| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2024 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod against wall and ground |
| Difficulty | Standard +0.3 Part (a) is a standard M2 limiting equilibrium problem requiring resolving forces and taking moments about one point to find tan α, using the friction law F = μR. Part (b) extends this by adding a horizontal force and finding the critical value where the rod is about to slip, requiring similar techniques but with an additional parameter. Both parts follow well-established M2 procedures with no novel insight required, making this slightly easier than average. |
| Spec | 3.03u Static equilibrium: on rough surfaces3.04b Equilibrium: zero resultant moment and force6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| \(F = \frac{1}{3}R\) | B1 | For a correct statement seen anywhere e.g. on a diagram |
| Either: Horizontal forces: \(S = F \left(= \frac{1}{3}mg\right)\) | B1 | |
| Equation for \(M(A)\) | M1 | Need correct terms, condone sign errors and sin/cos confusion. Condone \(a\) missing throughout |
| \(S \times 2a\cos\alpha = mga\sin\alpha\) | A1 | Correct unsimplified |
| Or: \(R = mg\) | B1 | |
| Equation for \(M(B)\) | M1 | Need correct terms, condone sign errors and sin/cos confusion. Condone \(a\) missing throughout |
| \(F \times 2a\cos\alpha + mga\sin\alpha = R \times 2a\sin\alpha\) | A1 | Correct unsimplified |
| Or: \(S = F \left(= \frac{1}{3}mg\right)\) | B1 | |
| Equation for \(M(G)\) | M1 | Need correct terms, condone sign errors and sin/cos confusion. Condone \(a\) missing throughout |
| \(Fa\cos\alpha + Sa\cos\alpha = mga\sin\alpha\) | A1 | Correct unsimplified |
| SC: \(S = F\left(=\frac{1}{3}mg\right)\) or \(R = mg\) and no moments equation | B1 | And no further marks |
| Solve for \(\tan\alpha\) | M1 | |
| \(\tan\alpha = \frac{2}{3}\) * | A1* | Obtain given answer from correct working |
| SC: Candidate never uses \(g\): B1B0M1A0M1A0 | ||
| Total: 6 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| Either: Use of \(R = mg\) and \(M(A)\) | M1 | |
| \(N = S = \frac{1}{3}mg\) | A1 | Correct only |
| Resolve horizontally: \(kmg = \frac{1}{3}R + N\) and solve for \(k\) | DM1 | Dependent on the moments equation; need correct terms, condone sign errors |
| \(k = \frac{2}{3}\) | A1 | Correct equation |
| Or: \(M(A)\) and | M1 | Need correct terms, condone sign errors and sin/cos confusion. Condone \(a\) missing throughout |
| \(mga\sin\alpha = N \times 2a\cos\alpha\) | A1 | Correct unsimplified equation |
| Resolve horizontally: \(kmg = \frac{1}{3}R + N\) and use \(R = mg\) and \(\tan\alpha = \frac{2}{3}\) to solve for \(k\) | DM1 | Dependent on the moments equation; need correct terms, condone sign errors. OR could use a second moments equation |
| \(k = \frac{2}{3}\) | A1 | Correct only |
| Or: \(M(B)\) | M1 | Need correct terms, condone sign errors and sin/cos confusion. Condone \(a\) missing throughout |
| \(mga\sin\alpha + kmg \times 2a\cos\alpha = R \times 2a\sin\alpha + \frac{1}{3}R \times 2a\cos\alpha\) | A1 | |
| Use of \(R = mg\) and \(\tan\alpha = \frac{2}{3}\) to solve for \(k\) | DM1 | Dependent on the moments equation. OR could use a second moments equation |
| \(k = \frac{2}{3}\) | A1 | Correct only |
| Or: \(M(G)\) | M1 | Need correct terms, condone sign errors and sin/cos confusion |
| \(Na\cos\alpha + kmga\cos\alpha = Ra\sin\alpha + Fa\cos\alpha\) | A1 | Correct unsimplified equation |
| Resolve horizontally: \(kmg = \frac{1}{3}R + N\) and use \(R = mg\) and \(\tan\alpha = \frac{2}{3}\) to solve for \(k\) | DM1 | Dependent on the moments equation; need correct terms, condone sign errors. OR could use a second moments equation |
| \(k = \frac{2}{3}\) | A1 | Correct only |
| Total: 4 marks |
## Question 6(a):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $F = \frac{1}{3}R$ | B1 | For a correct statement seen anywhere e.g. on a diagram |
| **Either:** Horizontal forces: $S = F \left(= \frac{1}{3}mg\right)$ | B1 | |
| Equation for $M(A)$ | M1 | Need correct terms, condone sign errors and sin/cos confusion. Condone $a$ missing throughout |
| $S \times 2a\cos\alpha = mga\sin\alpha$ | A1 | Correct unsimplified |
| **Or:** $R = mg$ | B1 | |
| Equation for $M(B)$ | M1 | Need correct terms, condone sign errors and sin/cos confusion. Condone $a$ missing throughout |
| $F \times 2a\cos\alpha + mga\sin\alpha = R \times 2a\sin\alpha$ | A1 | Correct unsimplified |
| **Or:** $S = F \left(= \frac{1}{3}mg\right)$ | B1 | |
| Equation for $M(G)$ | M1 | Need correct terms, condone sign errors and sin/cos confusion. Condone $a$ missing throughout |
| $Fa\cos\alpha + Sa\cos\alpha = mga\sin\alpha$ | A1 | Correct unsimplified |
| **SC:** $S = F\left(=\frac{1}{3}mg\right)$ or $R = mg$ and no moments equation | B1 | And no further marks |
| Solve for $\tan\alpha$ | M1 | |
| $\tan\alpha = \frac{2}{3}$ * | A1* | Obtain given answer from correct working |
| **SC:** Candidate never uses $g$: B1B0M1A0M1A0 | | |
| **Total: 6 marks** | | |
---
## Question 6(b):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| **Either:** Use of $R = mg$ and $M(A)$ | M1 | |
| $N = S = \frac{1}{3}mg$ | A1 | Correct only |
| Resolve horizontally: $kmg = \frac{1}{3}R + N$ and solve for $k$ | DM1 | Dependent on the moments equation; need correct terms, condone sign errors |
| $k = \frac{2}{3}$ | A1 | Correct equation |
| **Or:** $M(A)$ and | M1 | Need correct terms, condone sign errors and sin/cos confusion. Condone $a$ missing throughout |
| $mga\sin\alpha = N \times 2a\cos\alpha$ | A1 | Correct unsimplified equation |
| Resolve horizontally: $kmg = \frac{1}{3}R + N$ and use $R = mg$ and $\tan\alpha = \frac{2}{3}$ to solve for $k$ | DM1 | Dependent on the moments equation; need correct terms, condone sign errors. OR could use a second moments equation |
| $k = \frac{2}{3}$ | A1 | Correct only |
| **Or:** $M(B)$ | M1 | Need correct terms, condone sign errors and sin/cos confusion. Condone $a$ missing throughout |
| $mga\sin\alpha + kmg \times 2a\cos\alpha = R \times 2a\sin\alpha + \frac{1}{3}R \times 2a\cos\alpha$ | A1 | |
| Use of $R = mg$ and $\tan\alpha = \frac{2}{3}$ to solve for $k$ | DM1 | Dependent on the moments equation. OR could use a second moments equation |
| $k = \frac{2}{3}$ | A1 | Correct only |
| **Or:** $M(G)$ | M1 | Need correct terms, condone sign errors and sin/cos confusion |
| $Na\cos\alpha + kmga\cos\alpha = Ra\sin\alpha + Fa\cos\alpha$ | A1 | Correct unsimplified equation |
| Resolve horizontally: $kmg = \frac{1}{3}R + N$ and use $R = mg$ and $\tan\alpha = \frac{2}{3}$ to solve for $k$ | DM1 | Dependent on the moments equation; need correct terms, condone sign errors. OR could use a second moments equation |
| $k = \frac{2}{3}$ | A1 | Correct only |
| **Total: 4 marks** | | |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b6e93edf-1b9f-4ea9-bb41-f46f380bc623-18_625_803_246_632}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A uniform rod, $A B$, of mass $m$ and length $2 a$, rests in limiting equilibrium with its end $A$ on rough horizontal ground and its end $B$ against a smooth vertical wall.\\
The vertical plane containing the rod is at right angles to the wall.\\
The rod is inclined to the wall at an angle $\alpha$, as shown in Figure 2.\\
The coefficient of friction between the rod and the ground is $\frac { 1 } { 3 }$
\begin{enumerate}[label=(\alph*)]
\item Show that $\tan \alpha = \frac { 2 } { 3 }$
With the rod in the same position, a horizontal force of magnitude $k m g$ is applied to the $\operatorname { rod }$ at $A$, towards the wall. The line of action of this force is at right angles to the wall.
The rod remains in equilibrium.
\item Find the largest possible value of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2024 Q6 [10]}}