Angle of inclination of suspended bodies — A/L Combined Mathematics

How do I practise Angle of inclination of suspended bodies?

1 question with a worked solution

Check before you answer: Vertical Line Alignment Principle: Stating explicitly that the suspension point O and centre of mass G lie on the same vertical line.

1. When a rigid body is freely suspended from a fixed pivot point P, what geometric condition must be satisfied by its center of gravity G in static equilibrium?
   1. A) The line PG must be inclined at exactly 45° to the horizontal
   2. B) The center of gravity G must lie vertically directly below the suspension point P
   3. C) The center of gravity G must lie on the horizontal plane passing through P
   4. D) The point P must coincide with the geometric centroid of the top surface
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   Correct answer: B) The center of gravity G must lie vertically directly below the suspension point P

   For equilibrium under two forces (the upward reaction at the suspension point P and the downward weight acting through center of gravity G), the two forces must be collinear. Hence, the line segment PG must be vertical, placing G directly below P.

   Show worked solution

   Step 1: Equilibrium under 2 forces (reaction at P and weight at G). Step 2: Forces must be collinear and opposite => Line PG is vertical, G is below P. Answer: The center of gravity G must lie vertically directly below the suspension point P.
   Modelled on: the A/L Combined Mathematics examination pattern

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Where does Angle of inclination of suspended bodies fit in the syllabus?

Year 2 · Term 2