A rod of length l is acted upon by a couple a | vedclass

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(C) The moment of a couple is given by the product of the magnitude of one of the forces and the perpendicular distance between the lines of action of

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$\frac{2 	au}{l}$

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(C) The moment of a couple is given by the product of the magnitude of one of the forces and the perpendicular distance between the lines of action of the two forces.Let the angle between the rod and the direction of the force be $	heta = 30^{\circ}$.The perpendicular distance between the two parallel forces $F$ acting at the ends of the rod of length $l$ is $d = l \sin 	heta$.The moment of the couple $	au$ is given by:$	au = F 	imes d = F 	imes l \sin 30^{\circ}$Given $\sin 30^{\circ} = 0.5 = \frac{1}{2}$.Substituting the values:$	au = F 	imes l 	imes \frac{1}{2}$$	au = \frac{F l}{2}$Rearranging to solve for $F$:$F = \frac{2 	au}{l}$

$A$ rod of length $l$ is acted upon by a couple as shown in the figure. The moment of the couple is $\tau \text{ Nm}$. If the force at each end of the rod is $F$,then the magnitude of each force is (given $\sin 30^{\circ} = \cos 60^{\circ} = 0.5$):

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