Question

# A particle is acted upon by four forces simultaneously, $30N$ due east $20N$ due north $15N$ due west $40N$ due south find the resultant force of the particle?

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Solution

## Step 1: Given dataFour forces act on a particle as shown in the figure.Step 2: To findThe resultant force on the particle.Step 3: Calculation for magnitude of resultant forceWe know, resultant of the forces acting in opposite direction can be found by taking their difference. Here, force in the north-south direction,${F}_{NS}=40-20=20N$ towards south.Force in the east-west direction,${F}_{EW}=30-15=15N$ towards east.Resultant force, $F=\sqrt{{F}_{NS}^{2}+{F}_{EW}^{2}+2{F}_{NS}{F}_{EW}\mathrm{cos}\theta }\phantom{\rule{0ex}{0ex}}F=\sqrt{{20}^{2}+{15}^{2}+2×20×15×\mathrm{cos}90°}\left(\theta =90°\because theforcesare\perp r\right)\phantom{\rule{0ex}{0ex}}F=\sqrt{400+225+\left(600×0\right)}\phantom{\rule{0ex}{0ex}}F=\sqrt{625}\phantom{\rule{0ex}{0ex}}F=25N$Step 4: Calculation for angle made by the resultant force$\mathrm{tan}\left(\theta \right)=\frac{20}{15}\phantom{\rule{0ex}{0ex}}\mathrm{tan}\left(\theta \right)={53}^{0}$Hence, the resultant force of the particle is $25\mathrm{N}$ and it will make ${53}^{°}$with the west.

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