The monkey B, shown in the figure (5−E20), is holding on to the tail of monkey A that is climbing up a rope. The masses of monkeys A and B are 5 kg and 2 kg, respectively. If A can tolerate a tension of 30 N in its tail, what force should it apply on the rope in order to carry monkey B with it? Take g = 10 m/s2.
Answers
Given :
Let the acceleration of monkey A upwards be a, so that a maximum tension of 30 N is produced in its tail.
T − 5g − 30 − 5a = 0 …(i)
30 − 2g − 2a = 0 …(ii)
From equations (i) and (ii), we have:
T = 105 N (max.)
and a = 5 m/s2
So, A can apply a maximum force of 105 N on the rope to carry monkey B with it.
For minimum force, there is no acceleration of A and B.
T1 = weight of monkey B
⇒ T1 = 20 N
Rewriting equation (i) for monkey A, we get:
T − 5g − 20 = 0
⇒ T = 70 N
∴ To carry monkey B with it, monkey A should apply a force of magnitude between 70 N and 105 N
Answer: minimum 70N and maximum 105N.
Explanation:
Given ,mass of monkey A =5kg and mass of monkey B is 2kg.
Here, it's also stated that A can tolerate the maximum tension T=30N.
In case of maximum tension:
If it goes up with acceleration a then ,
T-mg=ma
>a={T-mg}/m
>a={30 -2 ×10}/2 [2=mass of monkey B]
>a=5m/s²
So,
We can treat both monkeys as a system,
In such case,
Tension (force applied) will be max.
T-(m1+m2)g=(m1+m2)a
>T=(m1+m2)(g+a)
>T=(2+5)(10+5)=105N
For second case where tension is minimum ,i.e. acceleration is =0
So,
T=
(m1+m2)(g+a)>(2+5)×10=70N.
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