Question

MAEG 3030: Fluid Mechanics HW #1 (Due on 7/2/2020, Friday) 1. Consider the cylindrical weir of diameter 3m and length 6m. If the fluid on the left has a specific gravity of 1.6, and on the right has a specific gravity of 0.8. Find the magnitude and direction of the resultant force applied to the weir. (Note: density of water, 1000kg/m3) 1.5 m

Answer 1

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SOLUTION:

Froe body diagram 3 m 1.5m Given! length, l=6m Diameter, 38= d = 3m Specibic gravity v ob lebt side, Viett16 Specibic gravity of Right Side, rich 0-8 (i) Forces on left side of Cylinder Fax, force on Abc = S912 -> 0 Centre of gravity, ha -15m Cross sectional area in y-direction A=6x3= 18 ma from o 28 Fari= Px 9 x Area x width of gate - sex p. x 9ч буй = 1-6x1000x 9.81x7.5x18 I E = 493 799 N7 -sox SHO

5,- force on ABCOA -- P9AW :=Scx S1120 width, w=6m Avea, A= . (-5) from equation @ f = SGxf, px x Axw = 1-6x1000x9-81 *(15) X 6 Ify - 332845-4585 N citii forces on Right side ob the cylinder f = fxgxA_X], - - force on area #roco Area Az= L xd = 1.5x8=9 m2 centre of gravity, , - 15.0.75m from equation ® Foz = SG x P x 9x4₂x h -0.8X1000X9.21x9x 0.75 Face = 529746 Fy = fx 9x A₂ x W 9 Area, 4,5 ayat 7 (0,05), width w=6m. from ean's fig = S. Gox S.,.xgx A, xw =0.8x1000X9.81% nx(1.519 X 6 4 Fox = 83all_364 W]

Resultant in x-direction Faca F- For, = 423792-52974 = 370818 N Resultant in y direction fy = fy, t fyz = 332845.4585783211-364 Fy =4160 56.8225 N Resultant, R=Sg24 Fig? = (3708183+(416056-8225) 2 TR = 557323-3071N7 Direction, tano Az tanland etan' 416056-82251 370818 10 =0.84°