1. Problem 5.158 The square gate AB is held in the position shown by hinges along its top edge A and by a shear pin at B. For a depth of water d = 3.5 ft, determine the force exerted on the gate by the shear pin. A B d 1.8 ft 30 o

2. Solving Problems on Your Own The square gate AB is held in the position shown by hinges along its top edge A and by a shear pin at B. For a depth of water d = 3.5 ft, determine the force exerted on the gate by the shear pin. A B d 1.8 ft 30 o Assuming the submerged body has a width b, the load per unit length is w = b  gh, where h is the distance below the surface of the fluid. 1. First, determine the pressure distribution acting perpendicular the surface of the submerged body. The pressure distribution will be either triangular or trapezoidal. Problem 5.158

3. Solving Problems on Your Own The square gate AB is held in the position shown by hinges along its top edge A and by a shear pin at B. For a depth of water d = 3.5 ft, determine the force exerted on the gate by the shear pin. A B d 1.8 ft 30 o 2. Replace the pressure distribution with a resultant force, and construct the free-body diagram. 3. Write the equations of static equilibrium for the problem, and solve them. Problem 5.158

4. Problem 5.158 Solution A B Determine the pressure distribution acting perpendicular the surface of the submerged body. 1.7 ft (1.8 ft) cos 30 o PAPA PBPB P A = 1.7  g P B = (1.7 + 1.8 cos 30 o )  g

5. Problem 5.158 Solution A B (1.8 ft) cos 30 o Replace the pressure distribution with a resultant force, and construct the free-body diagram. AyAy AxAx FBFB 1.7  g (1.7 + 1.8 cos 30 o )  g L AB /3 P1P1 P2P2 The force of the water on the gate is P = Ap = A(  gh) 1212 1212 P 1 = (1.8 ft) 2 (62.4 lb/ft 3 )(1.7 ft) = 171.85 lb 1212 P 2 = (1.8 ft) 2 (62.4 lb/ft 3 )(1.7 + 1.8 cos 30 o )ft = 329.43 lb 1212

6. Problem 5.158 Solution A B (1.8 ft) cos 30 o AyAy AxAx FBFB 1.7  g (1.7 + 1.8 cos 30 o )  g L AB /3 P1P1 P2P2 P 1 = 171.85 lbP 2 = 329.43 lb Write the equations of static equilibrium for the problem, and solve them.  M A = 0: ( L AB )P 1 + ( L AB )P 2 1313 2323 - L AB F B = 0 + (171.85 lb) + (329.43 lb) - F B = 0 1313 2323 F B = 276.90 lb 30 o F B = 277 lb