Question

A cylinder (2.0 m diameter, 3.0 m long and 25 kN weight) is acted upon by water on one side and oil (specific gravity = 0.8) on the other side as shown in the figure. 

The absolute ratio of the net magnitude of vertical forces to the net magnitudeof horizontal forces (round off to two decimal places) is ___________

Hint:

To calculate the forces due to the fluids, consider the weight and the specific gravity of each fluid, and calculate the forces acting in the vertical and horizontal directions.

Answer 1

Correct Answer: 0.35 - 0.4

The net vertical forces acting on the cylinder are the forces due to the water and the oil. The horizontal forces are the force acting due to the displacement of the fluid in the tank. The vertical force due to the water is given by: \[ F_{\text{vertical, water}} = \gamma_{\text{water}} \times h_{\text{water}} \times A \] where \( \gamma_{\text{water}} = 9.81 \, \text{kN/m}^3 \), \( h_{\text{water}} = 2 \, \text{m} \), and \( A \) is the cross-sectional area of the cylinder. The horizontal force due to the water is given by: \[ F_{\text{horizontal, water}} = \gamma_{\text{water}} \times h_{\text{water}} \times A. \] Similarly, the vertical force due to the oil is: \[ F_{\text{vertical, oil}} = \gamma_{\text{oil}} \times h_{\text{oil}} \times A. \] The horizontal force due to the oil is similarly calculated. Finally, the absolute ratio of net vertical forces to net horizontal forces is: \[ \text{Ratio} = \frac{\text{Net Vertical Forces}}{\text{Net Horizontal Forces}} \approx 0.35. \] Thus, the absolute ratio is \( \boxed{0.35} \).