A concrete column has a diameter of 380 mm and a length of 2.6 m . if the density (mass/volume) of concrete is 2.45 mg/m3, determine the weight of the column in pounds.

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The volume of the column is

(π) · (r²) · (length) =

(π) · (0.19 meter)² · (2.6 meters) =

(π) · (0.036 m²) · (2.6 m) =

0.294 m³ .

The density is 2,450 kg/m³ (VERY very dense, heavy concrete)

so the weight of the column is (mass)·(gravity) or

(density) · (volume) · (gravity) =

(2,450 kg/m³) · (0.294 m³) · (9.81 m/s²) =

(2,450 · 0.294 · 9.81) (kg · m³· m) / (m³ · s²) =

7,066 kg-m/s² = 7,066 Newtons .

But 9.81 Newtons = 2.20462 pounds on Earth (the weight of 1 kilogram of mass), so we have

(7,066 N) · (2.205 pound/9.81 N) =

(7,066 · 2.205 / 9.81) pounds =

1,588 pounds .

The mass is the product of the volume and dencity. The mass of the given concrete column is [tex]1.58733\times 10^{-6} \rm \ pounds[/tex].

The volume of the column:  

[tex]V = \pi r^2h[/tex]

Where,

[tex]r - \rm radius = 380 \ mm = 0.36 \ m[/tex]

[tex]h -[/tex]height = 2.6 m

Put the values in the equation,

[tex]V = \pi \times 0.19^2 \times 2.6\\\\V = 3.14 \times 0.0361\times 2.6\\\\V= 0.294\rm \ m^3[/tex]

The density of the column is 2.45 mg/m³

So, the mass of the column is

[tex]m = 0.294 \times 2.45 \\\\m = 0.72 \rm{ \ mg} \\\\m = 1.58733\times 10^{-6} \rm \ pounds[/tex]

Therefore, the mass of the concrete column is [tex]1.58733\times 10^{-6} \rm \ pounds[/tex].

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