Question

A 2-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

Answer 1

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.