Question

In the following figure the masses of the blocks $A$ and $B$ are the same and each equal to $m$. The tensions in the strings $OA$ and $AB$ are $T_2$ and $T_1$ respectively. The system is in equilibrium with a constant horizontal force $mg$ on $B$. Then $T_2$ is:

• A. $mg$
• B. $\sqrt{2}mg$
• C. $\sqrt{3}mg$
• D. $\sqrt{5}mg$

Answer 1

The correct option is D $\sqrt{5}mg$

From the free body diagram of block B:
$\begin{array}{}{T}_1\cos {\theta }_1=mg& \text{(i)}\end{array}$
$\begin{array}{}{T}_1\sin {\theta }_1=mg& \text{(ii)}\end{array}$
By squaring and adding $T_1^2({\sin }^2{\theta }_1+{\cos }^2{\theta }_1)=2(mg{)}^2$
$\therefore T_1=\sqrt{2}mg$
also ${\theta }_1=45°$
From the free body diagram of block $A$
For vertical equilibrium $T_2\cos {\theta }_2=mg+T_1\cos {\theta }_1$
$T_2\cos {\theta }_2=mg+\sqrt{2}mg\cos {45}^o$
or $\begin{array}{}{T}_2\cos {\theta }_2=2mg& \text{(i)}\end{array}$
For horizontal equilibrium
$T_2\sin {\theta }_2=T_{1}sin{\theta }_1=\sqrt{2}mg\sin {45}^o$
$\begin{array}{}{T}_2\sin {\theta }_2=mg& \text{(ii)}\end{array}$
By squaring and adding $\left(i\right)$ and $\left(ii\right)$ equilibrium: $T_2^2=5(mg{)}^2$
or $T_2=\sqrt{5}mg$