Respuesta :
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 [tex]\frac{4}{3}[/tex] π r³ - g (ρ₁ [tex]\frac{3}{4}[/tex] [tex]\frac{4}{3}[/tex] π r³ + ρ₂ [tex]\frac{1}{4}[/tex] [tex]\frac{4}{3}[/tex] π r³)
T = [tex]\frac{4}{3}[/tex] ρ g r³ - g π r³ (ρ₁ + [tex]\frac{1}{3}[/tex] ρ₂)
The term of the floating force is
B_total = g π r³ (ρ₁ + [tex]\frac{1}{3}[/tex] 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 = [tex]\frac{4}{3}[/tex] 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.
