Cylindrical tank 2 m diameter and 4 m long, with its axis horizontal is half filled with water and half filled with oil of density 880kg / (m ^ 3) Determine the magnitude and position of the net hydrostatic force on one end of the tank.
To determine the magnitude and position of the net hydrostatic force on one end of the tank, we need to consider the pressure exerted by water and oil separately, and then combine their effects.
First, let's calculate the pressure exerted by the water:
The weight of the water equals the mass of water multiplied by the acceleration due to gravity (9.8 m/s^2). The mass of water can be calculated by multiplying the volume of water by its density.
The volume of water can be found by considering the dimensions of the cylindrical tank. Since the tank is half-filled with water, the volume of water is equal to half of the volume of the entire tank.
The volume of the entire tank is given by the formula: V = πr^2h, where r is the radius of the tank and h is the height of the cylindrical tank.
Given:
Diameter of the tank (d) = 2 m
Radius of the tank (r) = d/2 = 2/2 = 1 m
Length of the tank (h) = 4 m
The volume of the entire tank is:
V = π(1^2)(4) = 4π m^3
Therefore, the volume of water (V_w) is:
V_w = (1/2) * 4π = 2π m^3
To calculate the mass of water (m_w), we multiply the volume by the density of water (ρ_w = 1000 kg/m^3):
m_w = V_w * ρ_w = 2π * 1000 = 2000π kg
The weight of the water (W_w) is:
W_w = m_w * g = 2000π * 9.8 ≈ 19600π N
Now, let's calculate the pressure exerted by the oil:
The mass of the oil can be calculated using the same formula as above, but with the density of oil (ρ_o = 880 kg/m^3).
The volume of oil (V_o) is also equal to half of the volume of the entire tank:
V_o = (1/2) * 4π = 2π m^3
The mass of oil (m_o) is:
m_o = V_o * ρ_o = 2π * 880 = 1760π kg
The weight of the oil (W_o) is:
W_o = m_o * g = 1760π * 9.8 ≈ 17248π N
To find the net hydrostatic force on one end of the tank, we subtract the weight of the oil from the weight of the water:
F_net = W_w - W_o = 19600π - 17248π = 2352π N
So, the magnitude of the net hydrostatic force on one end of the tank is 2352π N.
To determine the position of the net hydrostatic force, we need to calculate the center of pressure, which is the point at which the force acts.
In a uniform horizontal tank, the center of pressure is located at the vertical distance (h_c) from the bottom of the tank, given by:
h_c = (h/2) + ((m_o * y_o) - (m_w * y_w)) / (ρ_w * h)
Where:
y_o = distance from the bottom of the tank to the center of mass of the oil (2 m)
y_w = distance from the bottom of the tank to the center of mass of the water (1 m)
Substituting the given values:
h_c = (4/2) + (((1760π * 2) - (2000π * 1)) / (1000 * 4))
= 2 + ((3520π - 2000π) / 4000)
≈ 2 + (1520π / 4000)
≈ 2 + (19π / 50)
≈ 2.38 m (rounded to 2 decimal places)
Therefore, the position of the net hydrostatic force on one end of the tank is approximately 2.38 meters from the bottom of the tank.