June 2024 Paper 2 Q17
17 A uniform rod is resting on two fixed supports at points \(A\) and \(B\).
\(A\) lies at a distance \(x\) metres from one end of the rod.
\(B\) lies at a distance \((x + 0.1)\) metres from the other end of the rod.
The rod has length \(2L\) metres and mass \(m\) kilograms.
The rod lies horizontally in equilibrium as shown in the diagram below.
The reaction force of the support on the rod at \(B\) is twice the reaction force of the support on the rod at \(A\).
Show that
\[L - x = k\]where \(k\) is a constant to be found. [4 marks]
| Scheme | Marks | AO |
|---|---|---|
| Forms a dimensionally correct moment in \(L\) and \(x\) May see : \((L - x)R\) \((L - (x + 0.1))2R\) \((L - x)mg\) \((2L - (2x + 0.1))2R\) OE | B1 | 3.3 |
| Forms dimensionally correct moments equation with at least one term correct May also see: \((L - x)mg = (2L - (2x + 0.1))2R\) Or \((L - (x + 0.1))mg = (2L - (2x + 0.1))R\) Condone missing brackets | M1 | 1.1a |
| Obtains a fully correct equation in \(L\) and \(x\) only | A1 | 1.1b |
| Completes reasoned argument to obtain \(L - x\) = 0.2 Must show expansion of all brackets before the final answer | R1 | 2.1 |
| (4 marks) |
Typical solution
The reaction at A is \(R\)
\[(L - x)R = (L - (x + 0.1))2R\]\[L - x = 2(L - (x + 0.1))\]\[L - x = 2L - 2x - 0.2\]\[L - x = 0.2\]