Two blocks of mass $M_1=20 \mathrm{~kg}$ and $M_2=12 \mathrm{~kg}$ are connected by a metal rod of mass $8…
Two blocks of mass $M_1=20 \mathrm{~kg}$ and $M_2=12 \mathrm{~kg}$ are connected by a metal rod of mass $8 \mathrm{~kg}$. The system is pulled vertically up by applying a force of $480 \mathrm{~N}$ as shown. The tension at the mid-point of the rod is:
$144 \mathrm{~N}$
$96 \mathrm{~N}$
$240 \mathrm{~N}$
$192 \mathrm{~N}$
Solution
Acceleration produced in upward direction
$
$
\begin{aligned}
& \mathrm{a}=\frac{\mathrm{F}}{\mathrm{M}_1+\mathrm{M}_2+\text { Mass of metal rod }} \\
& =\frac{480}{20+12+8}=12 \mathrm{~ms}^{-2}
\end{aligned}
$
$
Tension at the mid point
$
$
\begin{aligned}
\mathrm{T} & =\left(\mathrm{M}_2+\frac{\text { Mass of rod }}{2}\right) \mathrm{a} \\
& =(12+4) \times 12=192 \mathrm{~N}
\end{aligned}
$
$
