User:VfibAfib/sandbox

overall force for a cube of pure weighing $$10kg

$$ falling at a rate of $$9.8m/s^2

$$ through air

Force of the cube

apply newton's second law of motion

$$F = ma

$$

where $$F

$$ is the force, $$m

$$ being the mass of the object, and $$a

$$ the acceleration of the object

plug in the values

$$F_{c} = 10\times9.8

$$

$$F_{c}=98N

$$

However this does not account for air resistance, we will need to take the drag formula to compute the air resistance.

$${\displaystyle F_{D}\,=\,{\tfrac {1}{2}}\,\rho \,v^{2}\,C_{D}\,A} $$

Values

$$F_{D} $$ being the drag force

$$\rho $$ being the density of the medium that the object is travelling through

$$v^{2} $$ being the velocity of the object

$$C_{D} $$ being the drag coefficient

and $$A $$ being the cross section of the object

Lets start off by calculating the value for $$A $$.

This may be tricky at first as we do not know the dimensions of the cube, however we can calculate its volume first.

The volume of the cube will be equal to the product of the division $$\nu = \frac{m}{\rho}

$$, the density of is $$7.874 g/cm^3

$$, and we will convert $$10kg

$$ to $$g

$$

$$10kg\times1000

$$ = $$10000 g

$$

all we need to do now is plug the values:

$$\nu = \frac{m}{\rho}

$$

$$\nu = \frac{10000g}{7.874g/cm^3}

$$

$$\nu = \approx1270cm^3

$$

Now, to find the $$A $$, we will need to find a side of the cube.

Provided that we have the volume and we know that; in order to find the volume of the cube it will be $$a^3

$$, then we will know that the inverse can be applied with $$\sqrt[3]{\nu}

$$, so if we were to plug the values it would be:

$$a

$$ = $$\sqrt[3]{\nu}

$$

$$a

$$ = $$\sqrt[3]{1270}

$$

$$a

$$ = $$10.83cm

$$

however now we need to raise $$a

$$ to $$a^2 $$

$$A $$ = $$10.83^2

$$

$$A $$ = $$117.28cm^2

$$

Now that we have the cross section of the cube, we are able to compute the rest of the values

Assuming that $$\rho $$ is air, then $$\rho $$ = $$0.001204g/cm^3 $$

Lets assume that after 1 second, $$v = 9.8m/s $$, then $$v^{2} $$ = $$96.04 $$

$$C_{D} $$ will be equal to $$1.05 $$ as it has been calculated already

Now that we have acquired all of our values, we can calculate the drag.

$${\displaystyle F_{D}\,=\,{\tfrac {1}{2}}\,\rho \,v^{2}\,C_{D}\,A} $$

Now, we plug our values

$$F_{D}= \frac{1}{2}0.001204\times 96.04 \times 1.05\times 117.28 $$

$$F_{D}= 0.000602\times 96.04 \times 1.05\times 117.28 $$

$$F_{D}= 0.05781608 \times 1.05\times 117.28 $$

$$F_{D}= 0.060706884 \times 117.28 $$

$$F_{D} = $$ $$\approx7.119N

$$

The rest is relatively easy, we can calculate $$F_{o\lambda} $$ by subtracting $$F_{D} $$ from $$F_{c}

$$.

$$F_{o\lambda}= F_{c}-F_{D}

$$

plug the values

$$F_{o\lambda}= 98-7.118

$$

$$F_{o\lambda}=90.8N

$$