According to Archimedes\' principle, the mass of a floating object equals the ma

Question

According to Archimedes' principle, the mass of a floating object equals the mass of the fluid displaced by the object. Use this principle to solve the following problems. (a) A wooden cylinder 30.0 cm high floats vertically in a tub of water (density = 1.00 g/cm^3). The top of the cylinder is 13.5 cm above the surface of the liquid. What is the density of the wood? (b) The same cylinder floats vertically in a liquid of unknown density. The top of the cylinder is 18.9 cm above the surface of the liquid. What is the liquid density? (c) Explain why knowing the length and width of the wooden objects is unnecessary in solving Parts (a) and (b).

Explanation / Answer

a)

First, do a balance according to Archimedes

Fbuoyant = Fweight

mass of object* g = mass of fluid * g

mass of object = mass of fluid displaced

Since mass = Density * Volume

D1*V1*g = D2*V2*g

cancel g

D1*V1 = D2*V2

D1 = density of wood, V1 = volume of wood (submerged)

D2 = density of water, V2 = volume of water displaced

h = 30 cm = 0.3 m

Dwater = 1 g/cm3 = 1000 kg/m3

Volume of water = submerged volume of wood

Vwater = hsubmerged*Acyl

Acylinder = 3.14*(R^2)

Vwater = hsubmerged*3.14*(R^2)

Vwood submerged = hsubmerged*3.14*(R^2)

then, substitute

D1*V1 = D2*V2

D1* htotal*3.14*(R^2) = D2*hsubmerged*3.14*(R^2)

note that we actually don't need the "AREA" of the cylinder, since they are both the same

D1*(htotal)= D2*hsubmerged

D1*(30cm) = (1 g/cm3)(30-13.5 cm)

D1 = 1*(30-13.5)/30 g/cm3

D1 = 0.55 g/cm3

this makes sense, since we expect wood to have a lower density than water (so it floats!)

b)

Apply Archemides principle

Fw = Fb

Fw = mass of object * g = D1*Vobj * g

Fb = mass of liquid *g = D2*Vsubm * g

then

D1*Vobj * g = D2*Vsubm * g

g cancels

D1*Vobj = D2*Vsubm

D1 = 0.55 g/cm3 (density of wood)

Vobj = hobj*Aobj

hobj = 30 cm

D2 = not known

Vsubm = hsub*Aobj

hsub = (30-18.9) = 11.1 cm

substitute in

D1*Vobj = D2*Vsubm

(0.55 g/cm3) * (30cm)(Aobj) = D2*(11.1cm)(Aobj)

Area of object cancels

(0.55 g/cm3) * (30cm) = D2*(11.1cm)

D2 = 0.55*(30/11.1) = 1.48648 g/cm3

the density of the liquid is 1.48648 g/cm3

this makes sense since we expected a heavier body, in order to "push" the cylinder higher

c)

we do not need those measurements since we can cancel them each other... We are using volumes so:

Volume of Submerged Object = Volume displaced of liquid

Volume of Submerged Object = hsubm*Aobj

then

Volume of Submerged Object = Volume displaced of liquid = hsubm*Aobj

Volume of total object = htotal*Aobj

The only thing that is needed is the hsubm = submerged height and the total height

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