| Exam Board | OCR |
|---|---|
| Module | H240/03 (Pure Mathematics and Mechanics) |
| Year | 2018 |
| Session | March |
| Marks | 9 |
| Topic | Moments |
| Type | Rod or block on rough surface in limiting equilibrium (no wall) |
| Difficulty | Standard +0.3 This is a standard M1 statics problem involving moments, resolving forces, and friction. Part (i) requires taking moments about A, resolving horizontally and vertically, then applying F ≤ μR to find the limiting friction coefficient—a routine multi-step procedure. Part (ii) is straightforward application of Pythagoras to the contact force components. While it has 9 marks total and requires careful setup, it follows a well-established method with no novel insight needed, making it slightly easier than average. |
| Spec | 3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Moments about \(A\): \(Wa\cos 30° + 2W(2a\cos 30°) = 2R_B a\) | M1 | Correct number of terms and attempt at component/perp. dist. for \(W\) and \(2W\) |
| \(R_B = \frac{5}{8}\sqrt{3}W\) | A1 | |
| Resolve vertically: \(R_A + R_B \cos 30° = W + 2W\) | M1 | Four terms, with attempt at component of the force at \(B\) |
| \(R_A = \frac{9}{8}W\) | A1ft | Consistent with their \(R_B\) |
| Resolve horizontally: \(F_A = R_B \sin 30°\) | B1 | ; \(F_A = \frac{5}{8}\sqrt{3}W\) |
| \(F_A \leq \mu R_A \Rightarrow \mu \geq \ldots\) | M1 | Dependent on all previous M marks; Allow equals here |
| \(\mu \geq \frac{5}{9}\sqrt{3}\) so the least value of \(\mu\) is \(\frac{5}{9}\sqrt{3}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Moments about \(B\): \(Wa\cos 30° + F_A(2a\sin 30°) = R_A(2a\cos 30°)\) | M1 | Correct number of terms and attempt at components/perp. distances |
| Resolve along \(AB\): \(R_A\cos 60° + F_A\cos 30° = W\cos 60° + 2W\cos 60°\) | M1 | Four terms, with components attempted |
| Both equations correct | A1 | Unsimplified |
| Solve simultaneous equations for \(R_A\) and \(F_A\) | M1 | Dependent on both previous M marks |
| \(R_A = \frac{9}{8}W\) and \(F_A = \frac{5}{8}\sqrt{3}W\) | A1 | Both correct, from correct equations |
| \(F_A \leq \mu R_A \Rightarrow \mu \geq \ldots\) | M1 | Dependent on all previous M marks; Allow equals here |
| \(\mu \geq \frac{5}{9}\sqrt{3}\) so the least value of \(\mu\) is \(\frac{5}{9}\sqrt{3}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(R_A^2 + F_A^2 = 39\) | M1 | Use of their \(R_A\) and \(F_A\) in terms of \(W\) in equation for magnitude of resultant |
| \(W = 4\) | A1 | cao |
## Part (i):
Moments about $A$: $Wa\cos 30° + 2W(2a\cos 30°) = 2R_B a$ | M1 | Correct number of terms and attempt at component/perp. dist. for $W$ and $2W$
$R_B = \frac{5}{8}\sqrt{3}W$ | A1 |
Resolve vertically: $R_A + R_B \cos 30° = W + 2W$ | M1 | Four terms, with attempt at component of the force at $B$
$R_A = \frac{9}{8}W$ | A1ft | Consistent with their $R_B$
Resolve horizontally: $F_A = R_B \sin 30°$ | B1 | ; $F_A = \frac{5}{8}\sqrt{3}W$
$F_A \leq \mu R_A \Rightarrow \mu \geq \ldots$ | M1 | Dependent on all previous M marks; Allow equals here
$\mu \geq \frac{5}{9}\sqrt{3}$ so the least value of $\mu$ is $\frac{5}{9}\sqrt{3}$ | A1 |
**Alternative solution:**
Moments about $B$: $Wa\cos 30° + F_A(2a\sin 30°) = R_A(2a\cos 30°)$ | M1 | Correct number of terms and attempt at components/perp. distances
Resolve along $AB$: $R_A\cos 60° + F_A\cos 30° = W\cos 60° + 2W\cos 60°$ | M1 | Four terms, with components attempted
Both equations correct | A1 | Unsimplified
Solve simultaneous equations for $R_A$ and $F_A$ | M1 | Dependent on both previous M marks
$R_A = \frac{9}{8}W$ and $F_A = \frac{5}{8}\sqrt{3}W$ | A1 | Both correct, from correct equations
$F_A \leq \mu R_A \Rightarrow \mu \geq \ldots$ | M1 | Dependent on all previous M marks; Allow equals here
$\mu \geq \frac{5}{9}\sqrt{3}$ so the least value of $\mu$ is $\frac{5}{9}\sqrt{3}$ | A1 |
## Part (ii):
$R_A^2 + F_A^2 = 39$ | M1 | Use of their $R_A$ and $F_A$ in terms of $W$ in equation for magnitude of resultant
$W = 4$ | A1 | cao\includegraphics{figure_10}
A uniform rod $AB$, of weight $W$ N and length $2a$ m, rests with the end $A$ on a rough horizontal table. A small object of weight $2W$ N is attached to the rod at $B$. The rod is maintained in equilibrium at an angle of $30°$ to the horizontal by a force acting at $B$ in a direction perpendicular to the rod in the same vertical plane as the rod (see diagram).
\begin{enumerate}[label=(\roman*)]
\item Find the least possible value of the coefficient of friction between the rod and the table. [7]
\item Given that the magnitude of the contact force at $A$ is $\sqrt{39}$ N, find the value of $W$. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR H240/03 2018 Q10 [9]}}