Two forces are acting on a body. One acts east, the other at 35° north of east. If the
two forces are equal in magnitude of 50 N, find the resultant using the Law of Sines
and the Law of Cosines. Please answer with full solution. Thanks

Respuesta :

  • A=B=50N
  • Angle=theta=35°

We know

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

[tex]\\ \sf\longmapsto R=\sqrt{50^2+50^2+2(50)(50)cos35}[/tex]

[tex]\\ \sf\longmapsto R=\sqrt{2500+2500+2(2500)\times (-0.9)}[/tex]

[tex]\\ \sf\longmapsto R=\sqrt{5000+5000(-0.9)}[/tex]

[tex]\\ \sf\longmapsto R=\sqrt{5000+(-4500)}[/tex]

[tex]\\ \sf\longmapsto R=\sqrt{5000-4500}[/tex]

[tex]\\ \sf\longmapsto R=\sqrt{-500}[/tex]

[tex]\\ \sf\longmapsto R=22.4i[/tex]

Resultant using the Law of Sines and the Law of Cosines will be R=95 N

What is force?

Force is an external agent applied on any object to displace it from its position. Force is a vector quantity, so with magnitude it also requires direction. Direction is necessary to examine the effect of the force and to find the equilibrium of the force.

The Magnitude of two forces =50 N

Angle between the forces = 35

By using the resultant formula

[tex]\rm R=\sqrt{A^2+B^2+2ABCos\theta}[/tex]

[tex]\rm R=\sqrt{50^2+50^2+2(50)(50)Cos35}[/tex]

[tex]\rm R=\sqrt{5000+5000(0.81)}[/tex]

[tex]\rm R=\sqrt{5000+4500}[/tex]

[tex]\rm R=95\ N[/tex]

Hence the Resultant using the Law of Sines and the Law of Cosines will be R=95 N

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