A2-ft-thick block constructed of wood (sg = 0.6) is submerged in oil (sg = 0.8), and has a 2-ft-thick aluminum (specific weight = 168 lb/ft3) plate attached to the bottom as indicated in fig. p2.146. (a) determine completely the force required to hold the block in the position shown. (b) loca

Refer to the figure shown below.

Assume that
g = 32 ft/s², acceleration due to gravity
ρ = 62.4 lb/ft³, density of water
A = 1 ft², the surface area of each of the blocks.

Calculate the weight of the wooden block (sg = 0.6).
W₁ = (0.6*62.4 lb/ft³)*(2 ft³) = 74.88 lb
Calculare the weight of the aluminum block (sg = 0.8)
W₂ = (0.8*62.4)*(2) = 99.84 lb

Use the Archimedes Principle to calculate buoyant forces.
For the wooden block,
B₁ = (62.4 lb/ft²)*(2 ft³) = 124.8 lb
Also, for the aluminum block,
B₂ = 124.8 lb

Let F = the force required to hold the blocks in position. Then
-F + B₁ + B₂ = W₁ + W₂
-F + 249.6 = 174.72
F = 74.88 lb

The answer for Part (a):   
74.88 lb, acting downward.

Forces W₁, B₁ act at the center of mass of the wooden block, and W₂, B₂ act at the center of mass of the aluminum block.
Let x =  the location of the overall force on the two blocks, measured downward from the top of the wooden block, as shown in the figure.
Then
(W₁ - B₁)*(1.0) + (W₂ - B₂)*(3.0) = (W₁+W₂-B₁-B₂)*x
-49.92 + (-24.96)*(3) = -74.88x
-124.8 = -74.88x
x = 1.668 ft

The answer for part (b): 
The resultant force should be applied at 1.67 ft below the top of the wooden block.