Question

Two particles A and B, each of mass $m$ are kept stationary by applying a horizontal force $F=mg$ on particle B as shown in figure. Then:

• A. $\tan \beta =2\tan \alpha$
• B. $2T_1=5T_2$
• C. $\sqrt{2}{T}_1=\sqrt{5}{T}_2$
• D. None of these

Answer 1

Answer: (A), (C)

$\tan \beta =2\tan \alpha$
$\sqrt{2}{T}_1=\sqrt{5}{T}_2$

Firstly draw free body diagrams (FBD) for particles A and B as shown in the image (here a $=\alpha$ and b $=\beta$).

Resolve these forces into their components along X-axis and Y-axis.

From FBD of A,

$T_1\sin \alpha =T_2\sin \beta ⟶\left(1\right)$

$T_1\cos \alpha =T_2\cos \beta +mg⟶\left(2\right)$

From FBD of B,

$T_2\cos \beta =mg⟶\left(3\right)$

$T_2\sin \beta =mg⟶\left(4\right)$

$\frac{\left(3\right)}{\left(4\right)}$ gives $\tan \beta =1$

Using $\left(4\right)$ in $\left(1\right)$ and $\left(3\right)$ in $\left(2\right)$,

$T\sin \alpha =mg$ and $T\cos \alpha =2mg$

$\Rightarrow \tan \alpha =1/2$

$\Rightarrow 2\tan \alpha =\tan \beta$

Now using $\alpha$ and $\beta$ values in $\left(1\right)$, $\sqrt{2}{T}_1=\sqrt{5}{T}_2$

Therefore options A and C are correct.