Block B lying on a table weighs w. The coefficient of static friction between the block and the table is $\mathrm{\mu }$. Assume that the cord between B and the knot is horizontal. The maximum weight of the block A for which the system will be stationary is                        

Let weight of A is w'. From the free body diagram, for equilibrium of the system,

$T \mathrm{cos\theta } = \mathrm{\mu }N\mathit{ }= \mathrm{\mu }w\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\left(\mathrm{i}\right)$

$T \mathrm{sin\theta } = w^{\mathit{\text{'}}}\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\left(\mathrm{ii}\right)$

where, T = tension in the thread lying between knot and the support.

On dividing Eq. (ii) by Eq. (i), we get

$\frac{T\sin \theta }{T\cos \theta }=\frac{w^{\text{'}}}{\mu w}\Rightarrow \tan \theta =\frac{w^{\text{'}}}{\mu w}$

$\Rightarrow w^{\text{'}}=\mu w\tan \theta$