| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod hinged to wall with string support |
| Difficulty | Standard +0.3 This is a standard M2 moments problem with a hinged rod in equilibrium. Part (a) requires taking moments about A and resolving vertically—routine techniques for this topic. Part (b) adds a simple step of finding the resultant force at a specific angle. The algebra is straightforward and the problem follows a familiar template for rod equilibrium questions. |
| Spec | 3.04b Equilibrium: zero resultant moment and force6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Resolving vertically: \(Y + P\cos\theta = W\) | M1, A1 | Needs all 3 terms. Condone sign errors and sin/cos confusion. Condone \(Wg\) |
| Moments about \(A\): \(Wl\cos\theta = 2lP\) | M1, A1 | Terms need to be correct structure, condone \(l\) implied if not seen |
| \(P = \frac{W\cos\theta}{2} \Rightarrow Y = W - \frac{W\cos^2\theta}{2} = \frac{W}{2}(2 - \cos^2\theta)\) \(\star\star\) | DM1, A1 | Substitute for \(P\) to obtain simplified \(Y\). Requires both preceding M marks. Obtain given result correctly |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\theta = 45° \Rightarrow Y = \frac{3W}{4}\) | B1 | |
| \(X = P\sin 45\) | M1 | Resolving horizontally. Accept in terms of \(\theta\) |
| \(= \frac{W\cos 45}{2} \cdot \sin 45 = \frac{W}{4}\) | DM1, A1 | Express \(X\) in terms of \(W\). Requires preceding M mark. Correct unsimplified but substituted |
| Resultant at \(A = \frac{W}{4}\sqrt{3^2 + 1^2} = \frac{W\sqrt{10}}{4}\) (\(0.79W\)) | DM1, A1 | Use of Pythagoras with \(X\), \(Y\) in terms of \(W\) only. Dependent on first M1. Or equivalent (0.79\(W\) or better) |
## Question 7:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Resolving vertically: $Y + P\cos\theta = W$ | M1, A1 | Needs all 3 terms. Condone sign errors and sin/cos confusion. Condone $Wg$ |
| Moments about $A$: $Wl\cos\theta = 2lP$ | M1, A1 | Terms need to be correct structure, condone $l$ implied if not seen |
| $P = \frac{W\cos\theta}{2} \Rightarrow Y = W - \frac{W\cos^2\theta}{2} = \frac{W}{2}(2 - \cos^2\theta)$ $\star\star$ | DM1, A1 | Substitute for $P$ to obtain simplified $Y$. Requires both preceding M marks. Obtain **given result** correctly |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\theta = 45° \Rightarrow Y = \frac{3W}{4}$ | B1 | |
| $X = P\sin 45$ | M1 | Resolving horizontally. Accept in terms of $\theta$ |
| $= \frac{W\cos 45}{2} \cdot \sin 45 = \frac{W}{4}$ | DM1, A1 | Express $X$ in terms of $W$. Requires preceding M mark. Correct unsimplified but substituted |
| Resultant at $A = \frac{W}{4}\sqrt{3^2 + 1^2} = \frac{W\sqrt{10}}{4}$ ($0.79W$) | DM1, A1 | Use of Pythagoras with $X$, $Y$ in terms of $W$ only. Dependent on first M1. Or equivalent (0.79$W$ or better) |
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{47420c50-c232-41e9-8c4d-a890d86ea933-12_837_565_226_694}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A uniform rod $A B$ of weight $W$ has its end $A$ freely hinged to a point on a fixed vertical wall. The rod is held in equilibrium, at angle $\theta$ to the horizontal, by a force of magnitude $P$. The force acts perpendicular to the rod at $B$ and in the same vertical plane as the rod, as shown in Figure 3. The rod is in a vertical plane perpendicular to the wall. The magnitude of the vertical component of the force exerted on the rod by the wall at $A$ is $Y$.
\begin{enumerate}[label=(\alph*)]
\item Show that $Y = \frac { W } { 2 } \left( 2 - \cos ^ { 2 } \theta \right)$.
Given that $\theta = 45 ^ { \circ }$
\item find the magnitude of the force exerted on the rod by the wall at $A$, giving your answer in terms of $W$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2014 Q7 [12]}}