Question

hey ! can anyone solve this ?​

Answer 1

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:

$\boxed{ \mathsf{ \frac{P}{ \sin( \theta _{1}) } = \frac{Q}{\sin( \theta _{2}) } = \frac{R}{\sin( \theta _{3}) } }}$

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 :

$\implies \: \mathsf{ \frac{R}{ \sin(150) } = \frac{P}{ \sin( \theta _{1} ) } }$

• sin (150) = sin(180- 30)

= sin 30

= 1/ 2

• P = 1.9318
• sin(θ₁) = 0.9659

$\implies \: \mathsf{ \frac{R}{ \frac{1}{2} } = \frac{1.9318}{ 0.9659 } }$

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

$\implies \: \mathsf{ \frac{2R}{1 } = \frac{2}{ 1 } }$

• Canceling 2 from both side

$\implies \mathsf{R \: = 1}$

that is option C.