Question

If the water face of a dam is in the shape of a trapezoid.

Height =(h), width at water level =(a+b) and width at base = (a)

How can you show the resultant force = 1/6wh^2 (3a+b) ?

Thanks

Answer 1

If the water face of a dam is in the shape of a trapezoid.

Height = (h), width at water level = (a + b) and width at base = (a)

How can you show the resultant force = 1/6wh^2 (3a + b)?

The force on $i$-th strip is $F_{i} = w * x_{i} * S_{i}$

Where $S_{i}$ is the area of ABCD, for practical purposes we suppose it to be a rectangle, so

$$S _ {i} = l _ {i} * \Delta x _ {i}$$ $$\frac {l _ {i} - a}{h - x _ {i}} = \frac {b}{h}$$ $$l _ {i} = a + b - \frac {b * x _ {i}}{h}, \text{ so } S _ {i} = (a + b - \frac {b * x _ {i}}{h}) * \Delta x _ {i} \text{ and } F _ {i} = w * x _ {i} * (a + b - \frac {b * x _ {i}}{h}) * \Delta x _ {i}$$

The resultant force is

$$F = \int_ {0} ^ {h} w x (a + b - \frac {b x}{h}) d x = w \left(\left(\frac {(a + b) x ^ {2}}{2} - \frac {b x ^ {3}}{3 h}\right)\right) \Bigg | _ {0} ^ {h} = w \left(\frac {(a + b) h ^ {2}}{2} - \frac {b h ^ {2}}{3}\right) = \frac {w h ^ {2}}{6} (3 a - b)$$