| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Ladder against wall |
| Difficulty | Standard +0.3 This is a standard statics problem requiring resolution of forces and taking moments about a point. The setup is straightforward (rod against wall and ground), and the solution follows a routine procedure: resolve vertically for reaction at A, take moments about A to find reaction at B, resolve horizontally for friction, then calculate coefficient. The given sin θ = 3/5 simplifies calculations. Slightly easier than average due to its textbook nature and clear methodology. |
| Spec | 3.04b Equilibrium: zero resultant moment and force6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks |
|---|---|
| (ii) | 3 |
| Answer | Marks |
|---|---|
| F = 13(.0) | M1 |
| Answer | Marks |
|---|---|
| [2] | P to centre of mass (= 0.3 m) |
Question 2:
--- 2 (i)
(ii) ---
2 (i)
(ii) | 3
Horizontal distance = 0.8 × × sin30
4
0.8
OR 0.8tan30cos30 – sin30
4
Mom. = (0.6sin30 × 20 =) 6 Nm AG
OR
0.8
Mom = 20cos30 × 0.8tan30 – 20sin30 ×
4
Mom = 6 Nm AG
6 = F × 0.8tan30
F = 13(.0) | M1
A1
[2]
M1
A1
M1
A1
[2] | P to centre of mass (= 0.3 m)
Resolves Wt // and perp axis and finds
moments of both components
Takes moments about P\includegraphics{figure_2}
A uniform rod $AB$ of mass $3m$ and length $4a$ rests in equilibrium in a vertical plane with the end $A$ on rough horizontal ground and the end $B$ against a smooth vertical wall. The rod makes an angle $\theta$ with the horizontal, where $\sin \theta = \frac{3}{5}$.
\begin{enumerate}[label=(\roman*)]
\item Find the normal reaction at $A$ and the normal reaction at $B$. [4]
\item Find the coefficient of friction between the rod and the ground. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2014 Q2 [6]}}