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Calc 2 Fluid Pressure problem setup

  • #1
658
2

Homework Statement



Calculate the fluid force of a right triangle with height 3ft and base 2ft submerged in water vertically and it's upper vertex at a depth of 4ft.


Homework Equations



F = w int(from a to b) y*f(y)dy

The Attempt at a Solution



I'm just not seeing how to set this one up. I setup my origin at the bottom of the triangle then the surface of the water would be y = 7. The triangle would be at y = 3. My height function would be 7 - y, right? Then would my other function be 2/3y + 2 ?
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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Homework Statement



Calculate the fluid force of a right triangle with height 3ft and base 2ft submerged in water vertically and it's upper vertex at a depth of 4ft.
This appears to me to be ambiguous. I interpret "submerged in water vertically" to mean that the normal to the triangle is horizontal but how is the triangle oriented in that plane?

Are we to assume that one of the legs is horizontal? If so, which? Or are we to assume that the hypotenuse is horizontal? Or some other orientation?

I suppose we could interpret the words "base" and "height" to mean that the 2 foot long side is horizontal and the 3 foot long side vertical. That appears to be consistent with what you are doing. (Though the words "base" and "height" are commonly used for right triangles without reference to the orientation of the triangle and are interchangable.)

I would set up a coordinate system with (0, 0) at the right angle. The two other vertices are then at (0, 3) and (2, 0). The hypotenuse is given by x/2+ y/3= 1 or, solving for x, x= 2(1- y/3)= 2- (2/3)y. I solved for x because that measures the width of the triangle at each y. Imagine dividing the triangle into many very thin horizontal rectangles, each of width "dy". The reason for using horizontal rectangles is that we can take each point in each rectangle as being at the same depth.

The length of each such rectangle is the x value, 2- (2/3)y and so the area is (2- (2/3)y)dy. The force on each such rectangle is that area times the pressure at that depth. the pressure at each y is the weigth of a thin column of water above that y which would be the density of water times the height of that column. Since y measures the distance above the bottom of the triangle, 3- y measures the distance below the vertex and that vertex is itself 4 ft below the surface of the water. The height of water above y is 4+ (3- y)= 7- y.

Homework Equations



F = w int(from a to b) y*f(y)dy

The Attempt at a Solution



I'm just not seeing how to set this one up. I setup my origin at the bottom of the triangle then the surface of the water would be y = 7. The triangle would be at y = 3. My height function would be 7 - y, right? Then would my other function be 2/3y + 2 ?
Yes, the height is 7- y. The other function, the width of triangle at that y is 2- (2/3)y, not (2/3)y+ 2. As a check, notice that, at the top vertex, where y= 3, your formula says the width of the triangle is 2+ 2= 4 while mine says that it is 2- 2= 0. Of course, 0 is correct because we are at a vertex.
 
  • #3
658
2
I totally agree that it's confusing in the wording. Unfortunately, they didn't give any picture associated with this problem but I did setup the problem the same way as you described. I'm gonna play with it and see what happens. Thank you for your time.
 

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