Question Number 33930 by mondodotto@gmail.com last updated on 27/Apr/18
$$\boldsymbol{\mathrm{Three}}\:\boldsymbol{\mathrm{forces}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{magnitude}}\:\mathrm{6}\boldsymbol{\mathrm{N}},\mathrm{2}\boldsymbol{\mathrm{N}} \\ $$$$\boldsymbol{\mathrm{and}}\:\mathrm{3}\boldsymbol{\mathrm{N}}\:\boldsymbol{\mathrm{act}}\:\boldsymbol{\mathrm{on}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{same}}\:\boldsymbol{\mathrm{point}}\:\boldsymbol{\mathrm{on}}\:\boldsymbol{\mathrm{the}} \\ $$$$\boldsymbol{\mathrm{north}},\boldsymbol{\mathrm{south}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{west}}\:\boldsymbol{\mathrm{directions}}\:\boldsymbol{\mathrm{respectively}}. \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{magnitude}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{direction}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}} \\ $$$$\boldsymbol{\mathrm{resultant}}\:\boldsymbol{\mathrm{force}}. \\ $$$$ \\ $$
Answered by MJS last updated on 27/Apr/18
$$\begin{pmatrix}{\mathrm{0}}\\{\mathrm{6}}\end{pmatrix}+\begin{pmatrix}{\mathrm{0}}\\{−\mathrm{2}}\end{pmatrix}+\begin{pmatrix}{−\mathrm{3}}\\{\mathrm{0}}\end{pmatrix}=\begin{pmatrix}{−\mathrm{3}}\\{\mathrm{4}}\end{pmatrix} \\ $$$$\mathrm{magnitude}=\mathrm{length}\:\mathrm{of}\:\mathrm{vector}=\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }=\mathrm{5} \\ $$$$\mathrm{direction}=\mathrm{angle}\:\mathrm{of}\:\mathrm{vector}\:\left(\frac{\pi}{\mathrm{2}}<\theta<\pi\right)= \\ $$$$=\pi+\mathrm{arctan}\:\frac{{y}}{{x}}=\pi+\mathrm{arctan}\:−\frac{\mathrm{4}}{\mathrm{3}}\approx\mathrm{2}.\mathrm{214}^{\mathrm{rad}} \approx\mathrm{126}.\mathrm{87}° \\ $$
Commented by mondodotto@gmail.com last updated on 27/Apr/18
$$\mathrm{thanx}\:\mathrm{be}\:\mathrm{blessed} \\ $$