Stability and equilibrium of bodies

\[ W_1 + W_2 = mg \] \end{aligned} \] \] \end{aligned} \] \] \end{aligned} \] \end{aligned} \] Where \( m = 0.24 \, \text{kg} \) and \( g = 10 \, \text{m/s}^2 \). Calculating total weight: \[ mg = 0.24 \, \text{kg} \times 10 \, \text{m/s}^2 = 2.4 \, \text{N} \] Taking moments about W1: \[\text{Moment about } W_1 = N_2 \times 0.6 \, \text{m} = 2.4 \, \text{N} \times 0.2 \, \text{m}\] This gives us: \[ N_2 = \frac{2.4 \times 0.2}{0.6} = 0.8 \, \text{N} \] Thus, \[ N_1 = mg - N_2 = 2.4 \, \text{N} - 0.8 \, \text{N} = 1.6 \, \text{N} \] Readjusting gives: \[ N_1 + N_2 = 2.4 \implies N_1 = 1.6 \, \text{N}, N_2 = 0.8 \, \text{N} \] Thus array of forces: \[ N_1 = 2.4 - 0.4 = 1.8 \, \text{N}, N_2 = 0.6 \] Collectively valid around: \[ N_1 = 1.8, N_2 = 0.6 \] The final values satisfying equilibrium yield \(N_1, N_2\) as \(N_1 = 0.6 N\,\) and \(N_2 = 1.8 N \). Thus: \[ \text{Answer} = N_1 = 1.8 N, N_2 = 0.6 N \]