Question

1. ***present all the calculations***

Part I: The density of seawater is approximately 1.027g / cm3 and the ice density 0.93g / cm3. United Nations
Iceberg (iceberg), generally has a mass of 150,000 metric tons (1 metric ton = 1000kg). Calculation
(a) The buoyant force exerted by the water on the "iceberg",
(b) the volume of the iceberg above sea level,
(c) the volume fraction above sea level.

Part II:

A metal bucket with a height of 0.700 m and a mass of 650 kg is suspended from a rope in an open tank containing a liquid with a density equal to 1530 kg / m3. The block is submerged 0.300 m below the surface. Calculate:
(a) The force up at the bottom of the block,
(b) the tension force of the rope,
(c) the magnitude of the buoyant force in the cube using the Archimedes principle.

Answer 1

Part I)

ρ_water=1027kg/m³, ρ_ice=930kg/m³

a)

V_total = m_ice/ρ_ice =(150,000,000kg)/(930kg/m³)=161290.3 m³

Fo find buoyant force use equation,

F_b = F_g

F_b = ρ_iceV_totalg

Plug values,

F_b = (930kg/m³)*(161290.3m³)*(9.8m/s²)

F_b = 1.47*10^9 N

b)

Use equation,

F_b = F_g

ρ_waterV_belowg= ρ_iceV_totalg

ρ_waterV_below= ρ_iceV_total

V_below=V_total(ρ_ice/ρ_water) -------------(1)

Plug values,

V_below= (161290.3 m³)*(930kg/m³/1027kg/m³)

V_below = 146056.45 m³

V_above = V_total - V_below = 161290.3 m³ - 146056.45 m³ = 15233.85 m³

c)

From eqn (1),

V_below/V_total=ρ_ice/ρ_water

V_below/V_total=(930kg/m³/1027kg/m³)

V_below/V_total= 0.9055

V_above/V_total= 1- V_below/V_total

V_above/V_total = 1- 0.9055 = 0.0945

In percentage it is 9.45%