Find the sum of the given vectors displaystyle{700}{N}angle{320}^{circ}+{400}{N}angle{20}^{circ} Rx = ? Ry = ? R = ? theta = ?

Step 1

The given expression is,
${\displaystyle 700N\mathrm{\angle }{230}^{\circ }+400N\mathrm{\angle }{20}^{\circ }}$
The angle of vector is measured whith positive x-asis.
Step 2
For $700N,\text{the angle made with positive x-axis is}{230}^{\circ }$
Then angle made by $700N\text{with negative x-axis is}{50}^{\circ }$
On resolving the vector, we get
${\displaystyle f_1=700N[(-\cos \left({50}^{\circ }\right))\hat{i}+(-\sin \left({50}^{\circ }\right))\hat{j}]}$
${\displaystyle =-\left(697.34N\right)\hat{i}-\left(536.23N\right)j}$
For $400N,$ on resolving we get
${\displaystyle F_2=400N[(\cos \left({20}^{\circ }\right))\hat{i}+(\sin \left({20}^{\circ }\right))\hat{j}]}$
${\displaystyle =\left(375.88N\right)\hat{i}+\left(136.81N\right)\hat{j}}$
Step 3
${\displaystyle R=F_1+F_2}$
${\displaystyle =[-\left(697.34N\right)\hat{i}-\left(536.23N\right)\hat{j}]+[\left(375.88N\right)\hat{i}+\left(136.81N\right)\hat{i}+\left(136.81N\right)\hat{j}]}$
${\displaystyle =\left[(-697.34+375.88)N\right]\hat{i}+\left[(-536.23+136.81)N\right]\hat{j}}$
${\displaystyle =(-321.46N)\hat{i}+(-399.42N)\hat{j}}$
Hence,
${\displaystyle R_x=-321}$
${\displaystyle R_y=-399}$
${\displaystyle R=\sqrt{R_x^2+R_y^2}}$
${\displaystyle =\sqrt{{(-321)}^2+{(-399)}^2}}$
$=513$
Step 4

${\displaystyle \tan \alpha =\frac{\left|R_y\right|}{\left|R_x\right|}}$
${\displaystyle \alpha ={\tan }^{-1},\frac{\left|R_y\right|}{\left|R_x\right|}}$
${\displaystyle ={\tan }^{-1}\left(\frac{399.42}{321.46}\right)}$
$=51$
Then angle made by resultant vector with positive x-axis is,
${\displaystyle \theta ={180}^{\circ }+\alpha }$
${\displaystyle ={180}^{\circ }+{51.17}^{\circ }}$
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