Question

A triangle with base 2 m and height 3 m is submerged vertically in water so that the tip is even with the surface. Express the hydrostatic force against one side of the plate as an integral and evaluate it. (Give your answer correct to the nearest whole number.)

Answer 1

Answer: Check the attached

Step-by-step explanation:

Answer 2

Answer:

The Force on one side of the pate is 58800N.

Step-by-step explanation:

The base of triangle is given as b=2m

The height of triangle is given as h=3m

Gravitational Acceleration is given as g=9.8 m/s^2

The density of water is given as ρ =1000 kg/m3

Now for a rectangular strip DE as shown in the attached figure, of width dy and length x is considered for which the area is

The area is given as

$dA=xdy$

Also for this the pressure at depth y is given as

$P=\rho g y$

As the triangles are similar so

$\frac{x}{y}=\frac{b}{h}\\\frac{x}{y}=\frac{2}{3}\\x=\frac{2y}{3}$

Now the differential force is given as

$dF=PdA\\dF=\rho gy \times xdy\\dF=\rho gy \times \frac{2y}{3}dy\\dF= \frac{2}{3}\rho gy^2dy$

Now for the total force, the integral for 0 to h=3 m is given as

$F=\int_{0}^{3}dF\\F=\int_{0}^{3}\frac{2}{3}\rho gy^2dy\\F=\frac{2}{3}\rho g\int_{0}^{3}y^2dy\\F=\frac{2}{3}\times 1000\times 9.8[\frac{y^3}{3}]_0^3\\F=6533.33[9]\\F=58799.97 \approx 58800 N$

So the Force on one side of the pate is 58800N.