Answer:
hydrostatic force is 327000.00 N
Explanation:
given data
base = 4 m
height = 5 m
density of water = 1000 kg/m3
to find out
hydrostatic force
solution
we know this is a triangle so we consider here a small strip PQ whoes area da in with length x and width dy
so area will be
da = base/H × height = base/H × y
so da = x dy = base/H × y dy
and we know pressure = ρ × g × h
here h = y
hydrostatic force = pressure × area
df = (ρ × g × h) × base/H × y dy
now integrate it from 0 to 5 height
f = ρ × g [tex]\int_{0}^{5} (h) (base/H)ydy[/tex]
f = ρ × g [tex]\int_{0}^{5} (y) (4/5)ydy[/tex]
f = ρ × g × 4/5 × [tex](y^{3} /3)^{5} _0[/tex]
now put value ρ = 1000 and g = 9.81
f = ρ × g × 4/5 × [tex](y^{3} /3)^{5} _0[/tex]
f = 1000 × 9.81 × 4/5 × (5³/3)
force = 7848 × 41.666667
force = 327000.00 N
hydrostatic force is 327000.00 N
Answer:
The hydro-static force is 326666.67 N.
Explanation:
Given that,
Base = 4 m
Height = 5 m
We need to calculate the area of strip
using similar triangle,
[tex]\dfrac{y}{x}=\dfrac{B}{H}\Rightarrow y=\dfrac{B}{H}x[/tex]
Area of strip, [tex]dA =y\ dx[/tex]
[tex]dA =\dfrac{b}{H}\times x\ dx[/tex]
Where, b = base
dx = width of strip
H = height
Put the value of base into the formula
[tex]dA=\dfrac{4}{5}\times x dx[/tex]
We need to calculate the force
Using formula of force
[tex]dF= P\times dA[/tex]
Where, P = pressure
A = area
Put the value into the formula
[tex]dF=\rho g h\times dA[/tex]
[tex]dF=\rho g x\times\dfrac{4}{5}xdx[/tex]
On integration
[tex]\int_{0}^{F}dF=\int_{0}^{5}{\rho g x}\times\dfrac{4}{5}\times xdx[/tex]
[tex]F = \rho g\int_{0}^{5}(\dfrac{4}{5}x^2)dx[/tex]
[tex]F=\rho g\times\dfrac{4}{5}(\dfrac{x^3}{3})_{0}^{5}[/tex]
[tex]F=1000\times9.8\times\dfrac{4}{5}\times\dfrac{5^3}{3}[/tex]
[tex]F=326666.67\ N[/tex]
Hence, The hydro-static force is 326666.67 N.
