Chapter 5: Applications of Integration
Section 5.9: Hydrostatic Force
The water behind a trapezoidal dam (dimensions 25 × 15 × 15 × 15 in feet, longest edge uppermost), is to the top edge. Find the total hydrostatic force on this structure if water weighs 62.5 lbs/ft3.
By the Pythagorean theorem,the height of the dam is 152−52=200=102. (See Figure 5.9.2(a).)
As per Figure 5.9.2(a), measure y positive upwards from the bottom of the dam. To find dA, the area of a horizontal strip at depth d=102−y, observe (from similar triangles) that
Adding 15 to twice this length gives
Figure 5.9.2(a) Hydrostatic force on a trapezoidal dam
so that dA=y2+15dy. The force is the product of the pressure P=62.5 d and the area, or
∫0102P d dA=1252∫0102102−y y2+15 ⅆy=3437503 ≐114583.33 lbs
Alternatively, obtain the hydrostatic force as the product of the area and the pressure at the centroid of the dam.
From Example 5.7.5, y&conjugate0; = 652/12, and A=2002.
Consequently, the hydrostatic force is h−y&conjugate0;A δ = 62.5×102−652/12×2002≐1.145833333⁢105
Expression palette: Definite-integral template
Enter the integral that gives the total hydrostatic force on the dam.
Context Panel: Evaluate and Display Inline
Context Panel: Approximate≻10
1252∫0102102−y y2+15 ⅆy = 3437503→at 10 digits1.145833333⁢105
Alternatively, F=h−y&conjugate0;A δ =
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