Question

P, Q and R are three coplanar forces acting at a point and are in equilibrium. Given P = 1.9318 kg wt, $\sin\theta_{1} =$0.9659, the value of R is ( in kg wt)

• A. 0.9659
• B. 2
• C. 1
• D. $\frac{1}{2}$

Answer 1

Answer: (C)

1

Answer 2

$\frac{P}{\sin\theta_{1}} = \frac{Q}{\sin\theta_{2}} = \frac{R}{\sin 150{^\circ}}$

$\Rightarrow \frac{1.93}{\sin\theta_{1}} = \frac{R}{\sin 150{^\circ}}$

$\Rightarrow R = \frac{1.93 \times \sin 150{^\circ}}{\sin\theta_{1}} = \frac{1.93 \times 0.5}{0.9659} = 1$