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 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.
