M2 June 2016 Q5
5.
A non-uniform rod \(AB\), of mass 5 kg and length 4 m, rests with one end \(A\) on rough horizontal ground. The centre of mass of the rod is \(d\) metres from \(A\). The rod is held in limiting equilibrium at an angle \(\theta\) to the horizontal by a force \(\mathbf{P}\), which acts in a direction perpendicular to the rod at \(B\), as shown in Figure 2. The line of action of \(\mathbf{P}\) lies in the same vertical plane as the rod.
Given that \(\tan\theta = \dfrac{5}{12}\) and that the coefficient of friction between the rod and the ground is \(\dfrac{1}{2}\),
| Scheme | Marks |
|---|---|
| M\((A)\): \(\ d\cos\theta \times 5g = 4P\) | M1 A1 |
| Resolving horizontally: \(\ P\sin\theta = F\) | B1 |
| Resolving vertically: \(\ P\cos\theta + R = 5g\) | M1 A1 DM1 |
| \(R = 5g - \dfrac{5gd\cos^2\theta}{4}\) | A1 |
| \(F = \dfrac{5gd\cos\theta\sin\theta}{4}\) | A1 |
| (8) |
Notes
M1 Terms must be dimensionally correct. Condone trig confusion
M1 Requires all 3 terms. Condone trig confusion and sign errors
A1 Correct equation
DM1 Substitute for \(P\) to find \(R\) or \(F\). Dependent on both previous M marks
A1 One force correct. Accept equivalent forms e.g. \(R = \dfrac{20g - 5gd + 20g\tan^2\theta}{4\left(1 + \tan^2\theta\right)}\)
A1 Both forces correct. Accept equivalent forms e.g. \(F = \dfrac{5gd\tan\theta}{4\sec^2\theta}\)
5a alt
| M\((B)\): \(5g\cos\theta \times (4 - d) + F\sin\theta \times 4 = R\cos\theta \times 4\) | M1 A1 |
| Resolve parallel to the rod: \(5g\sin\theta = R\sin\theta + F\cos\theta\) | M1 B1 A1 |
| \(\Rightarrow R = 5g - \dfrac{F\cos\theta}{\sin\theta}\) | |
| \(5g\cos\theta \times (4 - d) + F\sin\theta \times 4\) \(= 4\cos\theta\left(5g - \dfrac{F\cos\theta}{\sin\theta}\right)\) | DM1 |
| \(4F\left(\sin\theta + \dfrac{\cos^2\theta}{\sin\theta}\right)\) \(= 20g\cos\theta - 20g\cos\theta + 5gd\cos\theta\) | |
| \(F = \dfrac{5gd\cos\theta\sin\theta}{4}\) | A1 |
| \(R = 5g - \dfrac{5gd\cos^2\theta}{4}\) | A1 |
M1 Needs all three terms. Terms must be dimensionally correct. Condone trig confusion
A1 At most one error
M1 Requires all 3 terms. Condone trig confusion and sign errors
B1 At most one error
A1 Correct equation
DM1 Eliminate one variable to find \(F\) or \(R\). Dependent on both previous M marks
A1 One force correct
A1 Both forces correct
| Scheme | Marks |
|---|---|
| \(\mu = \dfrac{\frac{5gd\cos\theta\sin\theta}{4}}{5g - \frac{5gd\cos^2\theta}{4}}\) | M1 |
| \(\dfrac{1}{2}\left(5g - \dfrac{5gd\cos^2\theta}{4}\right) = \dfrac{5gd\cos\theta\sin\theta}{4}\) | A1 |
| \(4 \times 169 = 120d + 144d\) | M1 |
| \(d = \dfrac{169}{66}\) | A1 |
| (4) | |
| (12 marks) |
Notes
M1 Use of \(F = \mu R\)
A1 \(\left(4 - d\cos^2\theta = 2d\cos\theta\sin\theta\right)\)
M1 Use \(\tan\theta = \dfrac{5}{12}\) and solve for \(d\)
A1 (= 2.6 m or better)
5balt
| \(F = 5gd \times \dfrac{12}{13} \times \dfrac{5}{13} \times \dfrac{1}{4}\left(= \dfrac{75gd}{169}\right)\) | M1 |
| \(R = 5g - \dfrac{5gd}{4} \times \dfrac{144}{169}\) | A1 |
| \(75gd = \dfrac{1}{2}\left(5 \times 169g - 180gd\right)\) | M1 |
| \(150gd + 180gd = 845g,\ \ d = \dfrac{169}{66}\) | A1 |
M1 Use \(\tan\theta = \dfrac{5}{12}\)
A1 Both unsimplified expressions
M1 Use of \(F = \mu R\) and solve for \(d\)
A1 (= 2.6 m or better)
5balt
| \(R = 5g - \dfrac{12}{13}P,\ \ F = \dfrac{5}{13}P\) | M1 |
| \(\dfrac{5}{13}P = \dfrac{1}{2}\left(5g - \dfrac{12}{13}P\right)\) | M1 |
| \(\Rightarrow P = \dfrac{65}{22}g\) | A1 |
| \(d = \dfrac{4P}{5g\cos\theta} = \dfrac{169}{66}\) | A1 |
M1 Substitute trig in their equations from resolving.
M1 use \(F = \mu R\) and solve for \(d\)