A block of mass 8 kg is suspended by a rope | vedclass

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(B) Let $	heta$ be the angle that the string makes with the vertical in equilibrium.At equilibrium,the forces acting on the block are balanced:$1$. H

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$	an ^{-1}\left(\frac{1}{2}ight)$

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(B) Let $	heta$ be the angle that the string makes with the vertical in equilibrium.At equilibrium,the forces acting on the block are balanced:$1$. Horizontal force balance: $F = T \sin 	heta$ ... $(i)$$2$. Vertical force balance: $mg = T \cos 	heta$ ... (ii)Dividing equation $(i)$ by equation (ii),we get:$\frac{F}{mg} = \frac{T \sin 	heta}{T \cos 	heta} = 	an 	heta$Given $F = 40 \ N$,$m = 8 \ kg$,and $g = 10 \ m \ s^{-2}$:$	an 	heta = \frac{40}{8 	imes 10} = \frac{40}{80} = \frac{1}{2}$Therefore,$	heta = 	an ^{-1}\left(\frac{1}{2}ight)$.

$A$ block of mass $8 \ kg$ is suspended by a rope of length $3 \ m$ from the ceiling. $A$ force of $40 \ N$ is applied horizontally to the block. Then the angle that the rope makes with the vertical in equilibrium is (acceleration due to gravity $= 10 \ m \ s^{-2}$,neglect the mass of the rope).

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