Question 195: Laws of Motion

Question

The pulleys and strings shown in figure are smooth and of negligible mass. For the system to remain in equilibrium, angle $	heta$ should be

Accepted Answer

From the diagram,
$\mathrm{T}=\mathrm{Mg}$ and,
For the system to remain in equilibrium,
$$\begin{array}{ll} 
& 2 \mathrm{~T} \cos 	heta=\sqrt{2} \mathrm{Mg} \
	herefore & 2 \mathrm{~T} \cos 	heta=\mathrm{T} \sqrt{2} \
	herefore & \cos 	heta=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}} \
	herefore & \quad 	heta=\cos ^{-1}\left(\frac{1}{\sqrt{2}}ight)
\end{array}$$

Answer

$\cos ^{-1}(1)$

Answer

$\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)$

Answer

$\cos ^{-1}\left(\frac{1}{2}\right)$

Answer

$\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)$

Laws of Motion

The pulleys and strings shown in figure are smooth and of negligible mass. For the system to remain in equilibrium, angle \theta should be

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