water. That isn’t correct; ONLY THE DEPTH matters. If the tank were a mile long or, say, an inch long, the pressure on the semi-circular ends would be the same. That runs counter to my intuition (which is clearly bad).
This lead me to think about the following: what if one were to put in a baffle between the two ends and the baffle was the same shape as the semi-circular ends. What would be the force on that plate?
Of course, the net force would be zero; there are two sides of the plate.
Now drop an open cylinder into the tank (think: a can with the top and bottom cut away). Clearly: zero force on the sides, right? Two sides, right?
Now, drop a Mobius band into the tank. A Mobius band has but one side. What is the force on it?
The key here: the Mobius band is one sided, but it is LOCALLY two sided; one can break the surface into tiny rectangles and note that the net force on each rectangle is zero as it is locally two sided. Hence zero total force on the one sided object.