Kaplan Physics Discrete #18

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caduc3us

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Question: A cube of wood whose sides are each 10 cm weighs 16 N in air. When half submerged in an unknown liquid it weighs only 10 N. What is the density of the liquid? (Note: Assume g = 10 m/s^2.)

The answer is 1200 kg/m^3 which can be derived using the Fbuoy = pfluid vsub g where pfluid is the density of the fluid, vsub is the volume of the object that's submerged and g is gravity. Rearranging pfluid = Fbuoy/(vsub g) = 6N /( 5x10^-4 m^3 * 10 m/s^2) = 1200 kg/m^3. I have no problems solving the question using this method.

I've tried also using the equation vsub/v = pobject/pfluid where v is the volume of the object. Rearranged, pfluid = pobject v/vsub = pobject (2/1).

The mass of the wood is:
F=mg
m =16N/(10m/s^2) =1.6 kg

The volume of the wood is:
1x10cm^3 = 0.001m^3 =1x10^-3 m^3

The density of the wood (pobject)
1.6 kg / 1x10^-3 m^3 = 1600 kg/m^3

Plugging into pfluid = pobject * 2 = 3200 kg/m^3 which is way different then the other answer. Any help?
 
Last edited:
vsub/v = pobject/pfluid is valid only if the object is floating, i.e. its weight is measured to be zero when submerged. You can easily rearrange it to Vsub * ρfluid = V * ρobject, which are the buoyancy force and the weight of the object. In your case the submerged weight is 10 N (and the object will sink, left without any support).
 
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