If the tension in the horizontal wire shown i | vedclass

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(B) Let $T_{OA}$ be the tension in wire $OA$ and $T_{OB} = 30 	ext{ N}$ be the tension in the horizontal wire $OB$.At point $O$,the forces are in equ

Vedclass · Accepted Answer

$30 \sqrt{3} 	ext{ N}, 60 	ext{ N}$

Vedclass · Answer

(B) Let $T_{OA}$ be the tension in wire $OA$ and $T_{OB} = 30 	ext{ N}$ be the tension in the horizontal wire $OB$.At point $O$,the forces are in equilibrium.Resolving the tension $T_{OA}$ into horizontal and vertical components:Horizontal component: $T_{OA} \sin(30^{\circ}) = T_{OB} = 30 	ext{ N}$.$T_{OA} 	imes (1/2) = 30 	ext{ N} \implies T_{OA} = 60 	ext{ N}$.Vertical component: $T_{OA} \cos(30^{\circ}) = W$.$W = 60 	imes (\sqrt{3}/2) = 30 \sqrt{3} 	ext{ N}$.Thus,the weight $W$ is $30 \sqrt{3} 	ext{ N}$ and the tension in wire $OA$ is $60 	ext{ N}$.

If the tension in the horizontal wire shown in the figure is $30 \text{ N}$,then the weight $W$ and tension in the wire $OA$ are respectively:

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