We can begin by calculating the force due to the water pressure acting on the gate. The pressure at a depth h under the surface of the water is given by P = ρgh, where ρ is the density of water (1000 kg/m³), g is the acceleration due to gravity (9.81 m/s²), and h is the depth of the centroid of the gate.
The centroid of the quadrant of the circle lies at a radius of 4/3*pi from the center (you can calculate this using the formula for the centroid of a curve: (1/(pi R))*(integral from 0 to R of (R^2 - x^2)dx)). Therefore, the depth of the centroid is h = 1.5m - (4/3*pi).
Now we can calculate the pressure acting on the gate at the centroid:
P = ρgh = (1000 kg/m³)(9.81 m/s²)(1.5m - (4/3*pi))
P ≈ 6767.77 N/m²
The total force due to the water pressure is given by F = PA, where A is the area of the gate. Since the gate is a quadrant of a circle with a radius of 1.5m, its area is given by A = 1/4 * pi * (1.5m)² = 1.7671 * pi m². The width of the gate is 3m, so the total volume of the gate is 1.7671 * pi * 3 = 16.621 m². The total force due to the water pressure is then:
F = PA = (6767.77 N/m²)(16.621 m²) = 112540.23 N
The turning moment required to open the gate depends on the weight of the gate itself. Since the gate has a mass of 6000 kg, its weight is given by W = mg = (6000 kg)(9.81 m/s²) = 58860 N.
The weight of the gate acts downward at the centroid G, creating a clockwise turning moment about the pivot O. The magnitude of this turning moment is given by M = rW, where r is the perpendicular distance from the line of action of the force to the pivot. Since the gate is a quadrant of a circle, the distance between O and G is (4/3/pi)(1.5m) ≈ 0.6366 m.
Thus, the magnitude of the turning moment due to the weight of the gate is M = rW = (0.6366 m)(58860 N) = 37432.996 Nm.
The force due to the water pressure on the gate acts at a perpendicular distance of 2r/3 from the pivot O. Therefore, the magnitude of the turning moment due to the water pressure is M' = r'F = (2/3*r)(112540.23 N) = (2/3 * 1.5m)(112540.23 N) = 112540.23 Nm.
The net turning moment required to open the gate is the difference between the two moments:
Net turning moment = M' - M = (112540.23 Nm) - (37432.996 Nm) = 75107.234 Nm
The force exerted at the centroid is parallelogram, comprising the hydrostatic force and the vertical force due to the weight of gate.
Magnitude of the Resultant Force = sqrt(F^2 + W^2)
Magnitude of the Resultant Force = sqrt((112540.23 N)^2 + (58860 N)^2) = 125000 N (approximately)
The direction of the Resultant Force can be found using the tangent of the angle between the horizontal force and the Resultant Force:
tan(angle) = W / F
angle = arctan(W / F)
angle = arctan(58860 N / 112540.23 N) = 27.5 degrees (approximately)
So, the magnitude of the resultant force on the gate due to the water is approximately 125000 N and the direction of the force is 27.5 degrees from the horizontal. The turning moment required to open the gate is approximately 75107.234 Nm.
A sluice gate consists of a quadrant of a circle of radius 1.5m pivoted at its centre O. Its centre of gravity is at G, as Shown in fig 1. When the water is level with the pivot O, calculate the magnitude and direction of the resultant force on the gate due to the water and the turning moment required to open the gate. The width of the gate is 3m and it has a mass of 6000kg.
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