Answer :
Answer:
313.6 kN
Explanation:
Given,
base = 6 m
height = 2 m
vertex height = 4 m
the force on the strip at (6-y) depth
df = ρg(6-y) (3 (2-y) dy)
df = 3ρg (12-8y+y^2) dy
Integration
F = [tex]3\rho g\int\limits^2_{y=0} {12-8y+y^2} \, dy[/tex]
F = [tex]3\rho g (12y-8\frac{y^2}{2}+\frac{y^3}{3})^2_{0}[/tex]
F = 3ρg (32/3)
F = 32*1000*9.8
F = 313600 N
The hydrostatic force is the force due to the effect of pressure acting on an area of a submerged body's surface
The expression for the hydrostatic force acting against one side of the plate and the magnitude of the force are;
- [tex]The \ hydrostatic \ force \ as \ an \ integral, \ is \ \displaystyle 3 \cdot \rho \cdot g \cdot \int\limits^2_0 {\left (4 \cdot h + h^2 \right)} \, dh[/tex]
- The magnitude of the hydrostatic force is approximately 313,600 N
The reason the above values are correct are as follows:
The known parameters of the plate are;
Base length of the triangular plate, l = 6 m
Height of the triangular plate, h = 2 m
The depth of the highest vertex below the surface of the water = 4 m
Required:
To express the hydrostatic force against a side of the plate as an integral
Solution:
The area of small slit of the plate, dA = l × dh
The ratio of the length to height of the plate is given as follows;
[tex]\dfrac{h}{l} =\dfrac{2}{6} = \dfrac{1}{3}[/tex]
Therefore;
l = 3·h
Which gives;
dA = l × dh = 3·h × dh
dA = 3·h × dh
The force, dF acting on a small element dA, is given as follows;
dF = ρ·g·(4 + h)·dA
Therefore;
dF = ρ·g·(4 + h)·3·h × dh
dF = 3·ρ·g·(4·h + h²) × dh
Given that the height is 2 meters, the value of h ranges from 0 to 2, which gives;
[tex]F = \displaystyle \int\limits {dF} \, dh = 3 \cdot \rho \cdot g \cdot \int\limits^2_0 {\left (4 \cdot h + h^2 \right)} \, dh[/tex]
[tex]The \ hydrostatic \ force \ as \ an \ integral, \ is \ \displaystyle 3 \cdot \rho \cdot g \cdot \int\limits^2_0 {\left (4 \cdot h + h^2 \right)} \, dh[/tex]
Evaluating the integral gives;
[tex]3 \cdot \rho \cdot g \cdot \int\limits^2_0 {\left (4 \cdot h + h^2 \right)} \, dh = 3 \cdot \rho \cdot g \cdot \left [2\cdot h^2 + \dfrac{h^3}{3} \right ]^2_0[/tex]
Therefore, the hydrostatic force against one side of the plate as an integral is given as follows;
[tex]F = 3 \times 1,000 \times 9.8 \times \left [2\times 2^2 + \dfrac{2^3}{3} - 0\right ] \approx 313,600[/tex]
The hydrostatic force, against a side of the plate, F ≈ 313,600 N
Learn more about hydrostatic force here:
https://brainly.com/question/13622527
