The cross section of a 0.5 m wide vertical gate holding water and oil is shown in the figure. The unit weights of water and oil are 10 kN/m$^3$ and 7.5 kN/m$^3$, respectively. The horizontal hydrostatic force (in kN) acting on the vertical gate is _________ (rounded off to two decimal places).
To find the horizontal hydrostatic force on the gate, we need to calculate the forces due to oil and water separately, then sum them.
Force due to Water:
The pressure at the centroid: \( P_w = \gamma_w \cdot h_w \), where \( \gamma_w = 10 \text{ kN/m}^3 \) and \( h_w = 0.5 \text{ m} \).
So, \( P_w = 10 \times 0.5 = 5 \text{ kN/m}^2 \).
The force on water is: \( F_w = P_w \cdot A_w \), where \( A_w = \text{width} \times \text{height} = 0.5 \times 1 = 0.5 \text{ m}^2 \).
Thus, \( F_w = 5 \times 0.5 = 2.5 \text{ kN} \).
Force due to Oil:
The pressure at the centroid: \( P_o = \gamma_o \cdot h_o \), where \( \gamma_o = 7.5 \text{ kN/m}^3 \) and \( h_o = 0.25 + 1 = 1.25 \text{ m} \).
So, \( P_o = 7.5 \times 1.25 = 9.375 \text{ kN/m}^2 \).
The force on oil is: \( F_o = P_o \cdot A_o \), where \( A_o = \text{width} \times \text{height} = 0.5 \times 0.5 = 0.25 \text{ m}^2 \).
Thus, \( F_o = 9.375 \times 0.25 = 2.34375 \text{ kN} \).
Total Horizontal Hydrostatic Force:
The total force is: \( F_{\text{total}} = F_w + F_o = 2.5 + 2.34375 = 4.84375 \text{ kN} \).
Rounded to two decimal places, the hydrostatic force is \( 4.84 \text{ kN} \), which falls within the expected range.
Important Questions from Hydrostatic Force
The centre of pressure of a plane submerged surface
In the context of hydrostatics, the resultant hydrostatic force acting on a submerged plane surface passes through which of the following points?
The depth of the center of pressure on a vertical rectangular gate (4 m wide and 3 m high) with water up to top surface is
If a planar surface is immersed in a liquid, the resultant liquid pressure acts at a point called ___________.
The resultant of all normal pressure acts
