Physics - NLM Question with Solution | TestHub

Considering the equilibrium of A, we get \begin{array}{l} 10 \mathrm{a}=10 \mathrm{~g} \sin 30^{\circ}-\mathrm{T}-\mu \mathrm{N} \\ \text { where } \mathrm{N}=10 \mathrm{~g} \cos 30^{\circ} \end{array}
$\therefore 10\mathrm{a}=\frac{10}{2}\mathrm{g}-\mathrm{T}-\mu \times 10\mathrm{g}\cos 30^{\circ }$ but $\mathrm{a}=0,\mathrm{T}={\mathrm{m}}_{\mathrm{B}}\mathrm{g}$ $0=5\mathrm{g}-{\mathrm{m}}_{\mathrm{B}}\mathrm{g}-\frac{0.2\sqrt{3}}{2}\times 10\times \mathrm{g}$ $\Rightarrow {\mathrm{m}}_{\mathrm{B}}=3.268\approx 3.3\mathrm{kg}$