A-LevelMathematicsMechanics9709/52 · Oct/Nov 2019 · Q711 Marks
ABCD is a uniform lamina in the shape of a trapezium which has centre of mass G. The sides AD and BC are parallel and 1.8 m apart, with AD = 2.4 m and BC = 1.2 m [Figure X.X].
- (i)Show that the distance of G from AD is 0.8 m.[4]
- (ii)The lamina is freely suspended at A and hangs in equilibrium with AD making an angle of 30° with the vertical. Calculate the distance AG.[2]
- (iii)With the lamina still freely suspended at A a horizontal force of magnitude 7 N acting in the plane of the lamina is applied at D. The lamina is in equilibrium with AG making an angle of 10° with the downward vertical. Find the two possible values for the weight of the lamina.[5]
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Mark scheme
Marking points from the official mark scheme (Oct/Nov 2019). The mark or mark type is shown in brackets.
- [B1] Rectangle: Area = 1.2 × 1.8 = 2.16, y = 1.8/2 = 0.9
- [B1] Triangle(s): Area = 1.2x=1.08, y=1.8/3=0.6
- [M1] (2.16 + 1.08)Y = 2.16 × 0.9 + 1.08 × 0.6
- [A1] Y = 0.8 m
- [M1] AGsin30 = 0.8
- [A1] AG = 1.6 m
- [B1] AD makes an angle of 40° or 20° with the vertical
- [M1] WAGsin10 = 7 × 2.4cos40
- [A1] W=46.3 N
- [M1] WAGsin10 = 7 × 2.4cos20
- [A1] W=56.8 N
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