Question

P,Q and R are three co planar forces rating at a point and are in equilibrium:
Given $P=1.9318kgwt$ $sin=0.90591$, the value of $R$ ( in $kgwt$) is:

• A. 0.9059
• B. 2
• C. 1
• D. 1/2

Answer 1

Answer: (C)

1

$\frac{P}{\sin {\theta }_1}=\frac{Q}{\sin {\theta }_2}=\frac{R}{\sin {150}^o}$

$\Rightarrow \frac{1.93}{\sin {\theta }_1}=\frac{R}{\sin {150}^o}$

$\Rightarrow R=\frac{1.93\times \sin {150}^o}{\sin {\theta }_1}=\frac{1.93\times 0.5}{0.9669}=1$