Question

Q4 A square gate, with sides of length 1360 mm, is used to regulate the level of light machine tool lubricating oil (sg = 0.881) in a tank, as shown The gate is inclined at 44° to the horizontal. The depth of the oil in the tank is 1470 mm hned Determine: 1470 mm 1360 mm • Lc, the distance from the level of the surface of the oil to the centroid of the gate, measured along the angle of inclination of the gate. . Fr, the resultant of the fluid pressure on the gate. . 8, the direction of FR Lp, the distance from the level of the surface of the oil to the centre of pressure on the gate, measured along the angle of inclination of the gate. The tension T required in the vertical chain to begin to open the gate. You may assume that the weight of the gate is negligible compared with the force due to the liquid pressure and may be ignored.)

Answer 1

OCE S2S2 OB mm R_944.734 bommy TA (a) OB = fm 44 ) OB=OASA DA Tuzomm 90 B= 1360 Sin 44 OB=944.735 mm OC=1470-0B-525.265 mm Now Le = OCTOB = 525.265 225 +944-735 2 997.633 mm his o. O A=136m & M bressure frismon gate will look like po = pressure ato lg QC) (light) po density foil & Pw isdensity مگر = 881x9.81X0.525265 density = 881293 L asoc=525-2 f = 0.88|21000=881 kg/min 00-88 Keh PA of oily po = 4529,66 Nyom? ba= pressure at A - Log CAD )= 88179-813147 as AD- 1470 mm Resultant force due to sp=12704.6481m2 fluid pressure on gate per unit width = volume of pressure priem =10A) (POTPA) xl width Perkindiculas to klane of paper 3+x(1.3.6) (4539,66712704,64)X1

FR = 11926-124 N perunit with but it is quere gate given so width = 1.36m fe=11726.124 x 136 The = 15947.529N = 15.948 KN) Answer . (C) e, the direction offer Now since Fris acting perpendicular to gete then by geometry due to O Pascal's lave Fr. -90-44° = 46-0. VA FR do * = 46° ffris atting at O= 46° with S. respect to horigenten surface Answer le is distance of centroid of pressure prism from 0 I just see below u fe so from formula for centroid R so from of trepezium le= (pot2ph) 1.36 Spetta) 6-(4539.6C+ 2x12904.64) x 1-36 -0.787324m 74539.11+12704,64) 146-787.324mm distance of resultant promo."

+ Ch. O(= 525265mm distance of resultant I Losinta t ok (p=787.20 al from Surface;d =Octl Sin 44 = 525-265+ 787.324 Sin44 =1072.186mm (e) tension trequred AO'=BO =1360 font 4 =944-735mm : ke- Tola < F=15-948KN Now since Ois hinge in gate is hinged at o " A so for equilibrium net moment about o=0 et EMO =O FR XOP-TX (AO")= 15-948KN X (787.324)— TX_44.735) op=287-324 mm so > F= 13.291 KN Answer