M2 June 2007 Q3
3.
A uniform lamina \(ABCDEF\) is formed by taking a uniform sheet of card in the form of a square \(AXEF\), of side \(2a\), and removing the square \(BXDC\) of side \(a\), where \(B\) and \(D\) are the mid-points of \(AX\) and \(XE\) respectively, as shown in Figure 1.
The lamina is freely suspended from \(A\) and hangs in equilibrium.
| Scheme | Marks |
|---|---|
| M\((AF)\) \(4a^2 \cdot a - a^2 \cdot 3a/2 = 3a^2 \cdot \bar{x}\) | M1 A2,1,0 |
| \(\bar{x} = 5a/6\) | A1 |
| (4) |
Notes
M1 Taking moments about AF or a parallel axis, with mass proportional to area. Could be using a difference of two square pieces, as above, but will often use the sum of a rectangle and a square to make the L shape. Need correct number of terms but condone sign errors for M1.
A1 A1 All correct
A1 A0 At most one error
A1 \(5a/6\), (accept \(0.83a\) or better)
Condone consistent lack of \(a\)’s for the first three marks.
NB: Treating it as rods rather than as a lamina is M0
| Scheme | Marks |
|---|---|
| Symmetry \(\Rightarrow \bar{y} = 5a/6\), or work from the top to get \(7a/6\) | B1ft |
| \(\tan\theta = \dfrac{5a/6}{2a - 5a/6}\) \(\left(\dfrac{\bar{x}}{2a - \bar{y}}\right)\) | M1 A1ft |
| \(\theta \approx 35.5^\circ\) | A1 |
| (4) | |
| (8 marks) |
Notes
B1ft \(\bar{x} = \bar{y} =\) their \(5a/6\), or \(\bar{y} =\) distance from \(AB = 2a -\) their \(5a/6\). Could be implied by the working. Can be awarded for a clear statement of value in (a).
M1 Correct triangle identified and use of tan. \(\dfrac{2a - 5a/6}{5a/6}\) is OK for M1. Several candidates appear to be getting \(45^\circ\) without identifying a correct angle. This is M0 unless it clearly follows correctly from a previous error.
A1ft Tan \(\alpha\) expression correct for their \(5a/6\) and their \(\bar{y}\)
A1 35.5 (Q asks for 1d.p.)
NB: Must suspend from point A. Any other point is not a misread.