| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | June |
| Marks | 8 |
| 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 requiring taking moments about the hinge, resolving forces, and finding components. The setup is straightforward with clearly defined angles and forces, requiring routine application of equilibrium conditions (ΣM=0, ΣF=0) but no novel insight or complex geometry—slightly easier than average A-level. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| \(M(A), F.4\sin 40° = 5g.2\cos 25°\) | M1 | A complete method to find \(F\), e.g. take moments about \(A\). Condone sin/cos confusion. Requires correct ratio of lengths |
| A1 | Correct terms with at most one slip | |
| A1 | All correct | |
| \(F = 35\) | A1 | \(35\) or \(34.5\) (>3sf not acceptable due to use of \(9.8\), but only penalise once in a question) |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| \(F\cos 75° \pm Y = 5g\) | M1 | Resolve vertically. Need all three terms but condone sign errors. Must be attempting to work with their \(75°\) or \(15°\) |
| A1 | Correct equation (their \(F\)) | |
| A1 | \(40\) or \(40.1\) | |
| \(Y = 40\) ; UP | A1 | Apply ISW if the candidate goes on to find \(R\) |
| cso (the Q does specifically ask for the direction, so this must be clearly stated) | ||
| (4) | ||
| 8 |
| Answer | Marks | Guidance |
|---|---|---|
| OR1: \(\frac{4m\cos 25 \times Y}{-5g \times 2m\cos 25 + F \cos 15 \times 4m \sin 25}\) | M1, A1 | Taking moments about the point vertically below \(B\) and on the same horizontal level as \(A\) (Their \(F\)). Resolve parallel & perpendicular to the rod |
| OR2: \(R\cos\alpha = F\cos 40 + 5g\cos 65\) | ||
| \(R\sin\alpha + F\sin 40 = 5g\cos 25\) | Solve for \(R, \alpha\) | |
| \(R = 52.1, \alpha = 25.3°\) | M1A1 | |
| \(Y = R\sin(25 + \alpha)\) Etc. | Need a complete strategy to find \(Y\) for M1 | |
| (5) | ||
| 9 |
**(a)**
| $M(A), F.4\sin 40° = 5g.2\cos 25°$ | M1 | A complete method to find $F$, e.g. take moments about $A$. Condone sin/cos confusion. Requires correct ratio of lengths |
| | A1 | Correct terms with at most one slip |
| | A1 | All correct |
| $F = 35$ | A1 | $35$ or $34.5$ (>3sf not acceptable due to use of $9.8$, but only penalise once in a question) |
| | (4) | |
**(b)**
| $F\cos 75° \pm Y = 5g$ | M1 | Resolve vertically. Need all three terms but condone sign errors. Must be attempting to work with their $75°$ or $15°$ |
| | A1 | Correct equation (their $F$) |
| | A1 | $40$ or $40.1$ |
| $Y = 40$ ; UP | A1 | Apply ISW if the candidate goes on to find $R$ |
| | | cso (the Q does specifically ask for the direction, so this must be clearly stated) |
| | (4) |
| | 8 | |
**(b) Continued**
| OR1: $\frac{4m\cos 25 \times Y}{-5g \times 2m\cos 25 + F \cos 15 \times 4m \sin 25}$ | M1, A1 | Taking moments about the point vertically below $B$ and on the same horizontal level as $A$ (Their $F$). Resolve parallel & perpendicular to the rod |
| OR2: $R\cos\alpha = F\cos 40 + 5g\cos 65$ | | |
| $R\sin\alpha + F\sin 40 = 5g\cos 25$ | | Solve for $R, \alpha$ |
| $R = 52.1, \alpha = 25.3°$ | M1A1 | |
| $Y = R\sin(25 + \alpha)$ Etc. | | Need a complete strategy to find $Y$ for M1 |
| | (5) | |
| | 9 | |
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{12cd7355-f632-4a84-825f-a269851c6ec4-04_374_798_255_559}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A uniform rod $A B$, of mass 5 kg and length 4 m , has its end $A$ smoothly hinged at a fixed point. The rod is held in equilibrium at an angle of $25 ^ { \circ }$ above the horizontal by a force of magnitude $F$ newtons applied to its end $B$. The force acts in the vertical plane containing the rod and in a direction which makes an angle of $40 ^ { \circ }$ with the rod, as shown in Figure 1.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $F$.
\item Find the magnitude and direction of the vertical component of the force acting on the rod at $A$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2012 Q3 [8]}}