Equilibrant/Examples/100kg at 150, 75kg at 60, 50kg at -45/Proof 1
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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:
\(\ds \mathbf F_1\) | \(=\) | \(\ds \polar {100 \, \mathrm {kg}, 150 \degrees}\) | ||||||||||||
\(\ds \mathbf F_2\) | \(=\) | \(\ds \polar {75 \, \mathrm {kg}, 60 \degrees}\) | ||||||||||||
\(\ds \mathbf F_3\) | \(=\) | \(\ds \polar {50 \, \mathrm {kg}, -45 \degrees}\) |
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
\(\ds \mathbf E\) | \(=\) | \(\ds -\paren {\mathbf F_1 + \mathbf F_2 + \mathbf F_3}\) | Magnitude and Direction of Equilibrant | |||||||||||
\(\ds \) | \(=\) | \(\ds -\paren {\polar {100 \, \mathrm {kg}, 150 \degrees} + \polar {75 \, \mathrm {kg}, 60 \degrees} + \polar {50 \, \mathrm {kg}, -45 \degrees} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\paren {100 \, \mathrm {kg} \cis 150 \degrees + 75 \, \mathrm {kg} \cis 60 \degrees + 50 \, \mathrm {kg} \cis -45 \degrees}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\paren {\paren {100 \, \mathrm {kg} \cos 150 \degrees + 75 \, \mathrm {kg} \cos 60 \degrees + 50 \, \mathrm {kg} \cos -45 \degrees} + i \paren {100 \, \mathrm {kg} \sin 150 \degrees + 75 \, \mathrm {kg} \sin 60 \degrees + 50 \, \mathrm {kg} \sin -45 \degrees} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\paren {\paren {100 \dfrac {\sqrt 3} 2 \, \mathrm {kg} + 75 \paren {\dfrac 1 2} \, \mathrm {kg} + 50 \dfrac {\sqrt 3} 2 \, \mathrm {kg} } + i \paren {100 \, \mathrm {kg} \sin 150 \degrees + 75 \, \mathrm {kg} \sin 60 \degrees + 50 \, \mathrm {kg} \sin -45 \degrees} }\) | Cosine of $150 \degrees$, Cosine of $60 \degrees$, Cosine of $315 \degrees$ | |||||||||||
\(\ds \) | \(=\) | \(\ds -\paren {\paren {100 \dfrac {-\sqrt 3} 2 \, \mathrm {kg} + 75 \paren {\dfrac 1 2} \, \mathrm {kg} + 50 \dfrac {\sqrt 2} 2 \, \mathrm {kg} } + i \paren {100 \paren {\dfrac 1 2} \, \mathrm {kg} + 75 \dfrac {\sqrt 3} 2 \, \mathrm {kg} + 50 \dfrac {-\sqrt 2} 2 \, \mathrm {kg} } }\) | Sine of $150 \degrees$, Sine of $60 \degrees$, Sine of $315 \degrees$ | |||||||||||
\(\ds \) | \(=\) | \(\ds -\paren {\paren {-50 \sqrt 3 \, \mathrm {kg} + 37.5 \, \mathrm {kg} + 25 \sqrt 2 \, \mathrm {kg} } + i \paren {50 \, \mathrm {kg} + 37.5 \sqrt 3 \, \mathrm {kg} - 25 \sqrt 2 \, \mathrm {kg} } }\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds -\paren {\paren {-86.6 \, \mathrm {kg} + 37.5 \, \mathrm {kg} + 35.4 \, \mathrm {kg} } + i \paren {50 \, \mathrm {kg} + 65.0 \, \mathrm {kg} - 35.4 \, \mathrm {kg} } }\) | evaluating roots | |||||||||||
\(\ds \) | \(=\) | \(\ds -\paren {-13.7 \, \mathrm {kg} + i \, 79.6 \, \mathrm {kg} }\) | calculating | |||||||||||
\(\ds \) | \(=\) | \(\ds 13.7 \, \mathrm {kg} - i \, 79.6 \, \mathrm {kg}\) | calculating |
Then:
\(\ds \cmod {\mathbf E}\) | \(=\) | \(\ds \sqrt {\paren {13.7 \, \mathrm {kg} }^2 + \paren {-79.6 \, \mathrm {kg} }^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 80.8 \, \mathrm {kg}\) |
and:
\(\ds \map \arg {\mathbf E}\) | \(=\) | \(\ds \arctan \dfrac {-79.6} {13.75}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -80.2 \degrees\) |
Hence the result.
$\blacksquare$
Sources
- 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Supplementary Problems: Polar Form of Complex Numbers: $86 \ \text {(b)}$