Question

Two blocks of mass $M_1=20\text{kg}$ and $M_2=12\text{kg}$ are connected by a metal rod of mass $8\text{kg}$. The system is pulled vertically up by applying a force of $480\text{N}$ as shown. The tension at the mid-point of the rod is:

• A. $\text{144 N}$
• B. $\text{96 N}$
• C. $\text{240 N}$
• D. $\text{192 N}$

Answer 1

The correct option is D $\text{192 N}$

Acceleration produced in upward direction,

$a=\frac{F}{M_1+M_2+\text{Mass of metal rod}}$
$=\frac{480}{20+12+8}=12{\text{ms}}^{-2}$

Tension at the midpoint,

$T=(M_2+\frac{\text{Mass of rod}}{2})a$

$=(12+4)\times 12=192\text{N}$

Hence, $\left(D\right)$ is the correct answer.

$\text{Alternate solution:}$

FBD of system:



Using equilibrium condition,

$\Rightarrow 480-400=40\text{a}$

$\Rightarrow \text{a}=2{\text{m/s}}^2$

Now at midpoint,


$F_{\text{net}}=480-24\text{g}-T=24\text{a}$

$\Rightarrow 480-240-T=48$

$\therefore T=240-48=192\text{N}$

Hence, $\left(D\right)$ is the correct answer.