a) Consider a brass cylinder that is suspended in air. Air is also a fluid, with density ρair = 1.29 × 10−3 g/cm3 and yet you seldom consider the buoyant force exerted by air when you deal with an object like this. To see why determine the ratio, FB/Fgrav, of the buoyant force exerted by the air on the cylinder to the gravitational force exerted on the cylinder, whose density can be assumed to be 8.5 × 103 kg/m3 . Is the buoyant force exerted by the air on the brass cylinder significant compared to the gravitational force exerted on the cylinder? Hint: The answer does not depend on the volume of the cylinder. Try to do some algebraic manipulations to eliminate the volume; if these fail, select a numerical value for the volume and do the relevant numerical calculations.
b) Consider two objects, each of the same volume and which are rigidly attached together. One has density exactly half that of water and the other exactly double that of water. The objects are initially held at rest beneath the surface of the water. Will they subsequently sink or rise and partially protrude above the surface? Explain your answer. Hint: Since the two objects are rigidly attached together, this can be assessed by considering all forces acting on each of the two objects. Try comparing the magnitudes of the total force pushing up to that of the total force pulling down.
rho_air = 1.29*10^-3 g/cm^3 = 1.29 kg/m^3
rho_brass = 8.5*10^3 kg/m^3
let V is the volume of cyllinder.
F_grav = m*g
FB = weight of the displaced air
FB/Fgrav = rho_air*V*g/(rho_brass*V*g )
= 1.52*10^-4 <<<<<<------------Answer
b) They will sink.
Let V is the volume of each each object.
let rho_water is the density of water.
density of first object, rho1 = rho_water/2
density of second object, rho2 = 2*rho_water
buoyant froce acting on both objects, FB = rho_water*(2*V)*g
gravitational force acting on both objects, Fgrav = m1*g + m2*g
= rho1*V*g + rho2*V*g
= (rho_water/2)*V*g + 2*rho_water*V*g
clealry, F_grav > FB
so, They will sink.
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