Answer:
a) y ’= 0.068 m
, b) m_lead = 2,599 kg
Explanation:
a) To solve this exercise, let's use Archimedes' law, which establishes that the pressure of the water is equal to the weight of the dislodged liquid.
Let's write the forces in the equilibrium condition
B - W = 0
B = W
Thrust is
B = ρ_liquid g V_water
Body weight is
W = m g
the definition of density
ρ_body = m / V
we substitute
W = ρ_body g V_body
The volume of the body is
V_body = L³
The volume of water is
V_agua = y L²
we substitute
ρ_liquid g and L2 = rho_body g L3
ρ_liquid y = ρ_body L
y = L ρ_body /ρ_liquid
let's calculate
y = 651/100 0.195
y = 0.1269 m
this is the distance from the bottom of the cube, the distance from the top of the cube
y ’= 0.195 -y
y '= 0.195 - 0.1269
y ’= 0.068 m
b) In this case the equilibrium equation is
B - W_body - W_lead = 0
B = W_body + W_lead
ρ_liquido g V_liquido = ρ_body g V_body + m_lead g = 0
ρ_liquid L3 = ρ_body L3 + m_lead = 0
m_lead = L3 (ρ_liquid - ρ_body)
let's calculate
m_lead = 0.195 3 (1000 - 651)
m_lead = 2,599 kg
