Question

Two blocks of mass $m_1=20\text{kg}$ and $m_2=12\text{kg}$ are connected by a rope of mass $8\text{k}g$. The system pulled vertically up by applying a force of $480\text{N}$ as shown in figure. The tension at the mid-point of the rope will be:
(Consider rope to be uniform)

• A. $131\text{N}$
• B. $180\text{N}$
• C. $210\text{N}$
• D. $192\text{N}$

Answer 1

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

Considering mass $m_1$, rope, mass $m_1$ as a system and applying equation of dynamics in vertical direction.


By using second law of motion

$\Rightarrow 480-(m_1+m+m_2)g=(m_1+m+m_2)a$

$\Rightarrow 480-(20+8+12)10=(20+8+12)a$

$\Rightarrow 480-400=40a$

$\therefore a=\frac{80}{40}=2{\text{m/s}}^2$

Now drawing FBD of system with half part of rope:

Again applying $F_{net}=ma$

$\Rightarrow 480-T-(m_1+\frac{m}{2})g=(m_1+\frac{m}{2})a$

Substituting values of $m_1,m$ and $a$, we get,

$480-T-(20+4)10=24a$

$\Rightarrow 480-T-240=48$

$\Rightarrow T=480-240-48=192\text{N}$

Thus, tension at the mid-point of rope will be $192\text{N}$.

$\begin{array}{c}\text{Why this question}?\\ \text{Tip: For problems containing uniform rope with}\\ \text{mass, always try to obtain the acceleration}\\ \text{of system first. Thereafter we can apply}\\ \text{newton's}{2}^{nd}\text{law on part of system/rope to}\\ \text{find tension at desired point of rope.}\end{array}$