Explanation:
We define force as the product of mass and acceleration.
F = ma
It means that the object has zero net force when it is in rest state or it when it has no acceleration. However in the case of liquids. just like the above mentioned case, the water is at rest but it is still exerting a pressure on the walls of the swimming pool. That pressure exerted by the liquids in their rest state is known as hydro static force.
Given Data:
Width of the pool = w = 50 ft
length of the pool = l= 100 ft
Depth of the shallow end = h(s) = 4 ft
Depth of the deep end = h(d) = 10 ft.
weight density = ρg = 62.5 lb/ft
Solution:
a) Force on a shallow end:
[tex]F = \frac{pgwh}{2} (2x_{1}+h)[/tex]
[tex]F = \frac{(62.5)(50)(4)}{2}(2(0)+4)[/tex]
[tex]F = 25000 lb[/tex]
b) Force on deep end:
[tex]F = \frac{pgwh}{2} (2x_{1}+h)[/tex]
[tex]F = \frac{(62.5)(50)(10)}{2} (2(0)+10)[/tex]
[tex]F = 187500 lb[/tex]
c) Force on one of the sides:
As it is mentioned in the question that the bottom of the swimming pool is an inclined plane so sum of the forces on the rectangular part and triangular part will give us the force on one of the sides of the pool.
1) Force on the Rectangular part:
[tex]F = \frac{pg(l.h)}{2}(2(x_{1} )+ h)[/tex]
[tex]x_{1} = 0\\h_{s} = 4ft[/tex]
[tex]F = \frac{(62.5)(100)(2)}{2}(2(0)+4)[/tex]
[tex]F =25000lb[/tex]
2) Force on the triangular part:
[tex]F = \frac{pg(l.h)}{6} (3x_{1} +2h)[/tex]
here
h = h(d) - h(s)
h = 10-4
h = 6ft
[tex]x_{1} = 4ft\\[/tex]
[tex]F = \frac{62.5 (100)(6)}{6} (3(4)+2(6))[/tex]
[tex]F = 150000 lb[/tex]
now add both of these forces,
F = 25000lb + 150000lb
F = 175000lb
d) Force on the bottom:
[tex]F = \frac{pgw\sqrt{l^{2} + ((h_{d}) - h(s)) } (h_{d}+h_{s}) }{2}[/tex]
[tex]F = \frac{62.5(50)\sqrt{100^{2}(10-4) } (10+4) }{2}[/tex]
[tex]F = 2187937.5 lb[/tex]
