| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2006 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Two jointed rods in equilibrium |
| Difficulty | Challenging +1.2 This is a standard M3 statics problem involving two connected rods with multiple forces and moments. While it requires systematic application of equilibrium conditions (resolving forces and taking moments) across two bodies, the setup is straightforward with clearly defined geometry and weights. The multi-part structure guides students through the solution methodically. It's harder than basic single-body equilibrium problems but doesn't require novel insight—just careful bookkeeping and standard techniques, making it moderately above average difficulty. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (i) \(1.8F = 0.9\times40 + 1.4\times9\) | M1 | For any moment equation that includes \(F\) and all other relevant forces |
| A2,1,0 | \(-1\) each error | |
| Magnitude is \(27\) N | A1 | [4] AG |
| (ii) Vertical component is \(22\) N downwards | B1 | |
| \(1.2X = (40+9-27)\times(3.8-1.8) + 64\times1\) | M1, A2,1,0 ft | For any moment equation that includes \(X\) and all other relevant forces; \(-1\) each error; ft wrong vert. comp. |
| \((1.2X = 44 + 64)\) | ||
| Horizontal component is \(90\) N to the left | A1 | [5] |
| (iii) \(\mu = 27/[90]\) | M1 | For use of \(\mu = F/R\) |
| Coefficient of friction is \(0.3\) | A1 ft | [2] ft wrong answer in (ii) |
## Question 5:
| Answer/Working | Marks | Guidance |
|---|---|---|
| **(i)** $1.8F = 0.9\times40 + 1.4\times9$ | M1 | For any moment equation that includes $F$ and all other relevant forces |
| | A2,1,0 | $-1$ each error |
| Magnitude is $27$ N | A1 | [4] AG |
| **(ii)** Vertical component is $22$ N downwards | B1 | |
| $1.2X = (40+9-27)\times(3.8-1.8) + 64\times1$ | M1, A2,1,0 ft | For any moment equation that includes $X$ and all other relevant forces; $-1$ each error; ft wrong vert. comp. |
| $(1.2X = 44 + 64)$ | | |
| Horizontal component is $90$ N to the left | A1 | [5] |
| **(iii)** $\mu = 27/[90]$ | M1 | For use of $\mu = F/R$ |
| Coefficient of friction is $0.3$ | A1 ft | [2] ft wrong answer in (ii) |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{5bb3bd29-a2eb-4124-802c-fb17b68c50e4-3_462_1109_283_569}
Two uniform rods $A B$ and $B C$ have weights 64 N and 40 N respectively. The rods are freely jointed to each other at $B$. The rod $A B$ is freely jointed to a fixed point on horizontal ground at $A$ and the rod $B C$ rests against a vertical wall at $C$. The rod $B C$ is 1.8 m long and is horizontal. A particle of weight 9 N is attached to the rod $B C$ at the point 0.4 m from $C$. The point $A$ is 1.2 m below the level of $B C$ and 3.8 m from the wall (see diagram). The system is in equilibrium.\\
(i) Show that the magnitude of the frictional force at $C$ is 27 N .\\
(ii) Calculate the horizontal and vertical components of the force exerted on $A B$ at $B$.\\
(iii) Given that friction is limiting at $C$, find the coefficient of friction between the $\operatorname { rod } B C$ and the wall.
\hfill \mbox{\textit{OCR M3 2006 Q5 [11]}}