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\includegraphics[max width=\textwidth, alt={}, center]{f8633b64-b20c-4471-9641-ccc3e6854f2c-3_784_556_260_790} A uniform object is made by drilling a cylindrical hole through a rectangular block. The axis of the cylindrical hole is perpendicular to the cross-section \(A B C D\) through the centre of mass of the object. \(A B = C D = 0.7 \mathrm {~m} , B C = A D = 0.4 \mathrm {~m}\), and the centre of the hole is 0.1 m from \(A B\) and 0.2 m from \(A D\) (see diagram). The hole has a cross-section of area \(0.03 \mathrm {~m} ^ { 2 }\).

1. Show that the distance of the centre of mass of the object from \(A B\) is 0.212 m , and calculate the distance of the centre of mass from \(A D\). The object has weight 70 N and is placed on a rough horizontal surface, with \(A D\) in contact with the surface. A vertically upwards force of magnitude \(F \mathrm {~N}\) acts on the object at \(C\). The object is on the point of toppling.
2. Find the value of \(F\). The force acting at \(C\) is removed, and the object is placed on a rough plane inclined at an angle \(\theta ^ { \circ }\) to the horizontal. \(A D\) lies along a line of greatest slope, with \(A\) higher than \(D\). The plane is sufficiently rough to prevent sliding, and the object does not topple.
3. Find the greatest possible value of \(\theta\).