Question

One end of massless rope, which passes over a massless and frictionless pulley $P$ is tied to a hook while the other is free. Maximum tension the rope bear is $360 \,N$. With what value of maximum safe acceleration (in $ms ^{-2}$ ) can a man of $60\, kg$ climb on the rope ?

• A. 16
• B. 6
• C. 4
• D. 8

Answer 1

Answer: (C)

The free body diagram of the person can be drawn as

Let the person moves up with an acceleration ‘ a’
then $T - 60g = 60a$
$\Rightarrow a_{\max }=\frac{T_{\max }-}{60g}$
$=\frac{360-60 g}{60}=-v e$
Which means it is not possible to climb up on the rope. Even in this problem it is not possible to remain at rest on rope. No option is right. But if they will ask for the acceleration of climbing down then

$60 g-T=60 a$
$\Rightarrow 60 g-T_{\max }=60 a_{\min }$
$\therefore a_{\min }=\frac{60 g-360}{60}=4 m / s ^{2}$