Vectors Question 118
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)
[CET 1998]
Options:
A) 0.9659
B) 2
C) 1
D) $ \frac{1}{2} $
Correct Answer: C $ \frac{P}{\sin {\theta_1}}=\frac{Q}{\sin {\theta_2}}=\frac{R}{\sin 150{}^\circ } $
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Answer:
Solution:
$ \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 $
BETA
