| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2007 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Lamina hinged at point with string support |
| Difficulty | Challenging +1.2 This is a standard mechanics equilibrium problem requiring resolution of forces and taking moments about a point. While it involves a triangular lamina (slightly less routine than a rod), the setup is clearly defined with perpendicular rope to AC, making the geometry straightforward. Students must find the center of mass of a right-angled triangle (standard formula), apply three equilibrium equations, and solve simultaneously. The multi-part nature and need to handle both force components and moments places it above average difficulty, but the systematic approach is well-practiced in M2 courses. |
| 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 |
| Distance of centre of mass of triangle from wall is 0.4 m | B1 | |
| M1 | For taking moments about A | |
| \(4g \times 0.4 = T \times 1\) | A1ft | |
| Tension is 16N | A1 | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Horizontal component is 12.8N | B1ft | ft for \(0.8 \times\) candidate's \(T\) |
| \(Y + 0.6T = 4g\) | M1 | For resolving forces vertically |
| Vertical component is 30.4N | A1ft | 3 |
## Question 4:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Distance of centre of mass of triangle from wall is 0.4 m | B1 | |
| | M1 | For taking moments about A |
| $4g \times 0.4 = T \times 1$ | A1ft | |
| Tension is 16N | A1 | 4 | |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Horizontal component is 12.8N | B1ft | ft for $0.8 \times$ candidate's $T$ |
| $Y + 0.6T = 4g$ | M1 | For resolving forces vertically |
| Vertical component is 30.4N | A1ft | 3 | ft for $(40 - 0.6 \times \text{Candidate's } T)$, or for 27.2 following $X = 9.6$ and consistent sin/cos mix |
**Total: 7 marks**
---4\\
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A uniform triangular lamina $A B C$ is right-angled at $B$ and has sides $A B = 0.6 \mathrm {~m}$ and $B C = 0.8 \mathrm {~m}$. The mass of the lamina is 4 kg . One end of a light inextensible rope is attached to the lamina at $C$. The other end of the rope is attached to a fixed point $D$ on a vertical wall. The lamina is in equilibrium with $A$ in contact with the wall at a point vertically below $D$. The lamina is in a vertical plane perpendicular to the wall, and $A B$ is horizontal. The rope is taut and at right angles to $A C$ (see diagram). Find\\
(i) the tension in the rope,\\
(ii) the horizontal and vertical components of the force exerted at $A$ on the lamina by the wall.
\hfill \mbox{\textit{CAIE M2 2007 Q4 [7]}}