Question

A monkey of mass $40$ $kg$ climbs on a rope which can stand a maximum tension of $600$ $N$. In which of the following cases will the rope break. When the monkey

(a) Climbs up with an acceleration of $6$ $m{s}^{-2}$.

(b) Climbs down with an acceleration of $4$ $m{s}^{-2}$.

(c) Climbs up with a uniform speed of $5$ $m{s}^{-1}$.

(d) Falls down the rope nearly freely under gravity.

(Ignore the mass of the rope)

Answer 1

(a)

Mass of the monkey, m = 40 kg
Acceleration due to gravity, g = 10 m/s
Maximum tension that the rope can bear, Tmax = 600 N
Acceleration of the monkey, a = 6 m/s${}^2$ upward
Using Newtons second law of motion, we can write the equation of motion as:
T mg = ma
T = m(g + a)
= 40 (10 + 6)
= 640 N
Since T > T${}_{max}$, the rope will break in this case.

(b)
Acceleration of the monkey, a = 4 m/s${}^2$ downward
Using Newtons second law of motion, we can write the equation of motion as:
mg T = ma
T = m (g- a)
= 40(10-4)
= 240 N
Since T < T${}_{max}$, the rope will not break in this case.

(c)
The monkey is climbing with a uniform speed of 5 m/s. Therefore, its acceleration is zero, i.e., a = 0.
Using Newtons second law of motion, we can write the equation of motion as:
T - mg = ma
T-mg = 0
T = mg
= 40 $\times$ 10
= 400 N
Since T < T${}_{max}$, the rope will not break in this 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 Newtons second law of motion, we can write the equation of motion as:
mg + T = mg
T = m(g -g) = 0
Since T < T${}_{max}$, the rope will not break in this case.