A-LevelMathematicsMechanics9709/52 · Feb/Mar 2017 · Q47 Marks
The diagram shows a uniform lamina ABCD with AB = 0.75 m, AD = 0.6m and BC = 0.9 m. Angle BAD = angle ABC = 90°. [Figure X.X]
- (i)Show that the distance of the centre of mass of the lamina from AB is 0.38 m, and find the distance of the centre of mass from BC.[5]
- (ii)The lamina is freely suspended at B and hangs in equilibrium. Find the angle between BC and the vertical.[2]
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Mark scheme
Marking points from the official mark scheme (Feb/Mar 2017). The mark or mark type is shown in brackets.
- [B1] A = total area of the lamina (A = 0.6 x 0.75 + 0.3 x 0.75/2 (= 0.5625))
- [M1] Takes moments about AB
- [AG] [A1] x = 0.38 m (from AB)
- [M1] Takes moments about BC
- [A1] y = 0.35 m (from BC)
- [M1] tanθ= y/x where θ is the required angle
- [A1] θ=42.6°
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