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The resultant of these forces $\overrightarrow{O P}, \overrightarrow{O Q}, \overrightarrow{O R}, \overrightarrow{O S}$ and $\overrightarrow{{OT}}$ is approximately $\ldots \ldots {N}$.
[Take $\sqrt{3}=1.7, \sqrt{2}=1.4$ Given $\hat{{i}}$ and $\hat{{j}}$ unit vectors along ${x}, {y}$ axis $]$
$9.25 \hat{{i}}+5 \hat{{j}}$
$3 \hat{{i}}+15 \hat{{j}}$
$2.5 \hat{i}-14.5 \hat{{j}}$
$-1.5 \hat{{i}}-15.5 \hat{{j}}$
Solution
$\hat{F}_{x}=\left(10 \times \frac{\sqrt{3}}{2}+20\left(\frac{1}{2}\right)+20\left(\frac{1}{\sqrt{2}}\right)-15\left(\frac{1}{\sqrt{2}}\right)-15\left(\frac{\sqrt{3}}{2}\right)\right) \hat{i}$
$=9.25 \hat{i}$
$\vec{F}_{y}=\left(15\left(\frac{1}{2}\right)+20\left(\frac{\sqrt{3}}{2}\right)+10\left(\frac{1}{2}\right)-15\left(\frac{1}{\sqrt{2}}\right)-20\left(\frac{1}{\sqrt{2}}\right)\right) \hat{j}$
$=5 \hat{j}$
