| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2007 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod or block on rough surface in limiting equilibrium (no wall) |
| Difficulty | Standard +0.3 This is a standard two-rod statics problem requiring moments about hinges and resolution of forces. While it involves multiple steps (moments about A, force resolution, friction coefficient), the approach is methodical and follows textbook procedures for connected rods in equilibrium. The 60° angle simplifies calculations, and the limiting equilibrium setup is a familiar M3 scenario. Slightly easier than average due to its structured, sequential nature. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (i) \(160a = 2aY\) | M1 | For taking moments for AB about B |
| Component at B is \(80\text{N}\) | A1 | |
| Component at C is \(240\text{N}\) | B1ft | 3 ft \(160 + Y\) |
| (ii) | M1 | For taking moments for BC about B or C (using \(X = F\)) or for whole about A |
| \(160a\cos60° + 2aF\sin60° = 240\times2a\cos60°\) or \(80\times2a\cos60° + 160a\cos60° = 2aX\sin60°\) or \(240(2+2\cos60°)a = 160a + 160(2+\cos60°)a + 2aF\sin60°\) | A1ft | |
| Frictional force is \(92.4\text{N}\) | A1 | |
| Direction is to the left | B1 | 4 |
| (iii) \([92.4/240]\) | M1 | For using \(F = \mu R\) |
| Coefficient is \(0.385\) | A1ft | 2 |
# Question 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| **(i)** $160a = 2aY$ | M1 | For taking moments for AB about B |
| Component at B is $80\text{N}$ | A1 | |
| Component at C is $240\text{N}$ | B1ft | **3** ft $160 + Y$ |
| **(ii)** | M1 | For taking moments for BC about B or C (using $X = F$) or for whole about A |
| $160a\cos60° + 2aF\sin60° = 240\times2a\cos60°$ or $80\times2a\cos60° + 160a\cos60° = 2aX\sin60°$ or $240(2+2\cos60°)a = 160a + 160(2+\cos60°)a + 2aF\sin60°$ | A1ft | |
| Frictional force is $92.4\text{N}$ | A1 | |
| Direction is to the left | B1 | **4** |
| **(iii)** $[92.4/240]$ | M1 | For using $F = \mu R$ |
| Coefficient is $0.385$ | A1ft | **2** |
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\includegraphics[max width=\textwidth, alt={}, center]{f334f6e4-2a60-4647-8b37-e48937c85747-2_465_757_1146_694}
Two identical uniform rods, $A B$ and $B C$, are freely jointed to each other at $B$, and $A$ is freely jointed to a fixed point. The rods are in limiting equilibrium in a vertical plane, with $C$ resting on a rough horizontal surface. $A B$ is horizontal, and $B C$ is inclined at $60 ^ { \circ }$ to the horizontal. The weight of each rod is 160 N (see diagram).\\
(i) By taking moments for $A B$ about $A$, find the vertical component of the force on $A B$ at $B$. Hence or otherwise find the magnitude of the vertical component of the contact force on $B C$ at $C$. [3]\\
(ii) Calculate the magnitude of the frictional force on $B C$ at $C$ and state its direction.\\
(iii) Calculate the value of the coefficient of friction at $C$.
\hfill \mbox{\textit{OCR M3 2007 Q3 [9]}}