Respuesta :
Answer:
F = -4/5 W
Explanation:
We will do this exercise in parts, let's start with the data, The ball must be in balance
B -W = 0
B = mg (1)
where the thrust is given by Archimedes' principle
B = ρ g V_liquid
In the problem we are told that the volume of the submerged body is 1/5 of the volume of the body, let's use the subscript 2 for the ball and the subscript 1 for the liquid
V₁ = [tex]\frac{1}{5}[/tex] V₂
if we use the concept of density for the body
ρ₂ = m / V₂
m = ρ₂ V₂
We can see that mass and volume are directly proportional, so we can use a rule of three to find the submerged mass. If mass M is in volume V₂ what mass is there in volume V₂ /5
m₂ = [tex]\frac{V_{2} }{5} \frac{M}{V_{2} }[/tex]
m₂ = M / 5
we substitute in equation 1
B = M / 5 g
This is the thrust that is an upward vertical force, therefore to submerge the whole ball we must apply a downward vertical force equal to the rest of the weight, the equilibrium condition in this case is
B -W - F = 0
F = B -W
F = [tex]\frac{M}{5}[/tex] g - M g
F = -4/5 Mg
F = -4/5 W
The negative sign indicates that the force is vertical down
