| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod hinged to wall with rough contact at free end |
| Difficulty | Standard +0.3 This is a straightforward statics problem requiring moments about point A and resolving forces. The perpendicular rod simplifies geometry significantly, and the two-part structure (find tension, then friction coefficient) guides students through standard mechanics techniques with no novel insight required. |
| Spec | 3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(9 \times 0.4 = 0.6 \times \text{Tsin}30\) | M1 | Moments about A |
| \(T = 12\text{N}\) | A1 | [2] |
| (ii) \(\mu = \frac{(9 - 12\sin30)}{(12\cos30)}\) | M1 | For resolving horizontally and vertically |
| M1 | For using \(F = \mu R\) | |
| \(\mu = 0.289\) | A1 | [3] |
**(i)** $9 \times 0.4 = 0.6 \times \text{Tsin}30$ | M1 | Moments about A
$T = 12\text{N}$ | A1 | [2]
**(ii)** $\mu = \frac{(9 - 12\sin30)}{(12\cos30)}$ | M1 | For resolving horizontally and vertically
| M1 | For using $F = \mu R$
$\mu = 0.289$ | A1 | [3]
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\includegraphics[max width=\textwidth, alt={}, center]{d1f1f036-1676-443e-b733-ca1fe79972d4-2_334_679_258_731}
A non-uniform $\operatorname { rod } A B$, of length 0.6 m and weight 9 N , has its centre of mass 0.4 m from $A$. The end $A$ of the rod is in contact with a rough vertical wall. The rod is held in equilibrium, perpendicular to the wall, by means of a light string attached to $B$. The string is inclined at $30 ^ { \circ }$ to the horizontal. The tension in the string is $T \mathrm {~N}$ (see diagram).\\
(i) Calculate $T$.\\
(ii) Find the least possible value of the coefficient of friction at $A$.
\hfill \mbox{\textit{CAIE M2 2011 Q1 [5]}}