Question

In order to raise a mass of $100kg$, a man of mass $60kg$ fastens a rope to it and passes the rope over a smooth pulley. He climbs the rope with an acceleration of $\frac{5g}{4}$ relative to the rope. The tension in the rope is (take $g=10m{s}^{-2}$)

Answer 1

Step 1: Given Data

Given, the mass of the object $=100kg$

Let $T$ be the tension on the rope.

Given, the mass of the man $=60kg$

Given that the acceleration is $\frac{5g}{4}$ relative to the rope.

Let the acceleration of the rope be $a$ and the acceleration of the man be $a_m$.

Step 2: Equations of Motion

The mass undergoes a downward acceleration.

Similarly, it also undergoes an upward acceleration due to the pulling of the rope.

Hence, the net acceleration can be given as,

$\therefore a_m-\left(-a\right)=\frac{5g}{4}$

$\Rightarrow a_m=\frac{5g}{4}-a$

Hence, the equation of motion of the man can be written as

$T-60g=60\left(\frac{5g}{4}-a\right)$ $\left(1\right)$

Hence, the equation of motion of the object can be written as

$T-100g=100a$ $\left(2\right)$

Step 3: Solve the Equations

Upon subtracting equation $\left(2\right)$ from equation $\left(1\right)$ we get,

$40g=60\times \frac{5g}{4}-60a-100a$

$\Rightarrow 40g=75g-160a$

$\Rightarrow 160a=35g$

$\Rightarrow a=\frac{35g}{160}$

Step 4: Calculate the Tension

Upon substituting the values of $g$ and $a$ equation $\left(2\right)$ we get,

$T=100g+100a$

$\Rightarrow T=100\left(10+\frac{35\times 10}{160}\right)$

$=\frac{100}{16}\left(160+35\right)$

$=\frac{100}{16}\times 195$

$=1218.75\approx 1219N$

Hence, the tension of the rope is $1219N$.