The forces at the hinge would be in the \(xy\) plane. Let the \(x\) component be \(F_2\) and the \(y\) component \(F_1\).

* If the lower rod was not there the plates would have rotated inwards towards each other.

We get forces on the lower rod to be same as \(F_2\) as the plates don’t accelerate in the horizontal direction.

Analysis of Forces

From the free body diagram of the upper rod:

\[ F = 2F_1 \Rightarrow F_1 = \frac{F}{2} \]

Taking Torque about the Center of Mass (C.O.M.) of the plates:

\[ F_1 \times \frac{l}{2} = 2 \times F_2 \times \frac{l}{2} \] \[ \Rightarrow F_2 = \frac{F_1}{2} = \frac{F}{4} \]

a) Resultant Forces

b) Accelerating Case (\(F = 4ma\))

For the upper rod:

\[ F – 2F_1 = ma \]

Given that the total force accelerating the system is \(F = 4ma\) (since there are 4 masses involved implicitly or given in problem context):

\[ 4ma – 2F_1 = ma \] \[ 2F_1 = 3ma \Rightarrow F_1 = \frac{3ma}{2} \]

Substituting \(a = \frac{F}{4m}\):

\[ F_1 = \frac{3}{2} \times \frac{F}{4} = \frac{3F}{8} \]

Taking Torque about the centre:

\[ 2 \times F_2 \frac{l}{2} + F_3 \frac{l}{2} = F_1 \frac{l}{2} \]

For the bottom rod:

\[ 2F_3 = ma = \frac{F}{4} \] \[ \Rightarrow F_3 = \frac{F}{8} \]

Substituting \(F_3\) back into the torque equation:

\[ 2F_2 = F_1 – F_3 = \frac{3F}{8} – \frac{F}{8} \] \[ 2F_2 = \frac{2F}{8} \Rightarrow F_2 = \frac{F}{8} \]

Calculated Resultants for Case (b)

Force at A (\(F_A\)):

\[ F_A = \sqrt{F_1^2 + F_2^2} = \sqrt{\left(\frac{3F}{8}\right)^2 + \left(\frac{F}{8}\right)^2} \] \[ F_A = \sqrt{\frac{9F^2}{64} + \frac{F^2}{64}} = \sqrt{\frac{10F^2}{64}} = \frac{\sqrt{10}F}{8} \]

Force at B (\(F_B\)):

\[ F_B = \sqrt{F_2^2 + F_3^2} = \sqrt{\left(\frac{F}{8}\right)^2 + \left(\frac{F}{8}\right)^2} \] \[ F_B = \frac{\sqrt{2}F}{8} \]