Question

A solid sphere made of plastic density of 1350 kg/m3 has a radius of 35.0 cm. It is suspended by a massless cord. 75% of its volume is in water and 25% of it is in oil. The density of water is 1000 kg/m3 and the density of the oil is 850 kg/m3. What is the total buoyant foce on the sphere

Answer 1

Answer:

B_total = 1694 N ,  T = - 937.7 N

Explanation:

This is an exercise of the Archimedes principle that establishes that the thrust of a liquid is equal to the weight of the dislodged volume

          B = ρ g V_body

For this case we will assume that the weight of the body is in equilibrium with the thrust of the liquids and the tension of the rope

          B₁ + B₂ - W + T = 0

         

suppose liquid 1 is water and liquid 2 is oil

          ρ₁ g V₁ + ρ₂ g V₂ = W

for body weight let's use the definition of density

         ρ = m / V

         m = ρ V

         

we substitute

         T = ρ g V - (ρ₁ g V₁ + ρ₂ g V₂)

In the problem we are told that the volume in the water is 75% and the volume in the oil is 25% of the body's volume.

         T = ρ g V - g (ρ₁ 0.75 V + ρ₂ 0.25 V)

         

let's calculate the volumes

Body

         V = 4/3 π r³

liquid 1 (water)

         V₁ = 0.75 V = ¾ V

liquid 2 (oil)

         V₂ = 0.25 V = ¼ V

we substitute

        T = ρ g $\frac{4}{3}$ π r³ - g (ρ₁ $\frac{3}{4}$  $\frac{4}{3}$ π r³ + ρ₂  $\frac{1}{4}$ $\frac{4}{3}$ π r³)

        T = $\frac{4}{3}$ ρ g r³ - g π r³ (ρ₁ + $\frac{1}{3}$ ρ₂)

The term of the floating force is

        B_total = g π r³ (ρ₁ + $\frac{1}{3}$ ρ₂)

let's calculate its value

        B_total = 9.8 π 0.35³ (1000 + ⅓ 850)

        B_total = 1694 N

therefore the tension in the string is

         T = $\frac{4}{3}$ 1350 9.8 0.35³ - 1694

         T = 756.3 - 1694

         T = - 937.7 N

The negative sign indicates that the system tends to come out of the liquid, for which a downward force must be applied to keep it in position.