Question

Two men of masses $m_1$ and $m_2$ hold on the opposite ends of a rope passing over a frictionless pulley and are initially separated by $5m$ . Both the men are at rest, the man of mass $m_1$ climbs up the rope with an acceleration of $1.2m/s^2$ relative to the rope. The man of mass $m_2$ climbs up the rope with an acceleration of $2m/s^2$ relative to the rope. If $m_1=40kg$ and $m_2=60kg$,

• A. tension in the rope is $556.8N$
• B. tension in the rope is $586.8N$
• C. acceleration of rope is $2.72m/s^2$
• D. acceleration of rope is $3.72m/s^2$

Answer 1

Answer: (A), (C)

acceleration of rope is $2.72m/s^2$

Let $a_{m_{1}G}$ and $a_{m_{2}G}$ be the accelerations of man $m_1$ and man $m_2$ with respect to the ground.
And $a_{m_{1}R}$ and $a_{m_{2}R}$ be the accelerations of man $m_1$ and man $m_2$ with respect to the rope.
$a_{RG}$ is the acceleration of the rope with respect to the ground.
​​​​​​​
For man of mass $m_1$
$a_{m_{1}G}=a_{m_{1}R}+a_{RG}$
$\Rightarrow a_{m_{1}G}=(1.2+a)$
For man of mass $m_2$,
$a_{m_{2}G}=a_{m_{2}R}+a_{RG}$
$=(2-a)$


So now,
$T-mg=m_1(1.2+a).......\left(i\right)$
$T-mg=m_2(2-a)........\left(ii\right)$

Solve eq. $\left(i\right)$ & $\left(ii\right)$ and put $m_1=40kg$ and $m_2=60kg$, you get
$a=2.72m/s^2$
$T=556.8N$