| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2003 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod with end on ground or wall supported by string |
| Difficulty | Standard +0.3 This is a standard moments equilibrium problem requiring resolution of forces and taking moments about a point. While it involves multiple steps (finding tension via moments, then resolving horizontally and vertically for reaction components), the setup is straightforward with the rope perpendicular to the beam, making the geometry simple. It's slightly easier than average as it's a textbook application of equilibrium principles with no geometric complications or novel insight required. |
| Spec | 3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Rope is at 30° to wall, or beam is at 0° to the horizontal, or a correct trig. ratio used | B1 | |
| Taking moments about \(A\), or taking moments about \(P\) and resolving horizontally | M1 | |
| \(2.5T = 45g \times 3\cos 30°\) or \(5H = 45g \times 3\cos 30°\) and \(H = T\sin 30°\) | A1 ft | |
| Tension is 468 N | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Horizontal component is 234 N (ft \(\frac{1}{2}T\)) | B1 ft | |
| Resolving forces vertically (\(V = 45g - T\cos 30°\)) | M1 | |
| Magnitude of vertical component is 45 N | A1 ft | SR: angle incorrect (i) B0, M1, A1 ft A0, (ii) B1 ft (\(T\) and angle), M1, A0 |
# Question 3:
## Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Rope is at 30° to wall, or beam is at 0° to the horizontal, or a correct trig. ratio used | B1 | |
| Taking moments about $A$, or taking moments about $P$ and resolving horizontally | M1 | |
| $2.5T = 45g \times 3\cos 30°$ or $5H = 45g \times 3\cos 30°$ and $H = T\sin 30°$ | A1 ft | |
| Tension is 468 N | A1 | |
## Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Horizontal component is 234 N (ft $\frac{1}{2}T$) | B1 ft | |
| Resolving forces vertically ($V = 45g - T\cos 30°$) | M1 | |
| Magnitude of vertical component is 45 N | A1 ft | SR: angle incorrect **(i)** B0, M1, A1 ft A0, **(ii)** B1 ft ($T$ and angle), M1, A0 |
---3\\
\includegraphics[max width=\textwidth, alt={}, center]{7f8646df-a7d8-4ca1-a6ee-3ceab6bb83af-3_464_439_274_854}
A uniform beam $A B$ has length 6 m and mass 45 kg . One end of a light inextensible rope is attached to the beam at the point 2.5 m from $A$. The other end of the rope is attached to a fixed point $P$ on a vertical wall. The beam is in equilibrium with $A$ in contact with the wall at a point 5 m below $P$. The rope is taut and at right angles to $A B$ (see diagram). Find\\
(i) the tension in the rope,\\
(ii) the horizontal and vertical components of the force exerted by the wall on the beam at $A$.
\hfill \mbox{\textit{CAIE M2 2003 Q3 [7]}}