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Problem 2:

A long cylindrical rod is formed from wood (specific gravity = 0.7) and aluminum (specific gravity = 2.7) cylindrical sections. The lengths of the sections are 20.0 cm and 1.0 cm respectively. The rod is allowed to float in Lake Matoaka. How much of the cylinder's length is underwater? (See diagram).

Solution:

Archimedes' principle: the bouyant force (F_b) on the cylinder must balance the weight of the cylinder for it to float.

The weight of the cylinder is

    W = Mg  =  (V_woodwood + V_AlAl)g

where V_wood is the volume of the wood section and _wood is the density of wood, etc.

Call the cross-sectional area of the cylinder A. Then we have

   W =  [A(0.20 m)_wood + A(0.01 m)_Al]g   =   [A(0.20 m)(0.7_water) + A(0.01 m)(2.7_water)]g

where we have used the definition of specific gravity (=density relative to that of water).

The bouyant force is the weight of the displaced H₂O, so it is just

     F_b = (A L)_waterg

where L is the length of the cylinder that is under water. Equating W and F_b we get

(A L)_waterg  =  [A(0.20 m)(0.7_water) + A(0.01 m)(2.7_water)]g

or, dividing out the factors of A, g, and _water,

    L = [(0.20 m)(0.7) + (0.01 m)(2.7)]  =  0.167 m  =  16.7 cm

Problem 3
Test 3
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