M2 June 2018 Q4
4.
A uniform rod \(AB\), of mass \(m\) and length \(2a\), rests with its end \(A\) on rough horizontal ground. The rod is held in limiting equilibrium at an angle \(\theta\) to the horizontal by a light string attached to the rod at \(B\), as shown in Figure 3. The string is perpendicular to the rod and lies in the same vertical plane as the rod.
The coefficient of friction between the ground and the rod is \(\mu\).
Show that \(\mu = \dfrac{\cos\theta\sin\theta}{2 - \cos^2\theta}\) (10)
| Scheme | Marks |
|---|---|
| M\((A)\): \(\ 2aT = mga\cos\theta\ \ \ \ \left(T = \dfrac{1}{2}mg\cos\theta\right)\) M\((B)\): \(\ mga\cos\theta + Fr \times 2a\sin\theta = R \times 2a\cos\theta\) | M1A1 |
| Resolve \(\leftrightarrow\) : \(\ Fr = T\sin\theta\left(= \dfrac{1}{2}mg\cos\theta\sin\theta\right)\) | M1A1 |
| \(\updownarrow\) : \(\ R + T\cos\theta = mg\) | M1A1 |
| Use \(Fr = \mu R\) : \(\ \mu R = T\sin\theta\) | B1 |
| Form equation in \(\mu\) and \(\theta\): \(R = mg - \dfrac{1}{2}mg\cos\theta\cos\theta\) and \(\ \mu R = \dfrac{1}{2}mg\cos\theta\sin\theta\ \ \Rightarrow\) | DM1 |
| \(\mu = \dfrac{\frac{1}{2}mg\cos\theta\sin\theta}{mg - \frac{1}{2}mg\cos\theta\cos\theta}\) | DM1 |
| \(\mu = \dfrac{\cos\theta\sin\theta}{2 - \cos^2\theta}\) | A1 |
| (10) | |
| (10 marks) |
Notes
M1A1 First equation. Need all terms. Condone sign errors and sin/cos confusion
M1A1 Second equation. Need all terms. Condone sign errors and sin/cos confusion
M1A1 Third equation. Need all terms. Condone sign errors and sin/cos confusion
B1 Condone correct inequality
DM1 Eliminate T and R. Dependent on first 3 M marks
DM1 Solve for \(\mu\). Dependent on previous M
A1 Obtain given answer from correct working. Must explain if inequality becomes equality
Alt 1
| Moments (about \(B\)): \(mga\cos\theta + Fr \times 2a\sin\theta = R \times 2a\cos\theta\) | M1 A1 |
| Resolving (parallel to rod): \(Fr\cos\theta + R\sin\theta = mg\sin\theta\) | M2 A2 |
| Use of \(Fr = \mu R\) : \(mg\cos\theta + \mu R \times 2\sin\theta = R \times 2\cos\theta\) \(\mu R\cos\theta + R\sin\theta = mg\sin\theta\) | B1 |
| Form equation in \(\mu\) and \(\theta\): \(\dfrac{mg\sin\theta}{mg\cos\theta} = \dfrac{\mu R\cos\theta + R\sin\theta}{2R\cos\theta - 2\mu R\sin\theta}\) \(\dfrac{\sin\theta}{\cos\theta} = \dfrac{\mu\cos\theta + \sin\theta}{2\cos\theta - 2\mu\sin\theta}\) | DM1 |
| Solve for \(\mu\) : \(2\cos\theta\sin\theta - 2\mu\sin^2\theta = \mu\cos^2\theta + \cos\theta\sin\theta\) | DM1 |
| \(\mu = \dfrac{\sin\theta\cos\theta}{\cos^2\theta + 2\sin^2\theta} = \dfrac{\sin\theta\cos\theta}{2 - \cos^2\theta}\) | A1 |
| NB for alternatives using moments and resolving: e.g. Resolve \(\leftrightarrow\) : \(\ Fr = T\sin\theta\) \(M\)(centre): \(\ aT = a\cos\theta R - a\sin\theta Fr\) |
A1 Correct unsimplified
A2 -1 each error
A1 Obtain given answer from correct working
First equation M1A1. Sufficient equations to solve M2A2
Alt 2
 | |
| \(\tan(\theta + \alpha) = \dfrac{\tan\theta + \tan\alpha}{1 - \tan\theta\tan\alpha}\) | M1A1 |
| \(\tan\theta = \dfrac{a}{2a\tan\alpha}\ \Rightarrow \tan\alpha = \dfrac{1}{2\tan\theta}\) | M1 |
| \(\tan(\theta + \alpha) = \dfrac{\tan\theta + \frac{1}{2\tan\theta}}{1 - \tan\theta \times \frac{1}{2\tan\theta}}\) \(= 2\left(\dfrac{\sin\theta}{\cos\theta} + \dfrac{\cos\theta}{2\sin\theta}\right)\) | M1A1 A1 |
| \(F = \mu R\ \Rightarrow\) \(\mu = \dfrac{1}{\tan(\theta + \alpha)}\) | B1 DM1 |
| \(= \dfrac{1}{2}\left(\dfrac{2\sin\theta\cos\theta}{2\sin^2\theta + \cos^2\theta}\right) = \dfrac{\cos\theta\sin\theta}{2 - \cos^2\theta}\) | DM1 A1 |
3 concurrent forces
DM1 A1 Obtain given answer from correct working