4
\includegraphics[max width=\textwidth, alt={}, center]{84a2b2eb-a750-4047-864b-4a165fc66b2a-3_611_977_260_584}
\(A B C D E F\) is the cross-section through the centre of mass of a uniform solid prism. \(A B C F\) is a rectangle in which \(A B = C F = 1.6 \mathrm {~m}\), and \(B C = A F = 0.4 \mathrm {~m}\). \(C D E\) is a triangle in which \(C D = 1.8 \mathrm {~m}\), \(C E = 0.4 \mathrm {~m}\), and angle \(D C E = 90 ^ { \circ }\). The prism stands on a rough horizontal surface. A horizontal force of magnitude \(T \mathrm {~N}\) acts at \(B\) in the direction \(C B\) (see diagram). The prism is in equilibrium.
- Show that the distance of the centre of mass of the prism from \(A B\) is 0.488 m .
- Given that the weight of the prism is 100 N , find the greatest and least possible values of \(T\).