Posted by **hussain** on Sunday, January 30, 2011 at 10:35pm.

A dam 15 m high holds the water in reservoir . For safety reasons , the water level is maintained at a height 3 m below the top of the dam. At best of the dam , engineers have installed a vertical door in the shape of an equilateral triangle with sides of length 3 m , calculate the total hydrostatic force on the door. your solution must be clear and complete. you must define clearly in words each of the variables used and indicate their meaning on a digram . the density of water is p = 1000 kg/mˆ3. and the acceleration due to gravity is g= 9.8 m/sˆ2.

- calculus -
**MathMate**, Sunday, January 30, 2011 at 11:23pm
You will need to set up the expression for the pressure, and then proceed to integrate the expression throughout the height of the door.

We also have to assume that the door is at the *base* of the dam, and that the base of the triangular door is at the bottom of the dam.

We will denote elevation from the bottom of the dam by y, so that the water level is at y=15-3=12m.

The width w of the triangular door is diminishing with h, such that at y=0, w=3, and at y=top of the door h= 3√3/2, w=0.

Thus the width as a function of height is

w(y)=3(h-y)/h (in m) for 0<y<h

The hydrostatic pressure at an elevation y from the bottom is p(y)=ρg(H-y) (in N/m²)

The total hydrostatic force on a horizontal strip of door of height dy is therefore p(y)*w(y)dy

The total hydrostatic force for the whole door

= ∫pressure&dA

= ∫p(y)*w(y)dy for y=0 to h.

I get 425230 N. (to the nearest 10 N)

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