| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2006 |
| Session | June |
| 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 M3 statics problem requiring systematic application of equilibrium conditions (moments and forces) to a two-rod system. While it involves multiple steps and careful geometry, the techniques are routine for Further Maths students: taking moments about strategic points, resolving forces, and using symmetry. The 'show that' part provides scaffolding, and the methods are well-practiced at this level. |
| Spec | 3.04b Equilibrium: zero resultant moment and force6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (implied method) | M1 | For taking moments about C for the whole structure |
| \(1.4R = 0.35 \times 360 + 1.05 \times 200\) | A1 | |
| Magnitude is \(240\text{N}\) | A1 | AG |
| (implied method) | M1 | For taking moments about A for the rod AB |
| \(0.7 \times 240 = 0.35 \times 200 + 1.05T\) | A1 | |
| Tension is \(93.3\text{N}\) | A1 | 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (implied method) | M1 | For taking moments about A for AB and AC |
| \(0.7R_B = 70 + 1.05T\) and \(0.7R_C = 126 + 1.05T\) | A1 | |
| (implied method) | M1 | For eliminating T or for adding the equations, and then using \(R_B + R_C = 560\) |
| \(0.7(560 - R_B) - 0.7R_B = 126 - 70\) or \(0.7 \times 560 = 70 + 126 + 2.1T\) | A1 | For a correct equation in \(R_B\) only or T only |
| Magnitude is \(240\text{N}\) | A1 | AG |
| Tension is \(93.3\text{N}\) | A1 | 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Horizontal component is \(93.3\text{N}\) to the left | B1ft | |
| \(Y = 240 - 200\) | M1 | For resolving forces vertically |
| Vertical component is \(40\text{N}\) downwards | A1 | 3 |
# Question 3:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| (implied method) | M1 | For taking moments about C for the whole structure |
| $1.4R = 0.35 \times 360 + 1.05 \times 200$ | A1 | |
| Magnitude is $240\text{N}$ | A1 | AG |
| (implied method) | M1 | For taking moments about A for the rod AB |
| $0.7 \times 240 = 0.35 \times 200 + 1.05T$ | A1 | |
| Tension is $93.3\text{N}$ | A1 | 6 | |
**OR (i):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| (implied method) | M1 | For taking moments about A for AB and AC |
| $0.7R_B = 70 + 1.05T$ and $0.7R_C = 126 + 1.05T$ | A1 | |
| (implied method) | M1 | For eliminating T or for adding the equations, and then using $R_B + R_C = 560$ |
| $0.7(560 - R_B) - 0.7R_B = 126 - 70$ or $0.7 \times 560 = 70 + 126 + 2.1T$ | A1 | For a correct equation in $R_B$ only or T only |
| Magnitude is $240\text{N}$ | A1 | AG |
| Tension is $93.3\text{N}$ | A1 | 6 | |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Horizontal component is $93.3\text{N}$ to the left | B1ft | |
| $Y = 240 - 200$ | M1 | For resolving forces vertically |
| Vertical component is $40\text{N}$ downwards | A1 | 3 | |
---3\\
\includegraphics[max width=\textwidth, alt={}, center]{09d3e8ca-0062-4f62-8453-7acaff591db5-2_661_711_918_717}
Two uniform rods $A B$ and $A C$, of equal lengths, and of weights 200 N and 360 N respectively, are freely jointed at $A$. The mid-points of the rods are joined by a taut light inextensible string. The rods are in equilibrium in a vertical plane with $B$ and $C$ in contact with a smooth horizontal surface. The point $A$ is 2.1 m above the surface and $B C = 1.4 \mathrm {~m}$ (see diagram).\\
(i) Show that the force exerted on $A B$ at $B$ has magnitude 240 N and find the tension in the string.\\
(ii) Find the horizontal and vertical components of the force exerted on $A B$ at $A$ and state their directions.
\hfill \mbox{\textit{OCR M3 2006 Q3 [9]}}