AS June 2018 Q1
1.
A thin uniform rod, of total length \(30a\) and mass \(M\), is bent to form a frame. The frame is in the shape of a triangle \(ABC\), where \(AB = 12a\), \(BC = 5a\) and \(CA = 13a\), as shown in Figure 1.
The frame is freely suspended from \(A\). A horizontal force of magnitude \(kMg\), where \(k\) is a constant, is applied to the frame at \(B\). The line of action of the force lies in the vertical plane containing the frame. The frame hangs in equilibrium with \(AB\) vertical.
| Scheme | Marks | AO |
|---|---|---|
| Complete strategy to find \(d\) | M1 | 3.1b |
| \(\dfrac{5}{30}M \times \dfrac{5}{2}a + \dfrac{13}{30}M \times \dfrac{5}{2}a = M \times d\) | A1 | 1.1b |
| \(\left(\dfrac{25}{2}a + \dfrac{65}{2}a = 30d\right)\) | A1 | 1.1b |
| \(90a = 60d \Rightarrow d = \dfrac{3}{2}a\) * | A1* | 2.1 |
| (4) |
Notes
M1: Complete strategy to find \(d\) e.g. moments about \(AB\) or a parallel axis. Needs all relevant terms. Must be dimensionally correct.
Condone sign errors. \(M\)'s might cancel from the start.
A1: Unsimplified equation with at most one error
A1: Correct unsimplified equation
A1*: Obtain the given answer from a convincing argument
| Scheme | Marks | AO |
|---|---|---|
| Complete strategy to find \(k\), e.g. by use of a moments equation | M1 | 3.1b |
| \(Mg \times \dfrac{3}{2}a = kMg \times 12a\) | A1 | 1.1b |
| \(k = \dfrac{1}{8}\) | A1 | 1.1b |
| (3) | ||
| (7 marks) |
Notes
M1: Complete strategy to find \(k\) e.g. moments about \(A\).
Needs all relevant terms. Must be dimensionally correct.
Condone sign errors. Condone if \(a\), \(M\), \(g\) missing throughout
A1: Correct unsimplified equation in \(k\)
A1: Correct answer – any equivalent form
Alternative (b)
| Scheme | Marks | AO |
|---|---|---|
| Moments equation | M1 | |
| \(12a \times kM = \dfrac{13}{30}M \times 2.5a + \dfrac{5}{30}M \times 2.5a\) | A1 | |
| \(12k = \dfrac{45}{30}\), \(k = \dfrac{1}{8}\) | A1 | |
| (3) |