A) between 50 N and \[80N\]
B) between 30 N and \[80N\]
C) between 70 N and \[150N\]
D) between 60 N and \[90N\]
Correct Answer: C
Solution :
The monkey \[B\] (the child of monkey\[A\]) is stationary with respect to the monkey\[A\], and both move with some common acceleration\[,\]\[a\]. Now, the force with which monkey \[B\] should be pulled upwards so as to be lifted with acceleration a must be equal to, \[m(g+a)\] where m is the mass of monkey \[B\] (child). Since, the maximum tension in the tail of monkey\[A\]can be\[30\,\,N\], So, \[30=m(g+a)=2(10+a)\] or \[a=5m{{s}^{-2}}\] This is the maximum acceleration with which both can move upwards as a combined system. If acceleration is greater than this value, the tail of monkey A will break. Now, the force which the monkey \[A\] should apply on the rope so that the monkey \[A\] gets this force acted upon him as the reaction force must be sufficient to lift monkey \[A\] and the child, monkey \[B\] with the common acceleration of\[5\,\,m{{s}^{-2}}\] \[{{F}_{\max }}=(M+m)(g+a)=7\times 15=105\,\,N\] This is the maximum force with which the monkey A should pull the rope so that the combined system moves with maximum acceleration. The condition of minimum force acting on the combined system of the monkey and child will arise when both move up with zero acceleration. In that case\[,\]\[{{F}_{\min }}=(M+m)g=70\,\,N\]. Thus, the monkey A should pull the rope with the force of 70 N to move up with maximum acceleration, such that the tension in his tail is just equal to the breaking strength of the tail.
You need to login to perform this action.
You will be redirected in
3 sec
