Solution to Second Quiz Submerged Objects and Buoyancy

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Quiz two solutions ME 390, L. S. Caretto, Spring 2007 Page 2

Solution to Second Quiz – Submerged Objects and Buoyancy

A long vertical wall separates seawater from fresh water. If the seawater has a depth of 7 ft, what depth of fresh water is required so that the net force on the wall is zero? What is the moment at the bottom of the wall, per unit length of wall (if any), when the net force is zero?

As shown in the diagram to the right, the net force on the wall will be zero when the resultant force from the sea water, FRs, equals the resultant force for the fresh water, FRf. For both sides the resultant force is given by the product rgAycsinq. The location of the resultant force from the top of the liquid is given,for each side, by the equation yR = Ixc/(ycA) + yc.

For both fluids, q = 900 so sinq = 1. The area to which the force is applied has an indefinite length, L, into the page, so the area is (7 ft)L on the seawater side and hL on the fresh water side. The centroid distance, yc, is (7 ft)/2 = 3.5 ft on the seawater side and h/2 on the fresh water side. The specific weight of seawater is the product of its specific gravity and the specific weight of water: gs = SGsgw = (1.03)(62.4 lbf/ft3) = 64.3 lbf/ft3.

The force on the sea water side and the fresh water sides are computed as follows.

We can equate these two forces and solve for h.

h = 7.11 ft

The clockwise moment at the bottom of the wall is given by the sum, C = FRsyms – FRfymf, where we are using the magnitude of the forces and the moment arms are ymf = h – yRf and yms = 7 ft – yRs, where yrs and yrf are the distances from the surface of each liquid to the center of pressure for the resulting force. The equation for yR requires the moment of inertia about the centroid, Ixc. For both sides, Ixc = a3L/12 where a is the depth of the water. The values of yR for each fluid are:

This gives the moment arms for seawater and fresh water as yms= 7 ft – 4.67 ft =2.33 ft and ymf = 7.11 ft – 4.74 ft = 2.37 ft. The clockwise moment is then given by the equation C = FRsyms – FRfymf.

Inserting the value of h = 7.11 ft and dividing by L gives the moment per unit length of wall.

The negative sign for clockwise momentum indicates that it is in the counterclockwise direction which, from the diagram at the start of the solution, is the direction of the seawater. Although both the seawater and the fresh water exert the same force, the higher location of the resultant force in the water gives the net counterclockwise momentum.