Kinematics Question 154
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 it’s ( 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 } $ $ \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 $
Show Answer
Answer:
Solution:
