Question

Find the tension (in $N$) in the string connecting the $2m$ mass. (Strings are massless and inextensible)

• A. $\frac{15mg}{11}$
• B. $\frac{18mg}{13}$
• C. $\frac{18mg}{11}$
• D. $\frac{15mg}{13}$

Answer 1

The correct option is C $\frac{18mg}{11}$

The given system can be reframed as shown below as per intercept method

So, we have
a) ${\stackrel{..}{L}}_1+{\stackrel{..}{L}}_2=0$
$\Rightarrow a_1-a_p=0\Rightarrow a_p=a_1\dots \dots \left(1\right)$

b) ${\stackrel{..}{L}}_3+{\stackrel{..}{L}}_4=0\Rightarrow a_p+a_p-a_2=0$
$\Rightarrow a_2=2a_p\dots \dots \left(2\right)$
From (1) and (2) we get,
$a_2=2a_1$
From FBDs of the masses we have force equations as

$3mg-T_1=3m{a}_1\dots \dots \left(3\right)$
$T_2-2mg=2m{a}_2=2m\left(2a_1\right)=4m{a}_1\dots \left(4\right)$
and $T_1=2T_2\dots \dots \left(5\right)$
So, from above three equations we have
$-mg=11m{a}_1\Rightarrow a_1=\frac{-g}{11}$
Here, $-ve$ sign shows that the assumption that we made about the direction of acceleration of mass $3m$ is wrong i.e. the mass $3m$ will move upwards rather than downwards.
$\therefore$ $T_1=3m(g-a_1)\Rightarrow T_1=\frac{36mg}{11}$
As we have $T_1=2T_2$
$\Rightarrow T_2=\frac{T_1}{2}=\frac{18mg}{11}N$