As the elevator is going upward with acceleration \(\frac{9}{2}\)

so \(2\) log block with force \(2 \times \frac{9}{2}\) downward and \(4 \log\) block with force \(4 \times \frac{9}{2}\) downward as a prudes force.

\(\Rightarrow\) weight \(+\) pruedo \(- T = ma\) or \(4 g+2 g T=4 a+1)\)

\(\Rightarrow T -(2 g +9)=2 a\)

\(f = ma\)

\(T-3 g=2 a-(2)\)

\((1) \times 1(2) \times 2\)

\(69-T-(2 T-69)=+a+a\)

\(6 g-3 T+6 g=0\)

\(T =\frac{12 g }{3}=4 g =40\,N\)