M1 June 2012 Q2
2.

A non-uniform rod \(AB\) has length 3 m and mass 4.5 kg. The rod rests in equilibrium, in a horizontal position, on two smooth supports at \(P\) and at \(Q\), where \(AP = 0.8\) m and \(QB = 0.6\) m, as shown in Figure 1. The centre of mass of the rod is at \(G\). Given that the magnitude of the reaction of the support at \(P\) on the rod is twice the magnitude of the reaction of the support at \(Q\) on the rod, find

| Scheme | Marks |
|---|---|
| \(\uparrow\) \(2X + X = 4.5g\) | M1 A1 |
| Leading to \(X = \dfrac{3g}{2}\) or 14.7 or 15 (N) | A1 |
| (3) |
Notes
First M1 for a complete method for finding \(R_Q\), either by resolving vertically, or taking moments twice, with usual criteria (allow M1 even if \(R_P = 2R_Q\) not substituted)
First A1 for a correct equation in either \(R_Q\) or \(R_P\) ONLY.
Second A1 for 1.5g or 14.7 or 15 (A0 for a negative answer)
| Scheme | Marks |
|---|---|
| M\((A)\) \(4.5g \times AG = (2X) \times 0.8 + X \times 2.4\) | M1 A2 ft (1,0) |
| \(AG = \dfrac{4}{3}\) (m), 1.3, 1.33,... | A1 |
| (4) | |
| (7 marks) |
Notes
First M1 for taking moments about any point, with usual criteria.
A2 ft for a correct equation (A1A0 one error, A0A0 for two or more errors, ignoring consistent omission of g’s) in terms of \(X\) and their \(x\) (which may not be \(AG\) at this stage)
Third A1 for \(AG = 4/3\), 1.3, 1.33,….. (any number of decimal places, since g cancels) need ‘\(AG =\)’ or \(x\) marked on diagram
N.B. if \(R_Q = 2R_P\) throughout, mark as a misread as follows:
(a) M1A1A0 (resolution method) (b) M1A0A1A1, assuming all work follows through correctly..