A monkey of mass 40 kg climbs on a rope (Fig. 5.20) which can stand
Question.
A monkey of mass 40 kg climbs on a rope (Fig. 5.20) which can stand a maximum tension of 600 N. In which of the following cases will the rope break: the monkey

(a) climbs up with an acceleration of $6 \mathrm{~m} \mathrm{~s}^{-2}$

(b) climbs down with an acceleration of $4 \mathrm{~m} \mathrm{~s}^{-2}$

(c) climbs up with a uniform speed of $5 \mathrm{~m} \mathrm{~s}^{-1}$

(d) falls down the rope nearly freely under gravity?

(lgnore the mass of the rope).

solution:

Case (a)

Mass of the monkey, m = 40 kg

Acceleration due to gravity, g = 10 m/s

Maximum tension that the rope can bear, $T_{\max }=600 \mathrm{~N}$

Acceleration of the monkey, $a=6 \mathrm{~m} / \mathrm{s}^{2}$ upward

Using Newton’s second law of motion, we can write the equation of motion as:

$T-m g=m a$

$\therefore T=m(g+a)$

$=40(10+6)$

$=640 \mathrm{~N}$

Since $T>T_{\max }$, the rope will break in this case.

Case (b)

Acceleration of the monkey, $a=4 \mathrm{~m} / \mathrm{s}^{2}$ downward

Using Newton’s second law of motion, we can write the equation of motion as:

$m g-T=m a$

$\therefore T=m(g-a)$

$=40(10-4)$

$=240 \mathrm{~N}$

Since $T Case (c) The monkey is climbing with a uniform speed of 5 m/s. Therefore, its acceleration is zero, i.e., a = 0. Using Newton’s second law of motion, we can write the equation of motion as: T – mg = ma T – mg = 0$\therefore T=m g=40 \times 10$= 400 N Since$T
Case (d)

When the monkey falls freely under gravity, its will acceleration become equal to the acceleration due to gravity, i.e., a = g

Using Newton’s second law of motion, we can write the equation of motion as:

mg – T = mg

$\therefore T=m(g-g)=0$

Since \$T
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