Question

x value =1

yvalue =1

z value =2

A cylinder with a radius of R = 0.5 m and length L = 2/5 is a part of

engine oil SAE 30 tank at the left and to support a water tank at right as

shown. The height of SAE 30 is h = 1/2 m. The water tank has an inclined

gate ((LAB = 6 1cm. B = 52°). which is hinged at point A. If both tank has

2/5 length into the paper, calcualute (i) the direction and the magnitude of

average hydrostatic force caused by the engine oil on the, (ii) the force at

welding line (reaction at point B) and (ii) the height of the water in the

water tank. Note that both tanks are open and the water tank is welded with

the eylinder along its length.

water

>X

H

h = 1/2

B

SAE 30

R 0.5 m

A

welding line

note : i need solve after 30 min

1.3 A cylinder with a radius of R = 0.5 m and length L = 2/5 is a part of engine oil SAE 30 tank at the left and to support a water tank at right as shown. The height of SAE 30 is h = x/2 m. The water tank has an inclined gate ((LAB = 6 y cm, ß = 52°). which is hinged at point A. If both tank has z/5 length into the paper, calcualute (i) the direction and the magnitude of average hydrostatic force caused by the engine oil on the, (ii) the force at welding line (reaction at point B) and (iii) the height of the water in the water tank. Note that both tanks are open and the water tank is welded with the cylinder along its length. y water H h = x/2 BI SAE 30 LG B R = 0.5 m A welding line

Answer 1

R=0.5m I= z x=1, y = 1 Z=2 2 m (for both tank) (1 27 / ny h H₂O - Ļm LAB=LG - byem Wetting H h = 2 line = 61 cm = 0.6lm SAE30 01 La 1₂ m IR B = 52=520 A FR kg/m3 Density of SAE 30 oil (6)= 875.4 Density of water (Pw) - 1000 kg/m3 FH = Fx aputter Horizontal force fra fх - С. 474 x RXL Anea 875.4x 9.81 x 14 x 54 2/3 429.384 N 0.58 Vertical force Fv = log xh x R x L HER) XL 858.767N Weight of the fund of the control volume = wa ng egy logx (R² -.W- 875.4 x 9.81x (0.52 _ II x0,92 ) x 2 / 3 180.79 N Fy = Furwa 6 77, 976 N

2 The resultant force, for I Fun in fig? 802.51 N tand Fy FA 677.976 429.384 1.5789 0 - 57.65 the on The magnitua an of the nitude of the hydrostatic force orting cylinder is 802.51N and its line of action passes through the center cylinder raking angle 57. 65o with the horizontal ii) o- qo+ß) (90-B) A o - 90 + ß 57. 65° - 909 + 52° - 19.65° . The reaction et point B = The force at welding line => FB - Fr. Los co-90+B) FB = 802.514 los 19. 65" 755.775 N iii) Ev = ew gx (H - LABSIMB) XL ABLOSBAL 1473.67 2x (H - 0.48), N 1/3 & LABCOS BY LARSinBx L x Pw* FH = Fn = Pw 9x (H-LLAB SinB) &LAB SRBAL A Watakatakan 18 86.214x (H-0·24), Гр FY Fv+W 1473.672 *( H-0.289) :.FR = JFY 2 + fon2 (H-0.27) x V 1473.642+ 1886.214 23.93.64 x (H-0.24) N W xg= 354.1870 •C 0.24 А 2

3 tanda fy Fn 1473.672 1886.214 To hotel FB - 0.78 13 x=38° mand U ma IB X+B= go K ad A Take moment about a to find a (there neglect FB because we are interested to find the line of action of the resultant force only) FRX R = FvX LARCOSR + W x x * LAR Cossit FH X 1 x LARSinß =) 2393.64 x (H-0.24) xa 1473.672+ (H-0.48) x 0.61 x 20552 2 + 354.187 X X 0.61 x Coss2 + 1886.214x + x 0.61X Sin52° = 276.72 (H-0.48) + 44.34 + 3020226 276.72 (H+ 0.7724 0.1156* (H40.1124 H - 0.24 (FR and FB only) Now, take the moment about AA EMA 2=0 = ) - FBX LAB = FR x x 755.775%0.61 - 2393.64 xCH-0724) x 0.1156 x 0. 1156 X (H + 0.7724) (4-0.24) -- 276.72 x (H+0.7724) H = 0.8936 m * The height of water in 461.023 =) the water tank