# 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 monkeyA. climbs up with an acceleration of 6 m s-2B. climbs down with an acceleration of 4 m s-2C. climbs up with a uniform speed of 5 m s-1D. falls down the rope nearly freely under gravity?(Ignore the mass of the rope).

Given:

Mass of monkey = 40Kg

Maximum tension that rope can withstand, Tm = 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 2nd 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> Tm, 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 2nd 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< Tm, the rope will not break.

(B) Acceleration, a = 0ms-2

Climbing velocity, v = 5ms-1

Using Newton’s 2rd law we can write,

F = ma = T-mg

ma = T-mg

T = mg

T = 40Kg × 10 ms-2

T = 400 N

Since T<Tm, the rope will not break.

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

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

F = mg-T = mg

T = mg-mg

T = 0 N

Since, T< Tm, the rope will not break.

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