Physics - NLM Question with Solution | TestHub

Tendency of 2 kg block is of coming down and of 1 kg is of going up. FBD of 1 kg block FBD of 2 kg block

$1\times \mathrm{g}\times \sin \theta$ FBD of 2 kg block

$2\times 10\times 1/2$ $2\times \mathrm{g}\times \sin \theta$ Where ${\mathrm{F}}_1$ is friction between 1 kg and 2 kg block and ${\mathrm{F}}_2$ is friction between 2 kg and inclined surface. For 2 kg block $2\times 10\times \frac{1}{2}-\mathrm{T}-{\mathrm{F}}_1-{\mathrm{F}}_2=2\mathrm{a}$ For 1 kg block $\mathrm{T}-1\times 10\times \frac{1}{2}\times {\mathrm{F}}_1=\mathrm{a}$ ⇒ Add both

$$\begin{array}{rlr}& 10-5-2{\mathrm{F}}_1-{\mathrm{F}}_2=3\mathrm{a}& 5-2{\mathrm{F}}_1-{\mathrm{F}}_2=3\mathrm{a}\end{array}$$

The max. value ${\mathrm{F}}_1=0.1\times 1\times 10\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}$ ${\mathrm{F}}_{2\max }=3\times 10\times 0.2\times \frac{\sqrt{3}}{2}=3\sqrt{3}$

$$\begin{array}{rlr}& \mathrm{Max}\left(2{\mathrm{F}}_1+{\mathrm{F}}_2\right)=\sqrt{3}+3\sqrt{3}=4\sqrt{3}=5.1\mathrm{N}& \Rightarrow \text{ Static friction will act. }\Rightarrow \mathrm{a}=0\end{array}$$