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/s
2 , 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 – T = ma
∴ T = m (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 – mg = ma
T – mg = 0
∴ T = mg
= 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 = 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 < Tmax, the rope will not break in this case.
