| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2011 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Two jointed rods in equilibrium |
| Difficulty | Challenging +1.2 This is a standard two-rod statics problem requiring systematic application of equilibrium conditions (moments and resolving forces) across multiple parts. While it involves several steps and careful bookkeeping of forces at joints, the techniques are routine for M3 level—taking moments about strategic points, resolving vertically, and applying limiting friction. The geometry is given explicitly, and the question guides students through the solution with structured parts. More challenging than basic single-body equilibrium but less demanding than problems requiring geometric insight or non-standard approaches. |
| Spec | 3.03u Static equilibrium: on rough surfaces3.04b Equilibrium: zero resultant moment and force6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| \(2.8V = 1.4\times72\) | M1 | For taking moments about \(Q\) for \(PQ\) or for using symmetry |
| Vertical component at \(P\) is \(36 N\) | A1 | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(36 + N = 72 + 54\) | M1 | For resolving forces vertically on both rods |
| Normal component at \(R\) is \(90 N\) | A1, AG | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(1.44F = 1.2\times90 - 0.8\times54\) or \(72\times1.4 + 54\times3.6 + 1.44F = 90\times4\) with not more than 1 error in either case | M1 | For taking moments about \(Q\) for \(QR\) or about \(P\) for the whole structure (all terms needed) |
| Equation correct and leading to \(F = 45\) | A1 | |
| For using \(F = \mu R\) | M1 | |
| Coefficient is \(0.5\) | A1 | |
| [5] |
**Part i**
$2.8V = 1.4\times72$ | M1 | For taking moments about $Q$ for $PQ$ or for using symmetry
Vertical component at $P$ is $36 N$ | A1 |
| [2] |
**Part ii**
$36 + N = 72 + 54$ | M1 | For resolving forces vertically on both rods
Normal component at $R$ is $90 N$ | A1, AG |
| [2] |
**Part iii**
$1.44F = 1.2\times90 - 0.8\times54$ or $72\times1.4 + 54\times3.6 + 1.44F = 90\times4$ with not more than 1 error in either case | M1 | For taking moments about $Q$ for $QR$ or about $P$ for the whole structure (all terms needed)
Equation correct and leading to $F = 45$ | A1 |
For using $F = \mu R$ | M1 |
Coefficient is $0.5$ | A1 |
| [5] |3\\
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A uniform $\operatorname { rod } P Q$ has weight 72 N . A non-uniform $\operatorname { rod } Q R$ has weight 54 N and its centre of mass is at $C$, where $Q C = 2 C R$. The rods are freely jointed to each other at $Q$. The rod $P Q$ is freely jointed to a fixed point of a vertical wall at $P$ and the rod $Q R$ rests on horizontal ground at $R$. The rod $P Q$ is 2.8 m long and is horizontal. The point $R$ is 1.44 m below the level of $P Q$ and 4 m from the wall (see diagram).\\
(i) Find the vertical component of the force exerted by the wall on $P Q$.\\
(ii) Hence show that the normal component of the force exerted by the ground on $Q R$ is 90 N .\\
(iii) Given that the friction at $R$ is limiting, find the coefficient of friction between the rod $Q R$ and the ground.
\hfill \mbox{\textit{OCR M3 2011 Q3 [9]}}