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Find the resultant of two forces of 4.0N and 6.0N acting at an angle of 160° to each other.

Answer :

[tex]\star\sf \overrightarrow{A}=4N[/tex]

[tex]\star\sf\overrightarrow{B}=6N[/tex]

  • Angle=160

[tex]\boxed{\sf R=\sqrt{A^2+B^2+2ABcos\theta}}[/tex]

[tex]\\ \sf\longmapsto R=\sqrt{4^2+6^2+2(4)(6)cos160}[/tex]

[tex]\\ \sf\longmapsto R=\sqrt{16+36+48(-0.93)}[/tex]

[tex]\\ \sf\longmapsto R=\sqrt{52-44.6}[/tex]

[tex]\\ \sf\longmapsto R=\sqrt{7.4}[/tex]

[tex]\\ \sf\longmapsto R=2.7N[/tex]

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