Question

Three monkeys $A,B$ and $C$ with masses of $10,15$ & $8kg$ respectively are climbing up & down the rope suspended from $D$. At the instant represented, $A$ is descending the rope with an acceleration of $2m/s^2$ & $C$ is pulling himself up with acceleration of $1.5m/s^2$. Monkey $B$ is climbing up with a constant speed of $0.8m/s$. Calculate the tension $T$ in the rope at $D$. $(g=10m/s^{-2})$

Answer 1

From the FBD :

For monkey C, $T_1-m_{C}g=m_C\times 1.5$ ($a_C=1.5\frac{m}{s^2}$)

that gives : $T_1=m_c(1.5+10)=8\times 11.5=92N$ (g=10 is used)

For monkey B: $T_2=T_1+m_{B}g=92+150=242N$ (since B moves with constant speed , force should be balanced)

for monkey C: $T_3-T_2-m_{A}g=m_A\times -2$, ($a_A=2\frac{m}{s^2}$ in downward direction)

we get $T_3=T_2+m_A(10-2)=242+80=322N$

Here we can see that $T_3=322N$ is the required tension at point D