Pushing a block on a moving incline I

Consider the free body diagrams for the inclined plane and the wedge:

Note that the acceleration $a_1$ is the acceleration of the wedge along the incline and with respect to the incline. The force $m_1a_2$ acting on the wedge is a pseudo force, where $a_2$ is the horizontal acceleration of the incline with respect to the ground. Based on the FBDs, six equations can be obatined as such:

$f_1 = \mu_{12} N_1 \ \dots \ (1)$ $f_2 = \mu_{floor} N_2 \ \dots \ (2)$ $-f_1 \sin{\theta} + N_1 \cos{\theta} + m_2g = N_2 \ \dots \ (3)$ $f_1 \cos{\theta} + N_1 \sin{\theta} - f_2 = m_2a_2 \ \dots \ (4)$ $F \cos{\theta} - m_1g \sin{\theta} -f_1 -m_1a_2 \cos{\theta} = m_1a_1 \ \dots \ (5)$ $F \sin{\theta} + m_1g \cos{\theta}- m_1a_2 \sin{\theta} = N_1 \ \dots \ (6)$

Equation (3) is the equilibrium equation for the incline along the vertical. Equation (4) is the application of Newton's second law along the horizontal. Equation (5) is the application of Newton's law on the wedge along the incline while (6) is the equilibrium equation for the wedge perpendicular to the incline. Solving the equations gives answers for $a_1$ and $a_2$ . Recall that the acceleration of the wedge in the inertial frame of reference is asked for. By applying basic kinematic relations gives:

$\boxed{a_{res} = \sqrt{\left(a_1 \cos{\theta} + a_2\right)^2 + \left(a_1 \sin{\theta}\right)^2} \approx 22.83 \ m/s^2}$

Brilliant does not recognise the above value as the correct answer. It took me more than one try to realise that it recognises 22.8 as the answer. In case there is a mistake in my solution, I request that it be pointed out.

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Edit:

To solve the system of equations with $a_1$ , $a_2$ , $N_1$ and $N_2$ as unknown variables, I employ the use of linear algebra as that can be handled by a computer quite easily. I did not make an effort to compute a closed-form expression. The system of equations can be simplified a bit and re-written as:

Let: $\mu_1 = \mu_{12} \ ; \ \mu_2 = \mu_{floor}$

$\left[\begin{array}{cccc} \cos\left(\mathrm{\theta}\right)-\mu _{1}\,\sin\left(\mathrm{\theta}\right) & -1 & 0 & 0\\ \sin\left(\mathrm{\theta}\right)+\mu _{1}\,\cos\left(\mathrm{\theta}\right) & -\mu _{2} & 0 & -m_{2}\\ -\mu _{1} & 0 & -m_{1} & -m_{1}\,\cos\left(\mathrm{\theta}\right)\\ -1 & 0 & 0 & -m_{1}\,\sin\left(\mathrm{\theta}\right) \end{array}\right] \left[\begin{matrix} N_1\\N_2\\a_1\\a_2 \end{matrix}\right] = \left[\begin{array}{c} -\,m_{2}g\\ 0\\ -F\,\cos\left(\mathrm{\theta}\right)+\,m_{1}g\,\sin\left(\mathrm{\theta}\right)\\ -F\,\sin\left(\mathrm{\theta}\right)-\,m_{1}g\,\cos\left(\mathrm{\theta}\right) \end{array}\right]$

$\implies \left[\begin{matrix} N_1\\N_2\\a_1\\a_2 \end{matrix}\right] =\left[\begin{array}{cccc} \cos\left(\mathrm{\theta}\right)-\mu _{1}\,\sin\left(\mathrm{\theta}\right) & -1 & 0 & 0\\ \sin\left(\mathrm{\theta}\right)+\mu _{1}\,\cos\left(\mathrm{\theta}\right) & -\mu _{2} & 0 & -m_{2}\\ -\mu _{1} & 0 & -m_{1} & -m_{1}\,\cos\left(\mathrm{\theta}\right)\\ -1 & 0 & 0 & -m_{1}\,\sin\left(\mathrm{\theta}\right) \end{array}\right]^{-1}\left[\begin{array}{c} -\,m_{2}g\\ 0\\ -F\,\cos\left(\mathrm{\theta}\right)+\,m_{1}g\,\sin\left(\mathrm{\theta}\right)\\ -F\,\sin\left(\mathrm{\theta}\right)-\,m_{1}g\,\cos\left(\mathrm{\theta}\right) \end{array}\right]$

This is how I solved it earlier on a computer to obtain the accelerations, following which I got the arrived at the answer by computing the resultant.

Write down Newton 2nd law equations for $(x,y)$ both for the wedge and the incline. Note that we do not know a priori the direction of the acceleration of the wedge since the plane is also moving!... Note also that the acceleration of the wedge with respect to the ground is equal to the acceleration of the wedge with respect to the plane plus the acceleration of the plane with respect to the ground: $\bold a_1 = \bold a_{12} + \bold a_2$ . Hint: What direction makes the vector $\bold a_{12}$ with the horizontal?