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Question -

Amonkey of mass 40 kg climbs on a rope (Fig. 5.20) which can stand a maximumtension of 600 N. In which of the following cases will the rope break: themonkey

(a) climbs up with an acceleration of 6 m s–2

(b) climbs down with an acceleration of 4 ms–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 -

Case (a)

Mass of the monkey, m = 40kg

Acceleration due to gravity, g =10 m/s

Maximum tension that the rope canbear, Tmax = 600 N

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

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/s2 downward

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 speedof 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 × 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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