The resultant of these forces overrightarrow{ | vedclass

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(A) To find the resultant force,we resolve each force into its $x$ and $y$ components:$F_x = 10 \cos 30^\circ + 20 \sin 30^\circ + 20 \cos 45^\circ -

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$9.25 \hat{i} + 5 \hat{j}$

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(A) To find the resultant force,we resolve each force into its $x$ and $y$ components:$F_x = 10 \cos 30^\circ + 20 \sin 30^\circ + 20 \cos 45^\circ - 15 \cos 45^\circ - 15 \sin 60^\circ$$F_x = 10(0.85) + 20(0.5) + 20(0.7) - 15(0.7) - 15(0.85) = 8.5 + 10 + 14 - 10.5 - 12.75 = 9.25 	ext{ N}$$F_y = 10 \sin 30^\circ + 20 \cos 30^\circ + 15 \cos 60^\circ - 20 \sin 45^\circ - 15 \sin 45^\circ$$F_y = 10(0.5) + 20(0.85) + 15(0.5) - 20(0.7) - 15(0.7) = 5 + 17 + 7.5 - 14 - 10.5 = 5 	ext{ N}$Thus,the resultant force is $9.25 \hat{i} + 5 \hat{j} 	ext{ N}$.

The resultant of these forces $\overrightarrow{OP}, \overrightarrow{OQ}, \overrightarrow{OR}, \overrightarrow{OS}$ and $\overrightarrow{OT}$ is approximately $\ldots \ldots \text{N}$.
[Take $\sqrt{3}=1.7, \sqrt{2}=1.4$. Given $\hat{i}$ and $\hat{j}$ are unit vectors along $x, y$ axes.]

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The resultant of these forces $\overrightarrow{OP}, \overrightarrow{OQ}, \overrightarrow{OR}, \overrightarrow{OS}$ and $\overrightarrow{OT}$ is approximately $\ldots \ldots \text{N}$.[Take $\sqrt{3}=1.7, \sqrt{2}=1.4$. Given $\hat{i}$ and $\hat{j}$ are unit vectors along $x, y$ axes.]