| Exam Board | OCR |
|---|---|
| Module | H240/03 (Pure Mathematics and Mechanics) |
| Year | 2018 |
| Session | September |
| Marks | 8 |
| Topic | Moments |
| Type | Rod on smooth peg or cylinder |
| Difficulty | Standard +0.3 This is a standard moments problem requiring resolution of forces and taking moments about a point. While it involves multiple steps (finding normal reaction at B, then using limiting equilibrium for friction), the setup is straightforward with given angle, and the techniques are routine for M1 level. Slightly easier than average due to clear diagram and standard method. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03v Motion on rough surface: including inclined planes3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sin\theta = \frac{5}{13}\) and \(\cos\theta = \frac{12}{13}\) or \(\theta = 22.6\ldots\) | B1 | May be seen anywhere in solution |
| Moments about A: \(3 \times 10g \times \cos\theta = 6R_B\) | M1 | Allow sin/cos confusion |
| \(\Rightarrow R = \frac{60g}{13}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Vertically: \(R_A + R_B \cos\theta = 10g\) | M1* | Resolving vertically – allow sign errors and sin/cos confusion |
| A1 | Where \(R_A\) is the normal contact force acting on the plank at A | |
| Horizontally: \(F = R_B \sin\theta\) | B1 | |
| Using \(\mu = \frac{F}{R_A}\) | dep*M1 | With all angles replaced in their \(F\) and \(R_A\) |
| \(\mu = \frac{30}{97}\) or 0.309 | A1 | \(R_A = \frac{970g}{169}\), \(F = \frac{300g}{169}\), 0.309278... |
### (i)
$\sin\theta = \frac{5}{13}$ and $\cos\theta = \frac{12}{13}$ or $\theta = 22.6\ldots$ | B1 | May be seen anywhere in solution
Moments about A: $3 \times 10g \times \cos\theta = 6R_B$ | M1 | Allow sin/cos confusion
$\Rightarrow R = \frac{60g}{13}$ | A1 |
**[3 marks total]**
### (ii)
Vertically: $R_A + R_B \cos\theta = 10g$ | M1* | Resolving vertically – allow sign errors and sin/cos confusion
| A1 | Where $R_A$ is the normal contact force acting on the plank at A
Horizontally: $F = R_B \sin\theta$ | B1 |
Using $\mu = \frac{F}{R_A}$ | dep*M1 | With all angles replaced in their $F$ and $R_A$
$\mu = \frac{30}{97}$ or 0.309 | A1 | $R_A = \frac{970g}{169}$, $F = \frac{300g}{169}$, 0.309278...
**[5 marks total]**
---9\\
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The diagram shows a plank of wood $A B$, of mass 10 kg and length 6 m , resting with its end $A$ on rough horizontal ground and its end $B$ in contact with a fixed cylindrical oil drum. The plank is in a vertical plane perpendicular to the axis of the drum, and the line $A B$ is a tangent to the circular cross-section of the drum, with the point of contact at $B$. The plank is inclined at an angle $\theta$ to the horizontal, where $\tan \theta = \frac { 5 } { 12 }$.
The plank is modelled as a uniform rod and the oil drum is modelled as being smooth.\\
(i) Find, in terms of $g$, the normal contact force between the drum and the plank.\\
(ii) Given that the plank is in limiting equilibrium, find the coefficient of friction between the plank and the ground.
\hfill \mbox{\textit{OCR H240/03 2018 Q9 [8]}}