Respuesta :

Answer:

(c) 1

Explanation:

To solve such systems, "Lami's theorem" is used as it best relates the magnitudes of such coplanar, concurrent and non-collinear forces.

Statement:

When three forces acting at a point are in equilibrium, then each force is proportional to the sine of the angle between the other two forces.

In mathematical form:

[tex] \boxed{ \mathsf{ \frac{P}{ \sin( \theta _{1}) } = \frac{Q}{\sin( \theta _{2}) } = \frac{R}{\sin( \theta _{3}) } }}[/tex]

Solution:

According to the FBD, The given three forces are coplanar, concurrent(act at a same point), and in equilibrium.

Instead of θ₃, we have 150 and the value of sin(θ₁) is known.

Using Lami's :

[tex] \implies \: \mathsf{ \frac{R}{ \sin(150) } = \frac{P}{ \sin( \theta _{1} ) } }[/tex]

  • sin (150) = sin(180- 30)

= sin 30

= 1/ 2

  • P = 1.9318
  • sin(θ₁) = 0.9659

[tex] \implies \: \mathsf{ \frac{R}{ \frac{1}{2} } = \frac{1.9318}{ 0.9659 } }[/tex]

  • R is multiplied by the reciprocal of ½ that is 2,
  • upon solving the Right Hand Side, we get 2

[tex] \implies \: \mathsf{ \frac{2R}{1 } = \frac{2}{ 1 } }[/tex]

  • Canceling 2 from both side

[tex] \implies \mathsf{R \: = 1}[/tex]

that is option C.

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