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/s2upward
UsingNewton'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
UsingNewton'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.
UsingNewton's second law of motion, we can write the equation of motion as:
T- mg = ma
T- mg = 0
∴T = mg
= 40 x 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
UsingNewton'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.
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