Q. 334.3( 26 Votes )

# 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).

Answer :

Given:

Mass of monkey = 40Kg

Maximum tension that rope can withstand, T_{m} = 600 N

We need to find the case in which rope will break, i.e. the force should be more than 600 N.

(A) acceleration (Upwards) = 6ms^{-2}

The monkey is moving in upward direction,

From Newton’s 2^{nd} law we can write,

F = ma = T- mg

→ ma = T-mg

→ T = m(a + g)

→ T = 40 Kg( 6 ms^{-2} + 10 ms^{-2})

→ T = 640 N

Since T> T_{m}, the rope is destined for a failure, it will break.

(B) Acceleration (Downwards) = 4 ms^{-2}

The monkey is moving in upward direction,

From Newton’s 2^{nd} law we can write,

F = ma = mg-T

→ ma = mg-T

→ T = m (g - a)

→ T = 40 Kg (10 ms^{-2} - 4 ms^{-2})

→ T = 240 N

Since T< T_{m}, the rope will not break.

(B) Acceleration, a = 0ms^{-2}

Climbing velocity, v = 5ms^{-1}

Using Newton’s 2^{rd} law we can write,

F = ma = T-mg

→ ma = T-mg

→ T = mg

→ T = 40Kg × 10 ms^{-2}

→ T = 400 N

Since T<T_{m}, the rope will not break.

(C) If the monkey is free falling, i.e. a = g;

Using Newton’s 2^{nd} law of motion, we can write,

F = mg-T = mg

T = mg-mg

T = 0 N

Since, T< T_{m}, the rope will not break.

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