An object floats in liquid of density 1.2×10kgm/3 with one quarter of its volume above the liquid surface. determine the density of the object

900 kg/m³

Explanation:

There are two forces on the object: weight and buoyancy.

Newton's second law:

∑F = ma

B − W = 0

B = W

Weight is mass times gravity:

W = m₀ g

Mass is density times volume:

W = ρ₀ V₀ g

Buoyancy force is equal to the weight of displaced fluid:

B = ρ V g

Therefore:

ρ V g = ρ₀ V₀ g

ρ V = ρ₀ V₀

ρ₀ = ρ V / V₀

One quarter of the object's volume is above the liquid, which means three quarters of its volume is below the surface.

Given that V = ¾ V₀ and ρ = 1200 kg/m³:

ρ₀ = 1200 (3/4)

ρ₀ = 900 kg/m³

find the sum of the two horizontal vectors and the sum of the two vertical vectors

isoceles right angled triangle

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