Question

An equilateral triangular plate with sides 6 m is submerged vertically in water so that the base is even with the surface. Express the hydrostatic force against one side of the plate as an integral and evaluate it. (Round your answer to the nearest whole number. Use 9.8 m/s2 for the acceleration due to gravity. Recall that the weight density of water is 1000 kg/m3.)

Answer 1

Answer:

The hydro-static force against one side of the plate is 617400 N

Step-by-step explanation:

Here we have the depth given by

Depth of immersion = Height of triangle

Height, h of equilateral triangle of side L is given by;

$h = \sqrt{L^2 -(\frac{L}{2})^2 }$

Therefore height h = $\sqrt{6^2 -(\frac{6}{2})^2 } = 3\cdot\sqrt{3}$

If we take the width of a small section w(x) at a location of x m belo the surface, we have by similar triangles

$\frac{w(x)}{6} = \frac{3\cdot\sqrt{3}-x }{3\cdot\sqrt{3}}$

Therefore, w(x) = $6 -\frac{2\cdot x}{3\cdot \sqrt{3} }$

The    

$F = 1000 \times 9.8\times \int\limits^{3\cdot\sqrt{3}} _0 {x}[6-\frac{2x}{3\cdot \sqrt{3} } ] \, dx$

$F = 1000 \times 9.8\times [3\cdot x^{2} -\frac{x^{2} }{3\cdot\sqrt{3} } ]^{3\cdot \sqrt{3}}_0$ = 63×9800 = 617400 N