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Ruben asked in Science & MathematicsPhysics · 2 months ago

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

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  • 2 months 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

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  • 2 months 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

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