A 1 ~N pendulum bob is held at an angle theta | vedclass

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(C) The pendulum bob is in equilibrium under the action of three forces: the tension $T$ in the string,the horizontal force $F = 2 ~N$,and the weight

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$\sqrt{5}$

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(C) The pendulum bob is in equilibrium under the action of three forces: the tension $T$ in the string,the horizontal force $F = 2 ~N$,and the weight $W = 1 ~N$ acting vertically downwards.For equilibrium,the net force in both horizontal and vertical directions must be zero.Resolving the tension $T$ into components:Horizontal component: $T \sin 	heta = F = 2 ~N$ $(i)$Vertical component: $T \cos 	heta = W = 1 ~N$ (ii)Squaring and adding equations $(i)$ and (ii):$(T \sin 	heta)^2 + (T \cos 	heta)^2 = F^2 + W^2$$T^2 (\sin^2 	heta + \cos^2 	heta) = F^2 + W^2$$T^2 = F^2 + W^2$$T = \sqrt{F^2 + W^2} = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} ~N$.

$A$ $1 ~N$ pendulum bob is held at an angle $\theta$ from the vertical by a $2 ~N$ horizontal force $F$ as shown in the figure. The tension in the string supporting the pendulum bob (in newton) is

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