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A block of mass $8 \mathrm{~kg}$ is suspended by a rope of length $3 \mathrm{~m}$ from the ceiling. A force of $40 \mathrm{~N}$ is applied horizontally to the block. Then the angle that the rope makes with the vertical in equilibrium is
(acceleration due to gravity $=10 \mathrm{~m} \mathrm{~s}^{-2}$, neglect the mass of the rope)
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(acceleration due to gravity $=10 \mathrm{~m} \mathrm{~s}^{-2}$, neglect the mass of the rope)
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Verified Answer
The correct answer is:
$\tan ^{-1}\left(\frac{1}{2}\right)$
Let ' $\theta$ ' be the angle that string makes in equilibrium.
So, $\mathrm{F}=\mathrm{T} \sin \theta$...(i)
$\mathrm{mg}=\mathrm{T} \cos \theta$...(ii)
Dividing (ii) by (i), we get
$\begin{aligned} & \frac{\mathrm{F}}{\mathrm{mg}}=\tan \theta \Rightarrow \tan \theta=\frac{40}{80}=\frac{1}{2} \\ & \theta=\tan ^{-1}\left(\frac{1}{2}\right)\end{aligned}$
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