Question 2.6: Repeat Example 2.5 to sketch the pressure distribution on pl...

Part (a)

Point A is 9 ft deep, hence p_A = \gamma h_A = (64  lbf/ft^3)(9  ft) = 576  lbf/ft^2. Similarly, Point B is 15 ft deep, hence p_B = \gamma h_B = (64  lbf/ft^3)(15  ft) = 960  lbf/ft^2. This defines the linear pressure distribution in Fig. E2.6. The rectangle is 576 by 10 ft by 5 ft into the paper. The triangle is (960 – 576) = 384 lbf/ft^2 × 10 ft by 5 ft. The centroid of the rectangle is 5 ft down the plate from A. The centroid of the triangle is 6.67 ft down from A. The total force is the rectangle force plus the triangle force:

Part (b)

The moments of these forces about point A are

∑M_A = (28,800 lbf)(5 ft) + (9600 lbf)(6.67 ft) = 144,000 + 64,000 = 208,000 ft·lbf

Then         5  ft + l = \frac{M_A}{F} = \frac{208,000  ft \cdot lbf}{38,400  lbf} = 5.417  ft  hence  l = 0.417  ft

Comment: We obtain the same force and center of pressure as in Example 2.5 but with more understanding. However, this approach is awkward and laborious if the plate is not a rectangle. It would be difficult to solve Example 2.7 with the pressure distribution alone because the plate is a triangle. Thus moments of inertia can be a useful simplification.