The area of a square gate is (12.69m) provided in oil tank is hinged at its top edge figure below. Determine the necessary vertical pull to be applied at the lower edge to open the gate (N). Gasoline surface Negative pressure (8250N/m2) 1.8m Gaseline (S-0.77 hinge gate Vertical pull F4 145°
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- The area of a square gate is (12.69m) provided in oil tank is hinged at its top edge figure below. Determine the necessary vertical pull to be applied at the lower edge to open the gate (N). Gasoline surface Negative pressure (8250N/m?) 1.8m Gaseline (S-0.77 hinge -gate Vertical pull F 145°AIR . P = 7 KPä." 1.5 m OIL HINGE SG = 0.82 A TA 3m The gate shown in the figure is hinged at A and rests on a smooth floor at B. The gate is a square in shape with 3m x 3m in dimension, while it is submerged in oil (SG = 0.82) as shown. The air above the oil surface is under a pressure of 7KP.. If the gate weighs 5 KN, determine the vertical force "T" required to open the gate. (Notes: Hydrostatic force "F" is always perpendicular to the plane surface. Also, the h is the pressure head located at the center of gravity of the object, which means, what will be the height of the oil if I am going to convert the pressure at the top into oil) (A) 178.52 KN B 184.60 KN c) 202.66 KN (D) 196.36 KN (E) No answer among the choices GATE 3mNote: Please state what formula you are using. And please be clear and detailed with your solution. Atleast explain. It will help me, thanks. Problem: A Triangular gate having a horizontal base of 1.30 m and an altitude of 2 m is inclined 45o from the vertical with vertex pointing upward. The base of the gate is 2.70 m below the surface of oil (s = 0.80). What normal force must be applied at the vertex of the gate to open it? Show figure
- There is a circular gate at the end of the tank as illustrated. A 6.387 * 10–3 m³ wood block is tied to the end of the gate as illustrated. Density of the wood block is 0.6 g/cm³. Water is at 15.6 °C and Mercury is at 20 °C. What will be the reading h approximately on the mercury manometer on the left side when the gate opens up? Water 50 Wood Hinge 0.4 m Gate D=4 cm Mercury O a. 0.1 m O b.0.14 m c. 0.22 m d. 0.26 mAn a tank filled with mercury (density 13000 kg/m3)at depth 10 m , the force actining at the hatch with diameter 0.75 m (located at the bottom tank) is (Taken gravitational acceleration as 10 m/s2):Two vessel are connected to differential manometer using tubing being filled with water. The higher pressure vessel is than the other. (a) If the mercury reading is in meters of water? (b) If carbon tetrachloride (s = iwould be the manometer reading for the same pressure difference? mercury density= 13.6 Solution mercury, the connecting 1.38 m lower in elevation 110 mm, what is the pressure head difference 1.59 )were use instead of mercury, what 1.38 Air 4 water 3 m X Fig A 2 1 Air water gage liquid
- Please be clear with your solution. Make an explanation for the formula and solution you used. Problem: A Triangular gate having a horizontal base of 1 30 m and an altitude of 2 m is inclined 45 o from the vertical with vertex pointing upward The base of the gate is 2 70 m below the surface of oil s 0 80 What normal force must be applied at the vertex of the gate to open it? Show figureDetemine the horizontal and vertical components of the liquid pressure acting on the semicylindrical gate ABC, as shown in the figure. The width into the paper is 1 m. 1 m Liquid (s.g. = 0.88) 4 m B 4.2 m 1 m Fx=164.018 kN and Fy=37.229 kN O Fx=141.327 kN and Fy=31.171 kN Fx=257.430 kN and Fy=59.801 kN Fx=288.147 kN and Fy=69.098 kN Fx=213.112 kN and Fy=47.903 kN Fx-187.005 kN and Fy=42.182 kN Fx=234.789 kN and Fy=52.418 kN Fx=98.182 kN and Fy-20.002 kNO Fx-119.956 kN and Fy-25.550 kN Fx=269.894 kN and Fy-65.526 kN OPlease be xlear and detailed with your answer. Write an explanation in every formula and solution you used. Problem: A Triangular gate having a horizontal base of 1 30 m and an altitude of 2 m is inclined 45 o from the vertical with vertex pointing upward The base of the gate is 2 70 m below the surface of oil s 0 80 What normal force must be applied at the vertex of the gate to open it? Show figure
- A tank is filled with fluid under prcssurc anti the prcssurc gauge fixcd at the top indicatcs a prcssurc of 21 kPa. The curved surfacc of the top is quarter of a circular part of radius 2.6 m. calculate the magnitude (N) and direction (degree) of the resultant force on the cun'ed surface. (p=950kg./m³for fluid)5. For the setup shown in the figure, what is the pressure pA if the specific gravity of the oil is 0.82? (S.G. of Hg 13.6) Air Oil Орen 4 m H,0 320 mm Hg- Figure for Prob. No. 5A Horizontal Capillary Tube (Radius 10 Um) Con- Tains An Oil/Water Interface. The Pressure In The Oil Phase Is 14.5 Psi (100 KPa). Determine The Pressure In The Water Phase If The Contact Angle Is 30° And The IFT Is 30 Dynes/Cm (30 MN/M).