Find the total hydrostatic force on the face of a semi-circular dam

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Homework Statement



Find the total hydrostatic force on the face of a semi-circular dam with radius of 20 m, when its reservoir is full of water. The diameter of the semicircle is the top of the dam.

Homework Equations





The Attempt at a Solution



the width at any point would be 2sqrt(400-y^2) from x^2+y^2 = r^2
the density would be 1000 kg/m^3, this is the rough estimate for the density of water I am told to use in the class
gravity is 9.8 m/s^2, for some reason we don't use 9.81 in the class

I got F= 19600*integral[0,20] (20-y)sqrt(400-y^2)dy

were 19600 is just density times gravity times 2 = 1000*9.8*2
20-y is the depth at any given point y

I have no idea how to evaluate the indefinite integral
integral (20-y)sqrt(400-y^2)dy
when I tried doing so I spent a good amount of time trying to evaluate it using a trig sub but ended up doing something terribly wrong for some reason I got
integral (20-y)sqrt(400-y^2)dy = (ysqrt(400-y^2))/40 - 10 cos^(-1)(y/20) + sqrt(400-y^2)/3 - (y^2 sqrt(400-y^2))/1200 + c
I got frustrated with it and just plugged into my calculator
19600*integral[0,20] (20-y)sqrt(400-y^2)dy
and got about 70883765.35 N

This question however has reappeared on a practice test in which the problem was solved differently but I think it may have been solved wrong, the answer key my professor made had
F = integral[0,20] 9800sqrt(400-h^2)2h dh
as the integral to be evaluated were they just used h for the depth instead of (20-h). I thought you had to include this though in the integral and not just h because you want the depth at any point of any of the horizontal slices and not just h

So I thought that the problem was solved wrong on the answer key but sense my integral I had a really tough time evaluating by hand 19600*integral[0,20] (20-y)sqrt(400-y^2)dy I think I may have done something wrong and I was just suppose to use y instead of (20-y).

What's the correct integral to solve this problem? Thanks for any help!!!
 

Answers and Replies

  • #2
I get the same integral as your professor.

I worked it out in terms of depth, so if P(h) is the pressure as a function of depth:
P(h) = gdh [g=gravity, d=density, h=depth].

If w(h) is the width of the dam as a function of depth, then
w(h) = 2(r^2 -h^2)^(1/2) [r=radius of the semi-circle]

The force on a region of the dam, dh is pressure times area, so
dF = P(h)w(d)dh
so
F = [itex] \int_0^r P(h)w(d)dh [/itex]
or
F = [itex] \int_0^r (9.8)(1000)(h)(2)(400-h^2)^{1/2}dh [/itex]
 

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