Question

5000 kg A rectangular gate 5 m by 2 m is hinged at its based and inclined at 60° to the horizontal as shown in below figure. To keep the gate in a stable position, a counter weight of 5000 kgf is attached at the upper end of the gate as shown in the figure. (Note = Neglect the weight of the gate) Hence 1 kgf = 9.81 N a) Draw the free body diagram for the gate b) Draw the hydrostatic pressure distribution on the gate c) Determine the magnitude of the resultant hydrostatic force acting on the surface d) Determine the location of the resultant hydrostatic force acting on the surface e) Find the depth of water at which the gate begins to fall A609

Answer 1

folurim-; a free body diagram . sooo by FEW . b) San. 360° 0 Resultant hydrostatic ferre = f. F. 84 нь . Here, P = icro kg/m3 g = 9.81, A=2xh = 4h. - sinco T ķ :: Fa 1000*9*81x com 19620 mm. Li do let the resultant force 'f' acting at ht from the scrsale; but ht= ñ + In sin²o. AT area h = h ; Ju= moment of Inersia of immersed Ius einem here, 6= 2m, o ne za

i ha = ņ + unx Sin?60 mt þ-4-zm. downwards from 'D'. I evertically 5000kg fi e ) turing moment about a and equating it tozeso. we geti Sinco 5000 xg.81x AB - Fx 5000x5-81x AB = Fx Chat 49050 * 5 = 1962013 245250 = bagyon 245250 xha. 33 73= 56.25 > /h= 3. 8m)