Final answer:
The magnitude of the resultant force acting on the inclined 2-m-diameter circular gate is 95,200 N, and the point of application is at the center of pressure (CP), located 4 meters below the top of the gate.
Step-by-step explanation:
The resultant force acting on the inclined 2-m-diameter circular gate can be calculated using the principles of fluid mechanics. The force exerted by the water on the gate can be broken down into two components: the normal force (FN) and the frictional force (FF).
The normal force (F_N) can be calculated using the equation:
FN = ρ * A * g
where ρ is the density of water (1000 kg/m³), A is the area of the gate (2 m in diameter = π * (2 m)² = 4.54 m²), and g is the acceleration due to gravity (9.81 m/s²).
FN = ρ * A * g
FN = 1000 kg/m³ * 4.54 m² * 9.81 m/s²
FN = 45,400 N
The frictional force (FF) can be calculated using the equation:
FF = μ * FN
where μ is the coefficient of friction between the gate and the tank wall (typically around 0.05 to 0.1 for smooth surfaces).
FF = μ * FN
FF = 0.05 * 45,400 N
FF = 2,270 N
The resultant force (FR) can be calculated by adding the normal and frictional forces:
FR = FN + FF
FR= 45,400 N + 2,270 N
FR = 47,670 N
The point of application of the resultant force is at the center of pressure (CP), which is the point where the force is transmitted to the gate. The CP can be calculated using the equation:
CP = FR / 2 * (L + W)
where L is the length of the gate (2 m) and W is the width of the gate (2 m).
CP = 47,670 N / 2 * (2 m + 2 m)
CP = 23,800 N * 4 m
CP = 95,200 N
Complete question:
Find the magnitude and point of application of the resultant force acting on the inclined 2-m-diameter circular gate. The tank is closed and completely full of water. The pressure gage at the bottom of the tank reads 100 kPa gage pressure.