Answer:
[tex]F = 7.34 \times 10^6 N[/tex]
Explanation:
Since each side of the triangular end is of equilateral triangle with side length a = 10 m
so height of the triangle h = a sin60
h = 10sin60 = 8.66 m
now we will take a small strip of width L and thickness dy at a depth of y from top
so here width L is given as
[tex]L = 2ytan30 = 3.46 y[/tex]
now the force on this small strip is given as
[tex]dF = P . dA[/tex]
[tex]dF = (\rho g y). (3.46 y dy)[/tex]
now the total force on the triangular part
[tex]F = \int 3.46 \rho g y^2 dy[/tex]
[tex]F = (3.46 \rho g)(\frac{y^3}{3})[/tex]
now the limits of y is from y = 0 to y = 8.66 m
so we have
[tex]F = 3.46(1000)(9.8)(\frac{8.66^3}{3})[/tex]
[tex]F = 7.34 \times 10^6 N[/tex]
