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, 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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