As shown in the figure, monkey B is holding on to the tail of...

Let $a$ be the upward acceleration of the monkeys.

For monkey B, the equation of motion is:

$T - m_B g = m_B a$

Given $m_B = 2 \text{ kg}$ and $g = 10 \text{ m/s}^2$:

$T - 20 = 2a \Rightarrow a = \dfrac{T - 20}{2}$

For the system of monkeys A and B, the equation of motion is:

$F - (m_A + m_B)g = (m_A + m_B)a$

Given $m_A = 5 \text{ kg}$:

$F - 70 = 7a$

Substituting the value of $a$:

$F = 70 + 7 \left( \dfrac{T - 20}{2} \right) = 70 + 3.5(T - 20)$

To carry monkey B, the minimum upward acceleration is $a = 0$, which gives the minimum tension $T = 20 \text{ N}$.

The maximum tension monkey A's tail can tolerate is $T = 30 \text{ N}$.

For $T = 20 \text{ N}$, F_\min = 70 + 3.5(20 - 20) = 70 \text{ N}.

For $T = 30 \text{ N}$, F_\max = 70 + 3.5(30 - 20) = 105 \text{ N}.

Therefore, the force applied by monkey A should be between $70 \text{ N}$ and $105 \text{ N}$.