An aluminum weight of mass 200 g is attached by a short string to the bottom of

Question

An aluminum weight of mass 200 g is attached by a short string to the bottom of a boaat of density 0.3 g/cm3 which is supported by another string attached to a mass balance. When both the weight and the float are completely immersed in the water the balance reads 50 g. What is

a) the volume of the float

b) the mass of the float

c) the buoyant force of the water on the float

d) the buoyant force of the water on the aluminum weight

e) What would the balance read if only the weight were under water?

Explanation / Answer

V = M(aluminium)/density(aluminium) = 200/2.7 = 74cc

Hence the balancing equation:

DalValg+ DfloatVfloatg - DwaterVwaterg = 50g

Hence;

DalVal + DfloatVfloat - DwaterVwater = 50

200 + 0.3V - 74 - V =50

76 = 0.7 Vfloat

Vfloat = 76/0.7 = 108.5 cc

Mass = 0.3*108.5 =32.55g

Bouyant Force (float)= 1000*g*108.5*10^-6 = 1.0633N

Buoyant Force(aluminium) = 1000*g*74*10^-6 = 0.7252N

Net force =(1.96-0.7252)/9.8 + 32.55 =158gm approx

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