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

Question
Decimals
asked 2021-02-06
Find the sum of the given vectors
\(\displaystyle{700}{N}\angle{320}^{\circ}+{400}{N}\angle{20}^{\circ}\)
\(Rx = ?\)
\(Ry = ?\)
\(R = ?\)
\(\theta = ?\)

Answers (1)

2021-02-07
Step 1
image
The given expression is,
\(\displaystyle{700}{N}\angle{230}^{\circ}+{400}{N}\angle{20}^{\circ}\)
The angle of vector is measured whith positive x-asis.
Step 2
For \(700\ N,\ \text{the angle made with positive x-axis is}\ 230^{\circ}\)
Then angle made by \(700\ N\ \text{with negative x-axis is}\ 50^{\circ}\)
On resolving the vector, we get
\(\displaystyle{{f}_{{1}}=}{700}{N}{\left[{\left(- \cos{{\left({50}^{\circ}\right)}}\right)}\hat{{i}}+{\left(- \sin{{\left({50}^{\circ}\right)}}\right)}\hat{{j}}\right]}\)
\(\displaystyle=-{\left({697.34}{N}\right)}\hat{{i}}-{\left({536.23}{N}\right)}{j}\)
For \(400\ N,\) on resolving we get
\(\displaystyle{F}_{{2}}={400}{N}{\left[{\left( \cos{{\left({20}^{\circ}\right)}}\right)}\hat{{i}}+{\left( \sin{{\left({20}^{\circ}\right)}}\right)}\hat{{j}}\right]}\)
\(\displaystyle={\left({375.88}{N}\right)}\hat{{i}}+{\left({136.81}{N}\right)}\hat{{j}}\)
Step 3
\(\displaystyle{R}={F}_{{1}}+{F}_{{2}}\)
\(\displaystyle={\left[-{\left({697.34}{N}\right)}\hat{{i}}-{\left({536.23}{N}\right)}\hat{{j}}\right]}+{\left[{\left({375.88}{N}\right)}\hat{{i}}+{\left({136.81}{N}\right)}\hat{{i}}+{\left({136.81}{N}\right)}\hat{{j}}\right]}\)
\(\displaystyle={\left[{\left(-{697.34}+{375.88}\right)}{N}\right]}\hat{{i}}+{\left[{\left(-{536.23}+{136.81}\right)}{N}\right]}\hat{{j}}\)
\(\displaystyle={\left(-{321.46}{N}\right)}\hat{{i}}+{\left(-{399.42}{N}\right)}\hat{{j}}\)
Hence,
\(\displaystyle{R}_{{x}}=-{321}\)
\(\displaystyle{R}_{{y}}=-{399}\)
\(\displaystyle{R}=\sqrt{{{{R}_{{x}}^{{2}}}+{{R}_{{y}}^{{2}}}}}\)
\(\displaystyle=\sqrt{{{\left(-{321}\right)}^{2}+{\left(-{399}\right)}^{2}}}\)
\(= 513\)
Step 4
image
\(\displaystyle \tan{\alpha}=\frac{{\left|{{R}_{{y}}}\right|}}{{\left|{{R}_{{x}}}\right|}}\)
\(\displaystyle\alpha={{\tan}^{ -{{1}}},}\frac{{\left|{{R}_{{y}}}\right|}}{{\left|{{R}_{{x}}}\right|}}\)
\(\displaystyle={{\tan}^{ -{{1}}}{\left(\frac{399.42}{{321.46}}\right)}}\)
\(= 51\)
Then angle made by resultant vector with positive x-axis is,
\(\displaystyle\theta={180}^{\circ}+\alpha\)
\(\displaystyle={180}^{\circ}+{51.17}^{\circ}\)
\(= 231\)
0

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