Question

A monkey of mass 40 kg climbs on a rope (as shown in the figure) which can stand a maximum tension of 600 N. In which of the following cases will the rope brake: 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 s⁻¹
d. falls down the rope nearly freely under gravity?
(Ignore the mass of the rope).

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, Tmax = 600 N
Acceleration of the monkey, a = 6 m/s2upward
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 > Tmax, the rope will break in this case.

Case (b)
Acceleration of the monkey, a = 4 m/s2downward
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 < 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, we can write the equation of motion as:
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, 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 < Tmax, the rope will not break in this case.

Answer 2

Hey mate,

# Answer - Option (a)

## Explaination- For case (a) -
# Given-
m = 40 kg
g = 10 m/s
Tmax = 600 N
a = 6 m/s^2

# Solution-
By applying Newton’s second law of motion, we can calculate tension in the string as
T – mg = ma
T = m (g + a)
T = 40 (10 + 6)
T = 640 N

Since T > Tmax, the rope will break when acceleration is 6 m/s^2.

[ Note- In all other cases tension produced is less than maximum bearable tension, thus rope won't break. ]

Hope this helps you...