# Answer to Question #71302 in Other Physics for Courtney

Question #71302
A rock makes a dynamometer show 1.04N, when it’s lowered into a cup of water the dynamometer shows 0.68N. What is the rock’s density?
1
2017-11-25T09:15:07-0500
Given:
W_(in air)=1.04 N,
W_(in water)=0.68 N,
ρ=?

Buoyancy is the force acting opposite the direction of gravity that affects all objects submerged in a fluid.
Buoyancy = weight of displaced fluid.
The weight of the displaced fluid is directly proportional to the volume of the displaced fluid.
Archimedes’ Principle: The buoyant force on a submerged body equals the weight of the fluid it displaces:
BF =ρ_water gV_rock
The buoyant force is also equal to the weight of the object in the air minus the weight of the object in the fluid:
BF =W_(in air)-W_(in water)
The volume of rock is
V_rock=m/ρ=mg/ρg=W_(in air)/ρg
Equating the two expressions for the buoyant force we can solve for the density of rock:
ρ=(W_(in air) ρ_water)/(W_(in air)-W_(in water) )

Density of water is 1 g/cm3 or 1000 kg/m3.
Thus,
ρ=(1.04×1000)/(1.04-0.68)=2888.8 kg/m^3≈2890 kg/m^3≈2.89 g/cm^3

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