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Class 10 Class 12
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).

Case (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/s2 , upward

Using Newton’s second law of motion, equation of motion is

T – mg = ma

∴                          T = m(g + a

= 40 (10 + 6)

= 640 N

Since T > Tmax, 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, the equation of motion is,

mg – ma

∴                                        T = (g – a

= 40(10 – 4)

= 240 N

Since T < Tmax, 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, equation of motion is,

T – m= ma

T – mg = 0

∴                           m

= 40 × 10

= 400 N

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

Case (d)

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

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

mg – T = m

∴                                 T = m(g – g) = 0

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