Question

A semicircular plate with radius 7 m is submerged vertically in water so that the top is 3 m above 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:

F = 585844 N

Step-by-step explanation:

Given that:

A semicircular plate with radius 7 m is submerged vertically in water so that the top is 3 m above the surface.

The objective of this question is to express the hydrostatic force against one side of the plate as an integral and evaluate it.

To start with the equation of a circle:  a² + b² = r²

The equation of  circle with radius r = 7 can be expressed as:

a² + b² = 7²

a² + b² = 49

b² = 49 - a²

b = $\sqrt{49 -a}$

NOW;

The integral of the hydrostatic force with a semicircular plate with radius 7 m and the top is 3 m above the surface can be calculated as follows:

$\mathtt{F = 2 \rho g \int \limits^7_3 (a -3) \sqrt{49 -y^2} \ \ da}$

$\mathtt{F = 2 \rho g \begin {pmatrix}\dfrac{\sqrt{49 -a^2} \ (2a^2-9a - 98)-(441 \times sin^{-1} (\dfrac{a}{3})) }{6} \end{pmatrix}}$

where;

density of water is 1000 kg/m3

and acceleration due to gravity is 9.8 m/s

Solving the integral; we have:

F = 2 ×  1000 kg/m³ × 9.8 m/s × (29.89)

F = 585844 N