Question

Three monkeys A, B, and C weighing 20, 26, and 25 lb, respectively, are climbing up and down the rope suspended from D. At the instant represented, A is descending the rope with an acceleration of 4.2 ft/sec2, and C is pulling himself up with an acceleration of 5.4 ft/sec2. Monkey B is climbing up with a constant speed of 4.0 ft/sec. Treat the rope and monkeys as a complete system and calculate the tension T in the rope at D.

Answer 1

Answer:$2235.2lb-ft/s^2$

Explanation:

Given

Mass of monkey A=20lb

Mass of monkey B=26lb

Mass of monkey C=25lb

acceleration of monkey A=$4.2ft/s^2$

acceleration of monkey B=0

acceleration of monkey C=$-5.4ft/s^2$

Force Due to monkey A$\left ( F_A\right )=20\times 4.2 =84lb-ft/s^2\left ( downwards\right )$

Force Due to monkey A$\left ( F_B\right )=26\times 0 =0lb-ft/s^2$

Force Due to monkey A$\left ( F_C\right )=25\times 5.4 =135 lb-ft/s^2\left ( upward\right )$

In addition to it  Weights of monkeys will be acting downwards therefore net Downwards force is balanced by tension

T=$\left ( 20+26+25\right )32.2+84-135=2235.2 lb-ft/s^2$