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Laws of Motion

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Question : 33 of 40
Marks: +1, -0
A monkey of mass 40 kg climbs on a rope as shown in the figure 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^{-2}
(b) climbs down with an acceleration of 4 m s−2^{-2}
(c) climbs up with a uniform speed of 5 m s−1^{-1}
(d) falls down the rope nearly freely under gravity?
(Ignore the mass of the rope).
Solution:  
Here, mass of monkey m=40m = 40 kg
Maximum tension the rope can stand, T=600T = 600 N.
In each case, actual tension in the rope will be equal to apparent weight of monkey (R).
The rope will break when R exceeds T.
(a) When monkey climbs up with a = 6 m s−2^{-2},
R=m(g+a)R = m(g + a) =40(10+6)=640 N= 40(10 + 6) = 640\,\text{N} (which is greater than T)
Hence the rope will break.
(b) When monkey climbs down with a = 4 m s−2^{-2},
R=m(g−a)R = m(g - a) =40(10−4)=240 N= 40(10 - 4) = 240\,\text{N} (which is less than T)
∴ The rope will not break.
(c) When monkey climbs up with a uniform speed v=v = m s−1^{-1}, its acceleration a=0a = 0
∴R=mg=40×10=400 N\therefore R = mg = 40 \times 10 = 400\,\text{N} (which is less than T)
∴ The rope will not break.
(d) When monkey falls down the rope nearly freely under gravity,
a=ga = g
∴ R=m(g−a)=m(g−g)=R = m(g - a) = m(g - g) = zero
Hence the rope will not break.
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