1. Problem 5-c
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 d
1.8 ft B
30o

2. Solving Problems on Your Own
Problem 5-c
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 d
1.8 ft B
30o
Assuming the submerged body has a width b, the load per unit length is w = brgh, 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.

3. Solving Problems on Your Own
Problem 5-c
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 d
1.8 ft B
30o
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.

4. Problem 5-c Solution
Determine the pressure distribution
acting perpendicular the surface of the submerged body.
1.7 ft A PA
PA = 1.7 rg
PB = ( cos 30o)rg
(1.8 ft) cos 30o B PB

5. P1 = (1.8 ft)2(62.4 lb/ft3)(1.7 ft) = 171.85 lbAy
Problem 5-c Solution A Ax
Replace the pressure distribution with a resultant force, and construct the free-body diagram.
1.7 rg
(1.8 ft) cos 30o
LAB/3 P1 FB
LAB/3
The force of the water on the gate is P2 B
LAB/3 1 2 1 2
( cos 30o)rg
P = Ap = A(rgh) 1 2
P1 = (1.8 ft)2(62.4 lb/ft3)(1.7 ft) = lb 1 2
P2 = (1.8 ft)2(62.4 lb/ft3)( cos 30o)ft = lb

6. S MA = 0: ( LAB)P1 + ( LAB)P2 - LABFB = 0Ay
Problem 5-c Solution A Ax
Write the equations of
static equilibrium for the
problem, and solve them.
1.7 rg
(1.8 ft) cos 30o
LAB/3 P1 FB
S MA = 0:
LAB/3 + P2 B 1 3 2 3
( LAB)P1 + ( LAB)P2
LAB/3
( cos 30o)rg
- LABFB = 0
P1 = lb
P2 = lb 1 3 2 3
( lb) ( lb) - FB = 0
FB = lb
30o
FB = 277 lb