# Equilibrant/Examples/100kg at 150, 75kg at 60, 50kg at -45

## Example of Equilibrant

Three forces $\mathbf F_1, \mathbf F_2, \mathbf F_3$ act on a particle $B$ at a point $P$ embedded in the complex plane:

 $\displaystyle \mathbf F_1$ $=$ $\displaystyle \polar {100 \, \mathrm {kg}, 150 \degrees}$ $\displaystyle \mathbf F_2$ $=$ $\displaystyle \polar {75 \, \mathrm {kg}, 60 \degrees}$ $\displaystyle \mathbf F_3$ $=$ $\displaystyle \polar {50 \, \mathrm {kg}, -45 \degrees}$

The equilibrant $\mathbf E$ of $\mathbf F_1, \mathbf F_2, \mathbf F_3$ is:

$\mathbf E = \polar {80.8 \, \mathrm {kg}, -80.2 \degrees}$

## Proof 1

 $\displaystyle \mathbf E$ $=$ $\displaystyle -\paren {\mathbf F_1 + \mathbf F_2 + \mathbf F_3}$ Magnitude and Direction of Equilibrant $\displaystyle$ $=$ $\displaystyle -\paren {\polar {100 \, \mathrm {kg}, 150 \degrees} + \polar {75 \, \mathrm {kg}, 60 \degrees} + \polar {50 \, \mathrm {kg}, -45 \degrees} }$ $\displaystyle$ $=$ $\displaystyle -\paren {100 \, \mathrm {kg} \cis 150 \degrees + 75 \, \mathrm {kg} \cis 60 \degrees + 50 \, \mathrm {kg} \cis -45 \degrees}$ $\displaystyle$ $=$ $\displaystyle -\paren {\paren {100 \, \mathrm {kg} \cos 150 \degrees + 75 \, \mathrm {kg} \cos 60 \degrees + 50 \, \mathrm {kg} \cos -45 \degrees} + i \paren {100 \, \mathrm {kg} \sin 150 \degrees + 75 \, \mathrm {kg} \sin 60 \degrees + 50 \, \mathrm {kg} \sin -45 \degrees} }$ $\displaystyle$ $=$ $\displaystyle -\paren {\paren {100 \dfrac {\sqrt 3} 2 \, \mathrm {kg} + 75 \paren {\dfrac 1 2} \, \mathrm {kg} + 50 \dfrac {\sqrt 3} 2 \, \mathrm {kg} } + i \paren {100 \, \mathrm {kg} \sin 150 \degrees + 75 \, \mathrm {kg} \sin 60 \degrees + 50 \, \mathrm {kg} \sin -45 \degrees} }$ Cosine of $150 \degrees$, Cosine of $60 \degrees$, Cosine of $315 \degrees$ $\displaystyle$ $=$ $\displaystyle -\paren {\paren {100 \dfrac {-\sqrt 3} 2 \, \mathrm {kg} + 75 \paren {\dfrac 1 2} \, \mathrm {kg} + 50 \dfrac {\sqrt 2} 2 \, \mathrm {kg} } + i \paren {100 \paren {\dfrac 1 2} \, \mathrm {kg} + 75 \dfrac {\sqrt 3} 2 \, \mathrm {kg} + 50 \dfrac {-\sqrt 2} 2 \, \mathrm {kg} } }$ Sine of $150 \degrees$, Sine of $60 \degrees$, Sine of $315 \degrees$ $\displaystyle$ $=$ $\displaystyle -\paren {\paren {-50 \sqrt 3 \, \mathrm {kg} + 37.5 \, \mathrm {kg} + 25 \sqrt 2 \, \mathrm {kg} } + i \paren {50 \, \mathrm {kg} + 37.5 \sqrt 3 \, \mathrm {kg} - 25 \sqrt 2 \, \mathrm {kg} } }$ simplifying $\displaystyle$ $=$ $\displaystyle -\paren {\paren {-86.6 \, \mathrm {kg} + 37.5 \, \mathrm {kg} + 35.4 \, \mathrm {kg} } + i \paren {50 \, \mathrm {kg} + 65.0 \, \mathrm {kg} - 35.4 \, \mathrm {kg} } }$ evaluating roots $\displaystyle$ $=$ $\displaystyle -\paren {-13.7 \, \mathrm {kg} + i \, 79.6 \, \mathrm {kg} }$ calculating $\displaystyle$ $=$ $\displaystyle 13.7 \, \mathrm {kg} - i \, 79.6 \, \mathrm {kg}$ calculating

Then:

 $\displaystyle \cmod {\mathbf E}$ $=$ $\displaystyle \sqrt {\paren {13.7 \, \mathrm {kg} }^2 + \paren {-79.6 \, \mathrm {kg} }^2}$ $\displaystyle$ $=$ $\displaystyle 80.8 \, \mathrm {kg}$

and:

 $\displaystyle \map \arg {\mathbf E}$ $=$ $\displaystyle \arctan \dfrac {-79.6} {13.75}$ $\displaystyle$ $=$ $\displaystyle -80.2 \degrees$

Hence the result.

$\blacksquare$

## Proof 2

Using the Parallelogram Law to add $\mathbf F_1$ and $\mathbf F_2$ we arrive at $\mathbf F_1 + \mathbf F_2$.

Again using the Parallelogram Law to add $\mathbf F_1 + \mathbf F_2$ and $\mathbf F_3$ we arrive at $\mathbf F_1 + \mathbf F_2 + \mathbf F_3$.

The equilibrant $\mathbf E$ can then be found by taking the negative of $\mathbf F_1 + \mathbf F_2 + \mathbf F_3$.

$\blacksquare$