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 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).

Fig. 5.20

Answer 1

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 N

Acceleration of the monkey, a = 6 m/s² upward

Using Newton’s 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.

Case (b)

Acceleration of the monkey, a = 4 m/s² downward

Using Newton’s 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.

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

∴T = mg

= 40 × 10

= 400 N

Since T < T_max, the rope will not break in this case.

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

∴T = m(g – g) = 0

Since T < T_max, the rope will not break in this case.