| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2008 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Framework or multiple rod structures |
| Difficulty | Challenging +1.2 This is a multi-part moments problem requiring systematic application of equilibrium conditions to a composite rigid body system. While it involves several steps (finding centroids of rods, taking moments, resolving forces) and two different configurations, the techniques are standard for A-level mechanics. The right-angled triangle simplifies calculations, and the problem follows a predictable structure without requiring novel geometric insights or complex algebraic manipulation. |
| Spec | 3.04b Equilibrium: zero resultant moment and force6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([5T = 2(20 + 25)]\) | M1 | For taking moments about B |
| \(T = 18\) | A1 | Subtotal: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(5T = 2(20 + 25) + 60x4/3\) | B1ft | |
| \(T = 34\) | B1 | Subtotal: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([Y = (15+20+25) + 60 - 34x4/5]\) | M1 | For resolving forces vertically |
| Vertical component has magnitude 92.8N | A1ft | ft \(120 - 0.8T\). Subtotal: 2. Total: 6 |
## Question 4:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[5T = 2(20 + 25)]$ | M1 | For taking moments about B |
| $T = 18$ | A1 | **Subtotal: 2** |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $5T = 2(20 + 25) + 60x4/3$ | B1ft | |
| $T = 34$ | B1 | **Subtotal: 2** |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[Y = (15+20+25) + 60 - 34x4/5]$ | M1 | For resolving forces vertically |
| Vertical component has magnitude 92.8N | A1ft | ft $120 - 0.8T$. **Subtotal: 2. Total: 6** |
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Uniform rods $A B , A C$ and $B C$ have lengths $3 \mathrm {~m} , 4 \mathrm {~m}$ and 5 m respectively, and weights $15 \mathrm {~N} , 20 \mathrm {~N}$ and 25 N respectively. The rods are rigidly joined to form a right-angled triangular frame $A B C$. The frame is hinged at $B$ to a fixed point and is held in equilibrium, with $A C$ horizontal, by means of an inextensible string attached at $C$. The string is at right angles to $B C$ and the tension in the string is $T \mathrm {~N}$ (see diagram).\\
(i) Find the value of $T$.
A uniform triangular lamina $P Q R$, of weight 60 N , has the same size and shape as the frame $A B C$. The lamina is now attached to the frame with $P , Q$ and $R$ at $A , B$ and $C$ respectively. The composite body is held in equilibrium with $A , B$ and $C$ in the same positions as before. Find\\
(ii) the new value of $T$,\\
(iii) the magnitude of the vertical component of the force acting on the composite body at $B$.
\hfill \mbox{\textit{CAIE M2 2008 Q4 [6]}}