physics problem help?
A completely submerged chunk of metal sinks in water with an
acceleration equal to 1/5 of g ( where g=9.8 m/s2 ). Find the
specific gravity of this metal. Ignore all drag effects in the fluid.
2 Answers
- billrussell42Lv 71 year ago
The weight in air is w₀ = mg
the weight in water is w₁ = mg/5
buoyancy force is the difference = mg – mg/5 = (4/5)mg
Specific gravity is the ratio of the density of the material to the density of water
the buoyancy force is the weight of the displaced water = m₂g = V₂ρg where ρ is the density of water, V₂ is the volume of the displaced water.
buoyancy force = (4/5)mg = V₂ρg
(4/5)m = V₂ρ
specific gravity = (m/V) / ρ
chunk's volume is the same as the volume of the displaced water
V = V₂
specific gravity = (m/V) / ρ
(4/5)m = Vρ
V = (4/5)m/ρ = 4m/5p
specific gravity = (m/(4m/5p)) / ρ
specific gravity = ((m/ρ) / (4m/5p))
specific gravity = (m/ρ) (5p/4m)
specific gravity = 5/4
density of fresh water at 20C = 0.998 g/cm³ = 0.998 kg/L
= 998 kg/m³ = 8.33 lb/gal = 62.1 lb/ft³
- Steve4PhysicsLv 71 year ago
You can shorten the working but here it is in detail:
Mass of metal = m. Mass of water displaced = M.
Volume of metal = volume of water displaced = V.
Density of water = ρ
Resultant force on metal F = weight – upthrust:
F = mg – U
Using 'F = ma' with a = g/5 gives:
mg – U = mg/5
U = (4/5)mg
Upthrust = weight of water displaced. So weight of water displaced is (4/5)mg:
Mg = (4/5)mg
M= (4/5)m
Since volume = mass/density, volume of water displaced is:
V = M/ρ
. .= (4/5)m/ρ (equation 1)
Call metal's specific gravity ‘s’, then its density = ρs. Since volume = mass/density:
V = m/(ρs) (equation 2)
From equations 1 and 2:
(4/5)m/ρ = m/(ρs)
4/5 = 1/s
s = 5/4
. .= 1.25