Respuesta :
Answer:
44100 N
Explanation:
Each wall will have dimension of 4 m x 1.5 m
Whole force will act on central point of wall situated at a depth of 1.5 /2 = .75m
pressure at CM = h d g , h = .75 , d ( density of water = 10³ )
pressure at CM = .75 x 10³ x 9.8
= 7350 N / m²
Total force on each wall
= pressure x area
= 7350 x 4 x 1.5
= 44100 N Ans
b ) If h = 1.5 x 2 = 3
Pressure = hdg
1.5 x 10³ x 9.8
= 14700 N / m²
Force
= pressure x area
14700 x 3 x 4
= 176400 N
Which is 4 times 44100 N
So force will quadruple.
It is so because both area and height have become twice.
(a) The hydrostatic force on each wall is 44,100 N.
(b) When the height of the walls of the pool is doubled, the hydrostatic force on each wall quadruples because the area and height of the pool doubles.
The given parameters;
- length of pool, l = 4 m
- width, w = 4 m
- height, h = 1.5 m
- density of water, ρ = 1000 kg/m³
The pressure at the center of gravity of the pool is calculated as;
[tex]P = \rho g (\frac{h}{2} )\\\\P = (1000)(9.8)(0.75)\\\\P = 7350 \ N/m^2[/tex]
Area of the pool is calculated as;
A = 4 x 1.5 = 6 m²
The hydrostatic force on each wall is calculated as follows
F = PA
P = 7350 x 6
P = 44,100 N
When the height doubles;
h = 2 x 1.5 = 3 m
The pressure at the center of gravity of the pool is calculated as;
[tex]P = \rho g(\frac{h}{2} )\\\\P = (1000)(9.8)(1.5)\\\\P = 14,700 \ N/m^2[/tex]
Area of the pool
A = 4 x 3 = 12 m²
The hydrostatic force on each wall is calculated as follows
F = PA
F = 14700 x 12
F = 176,400 N
The ratio of the two hydrostatic forces;
[tex]= \frac{F_2 }{F_1} = \frac{176400}{44,100} = 4[/tex]
Thus, when the height of the walls of the pool is doubled, the hydrostatic force on each wall quadruples because the area and height of the pool doubles.
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