Question

length L = 3 m, width b = 4 The gate creates an angle on 30° horizontal. It is assumed that the gate can rotate as a stiff system without friction around A. Movable mass on the gate M = 5000 kg in the distance x from B Gate mass m = 6000 kg Density of water = 997 kg/m^3 Gravitational acceleration = 9,81 m/s^2 L = 3 m 0 = 30° Water M a) Find the distance x so that the gate is just about to open b) Find the vertical, horizontal reaction forces and the resulting force in A

Answer 1

Ay X = 1500 Mg mo FH F Los300 o Fu = Pg An ū = Lsin30° 2 4 ht. TV Ax ht Just A By open (Laos condition A = bx L singu 6 nom? 3 & Fv = weight of water undere gate Fu = pgv 99 7 (9-81)( L Sing0 x 260438°) 6 Fr= 76231.99 N 4 FH = 997) 09.81) [ 6)sing oo) Fr= 44012.56 N ñ + IGG in An at Lsins + (b)(2) ein 253 (6) (Lsina in = 1.25 = 2 2 12 ta

ΣΜΑ (FV)( .40336) +(5^) (ht) – mg £60431 - Mg (3-4) Cos 30º = 0 x = 1.95 m b & Fyzo (9+) Ay + Fv - Mg - mg 76231-99 = Ay (11000) 9.81 31678.01N) Efx (to | Ач also FH - Ax =0 | Ax 440 12.56 N Resultant reaction force at A, A = A +Ay 54227.31 IA = 54.227 KN