M2 June 2010 Q3
3.
A triangular frame is formed by cutting a uniform rod into 3 pieces which are then joined to form a triangle \(ABC\), where \(AB = AC = 10\) cm and \(BC = 12\) cm, as shown in Figure 1.
(a) Find the distance of the centre of mass of the frame from \(BC\). (5)
The frame has total mass \(M\). A particle of mass \(M\) is attached to the frame at the mid-point of \(BC\). The frame is then freely suspended from \(B\) and hangs in equilibrium.
(b) Find the size of the angle between \(BC\) and the vertical. (4)
| Scheme | Marks | |||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| B1 B1 | |||||||||||||||
| Moments about \(BC\): | ||||||||||||||||
| \(10 \times 4 + 10 \times 4 + 0 = 32\bar{x}\) | M1 A1 | |||||||||||||||
| \(\bar{x} = \dfrac{80}{32}\) | ||||||||||||||||
| \(\bar{x} = 2\frac{1}{2}\quad (2.5)\) | A1 | |||||||||||||||
| (5) |
| Scheme | Marks |
|---|---|
| Moments about \(B\): | |
| \(Mg \times 6\sin\theta = Mg \times \left(\bar{x}\cos\theta - 6\sin\theta\right)\) | M1 A1 A1 |
| \(12\sin\theta = \bar{x}\cos\theta\) | |
| \(\tan\theta = \dfrac{\bar{x}}{12}\) | |
| \(\theta = 11.768\ldots = 11.8^\circ\) | A1 |
| (4) | |
| (9 marks) |
Alternative method
| C of M of loaded frame at distance \(\frac{1}{2}\bar{x}\) from \(D\) along \(DA\) | B1 |
| \(\tan\theta = \dfrac{\frac{1}{2}\bar{x}}{6}\) | M1 A1 |
| \(\theta = 11.768\ldots = 11.8^\circ\) | A1 |