How to solve this Question regarding compound accelerations

Assuming there is no friction.

First of all lets define the coordinate systems. In the image there is the inertal system (lets denote it Oxyz) and there is the parallel axis accelerating system with the 10kg mass (let's denote it as P - from the pulley). In general, the following equations hold:

$$x_B = x_A + x_{B|A}, qquad ddot{x}_B = ddot{x}_A + ddot{x}_{B|A}$$

mass 2

On mass 2 you will have only 3 forces. The weight $m_2 g$, the tension of the rope $T$ and the reaction $N$ from the contact between mass 2 and mass 10. the following equation can be written:

$$m_2 ddot{x}_{2|O} = N $$ $$m_2 ddot{y}_{2|O} = T-m_2g $$

where:

• $ddot{x}_{2|O}$ the horizontal acceleration on mass 2 wrt the inertial system O. It is worth noting that $ddot{x}_{2|O} = ddot{x}_{10|O}$ ,because mass 2 moves horizontally attached to mass 10 (when they are moving in the same direction).
• $ddot{y}_{2|O}=4 [m/s2]$ the vertical acceleration on mass 2 wrt the inertial system O

therefore the above two equations can be rewritten (assuming gravity is g=10m/s2) as:

$$m_{2} ddot{x}_{10|O} = N $$ $$T = m_2 (g+ ddot{y}_{2|O}) =28 N$$

mass 3

On mass 3 only the wire force T is applied. In the system P ( which is moving with the pulley) we can see that: $$ ddot{x}_{3|P} =ddot{y}_{2|P} = 4[m/s^2] $$

Therefore, the acceleration of mass 3 wrt to O would be: $$ ddot{x}_{3|O} = ddot{x}_{P|O} + ddot{x}_{3|P} $$

where: $ddot{x}_{P|O} = ddot{x}_{10|O} = ddot{x}_{10} $

So on the inertial system the following equation holds $$ m_3 ddot{x}_{3|0} = -T $$ $$ m_3 (ddot{x}_{3|P} + ddot{x}_{10|O}) = -T $$ $$ m_3 (4[m/s^2] + ddot{x}_{10} ) = -T $$ $$ ddot{x}_{10} = -4[m/s^2] - frac{T}{m_3} $$ $$ ddot{x}_{10} = -frac{40}{3}[m/s^2]$$

mass 10

On the large mass the following equation describes the motion:

$$m_{10} ddot{x}_{10} = F -N-T$$

That can be rearranged to:

$$ F = m_{10} ddot{x}_{10} +N + T = m_{10} ddot{x}_{10} +m_{2} ddot{x}_{10|O} + T= m_{10} ddot{x}_{10} +m_{2} ddot{x}_{10} +T $$

$$ F = (m_{10} +m_2) ddot{x}_{10} + T=188 [N] $$