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(C) The resultant force of the system is $F_{res} = 5 \ N + 3 \ N = 8 \ N$,which acts downwards through point $R$.For the system to be in rotational e

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$\frac{3}{5} RQ$

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(C) The resultant force of the system is $F_{res} = 5 \ N + 3 \ N = 8 \ N$,which acts downwards through point $R$.For the system to be in rotational equilibrium about point $R$,the sum of the moments of the forces about $R$ must be zero.The force of $5 \ N$ at $P$ creates a clockwise moment about $R$,and the force of $3 \ N$ at $Q$ creates a counter-clockwise moment about $R$.Taking moments about $R$:$5 	imes PR - 3 	imes RQ = 0$$5 	imes PR = 3 	imes RQ$$PR = \frac{3}{5} RQ$

The resultant of the system in the figure is a force of $8 \ N$ parallel to the given forces,acting through point $R$. The value of $PR$ is equal to . . . . . .

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