| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2017 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Ladder against wall |
| Difficulty | Standard +0.3 This is a standard ladder equilibrium problem requiring resolution of forces and taking moments about a point. Part (a) involves basic moment equation, part (b) requires finding friction coefficient from limiting equilibrium (standard technique), and part (c) extends to include an additional mass. All steps follow textbook methods with no novel insight required, making it slightly easier than average for M2. |
| 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 |
| \(M(A): S \cdot 2a\cos 30° = mga\sin 30°\) | M1 | Correct number of terms. Terms must be dimensionally correct. Condone trig confusion |
| (correct equation) | A1 | At most one error. Consistent trig confusion is one error |
| (correct unsimplified equation) | A1 | Correct unsimplified equation |
| \(S = \dfrac{mg\sqrt{3}}{6}\) | A1 | Accept exact equivalent. Accept \(0.289mg\) or better |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(R = mg\); \(F = S\) | B1 | Resolve vertically and horizontally — must be stated or shown on diagram |
| \(\dfrac{mg\sqrt{3}}{6} \leq \mu mg\) | M1 | Use of \(F \leq \mu R\) (not for \(F = \mu R\) followed by a fudge of the inequality) |
| \(\dfrac{\sqrt{3}}{6} \leq \mu\) | A1 | Answer Given CSO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\uparrow: U = mg + kmg = mg(1+k)\) | B1 | Or equation in \(U\) and \(k\) from a second moments equation |
| \(M(A): T \times 2a \times \dfrac{\sqrt{3}}{2} = mga \times \dfrac{1}{2} + kmg \cdot 2a \times \dfrac{1}{2}\) | M1 | Need all three terms. Condone \(\mu = \dfrac{\sqrt{3}}{6}\). Terms must be dimensionally correct. Condone trig confusion. Condone sign errors |
| \(\Rightarrow 2T\cos 30° = mg\sin 30° + 2kmg\sin 30°\) | A1 | Correct unsimplified moments equation |
| \(\Rightarrow \dfrac{3}{5}U = \dfrac{1}{2}mg + kmg\) | A1 | Correct equation in \(U\) (and \(k\)), \(\mu\) correct if used |
| \(\Rightarrow \dfrac{3}{5}(1+k) = \dfrac{1}{2} + k\) | DM1 | Solve for \(k\). Dependent on preceding M |
| \(k = \dfrac{1}{4}\) | A1 |
## Question 7(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $M(A): S \cdot 2a\cos 30° = mga\sin 30°$ | M1 | Correct number of terms. Terms must be dimensionally correct. Condone trig confusion |
| (correct equation) | A1 | At most one error. Consistent trig confusion is one error |
| (correct unsimplified equation) | A1 | Correct unsimplified equation |
| $S = \dfrac{mg\sqrt{3}}{6}$ | A1 | Accept exact equivalent. Accept $0.289mg$ or better |
**Total: 4 marks**
---
## Question 7(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R = mg$; $F = S$ | B1 | Resolve vertically and horizontally — must be stated or shown on diagram |
| $\dfrac{mg\sqrt{3}}{6} \leq \mu mg$ | M1 | Use of $F \leq \mu R$ (not for $F = \mu R$ followed by a fudge of the inequality) |
| $\dfrac{\sqrt{3}}{6} \leq \mu$ | A1 | **Answer Given** CSO |
**Total: 3 marks**
---
## Question 7(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\uparrow: U = mg + kmg = mg(1+k)$ | B1 | Or equation in $U$ and $k$ from a second moments equation |
| $M(A): T \times 2a \times \dfrac{\sqrt{3}}{2} = mga \times \dfrac{1}{2} + kmg \cdot 2a \times \dfrac{1}{2}$ | M1 | Need all three terms. Condone $\mu = \dfrac{\sqrt{3}}{6}$. Terms must be dimensionally correct. Condone trig confusion. Condone sign errors |
| $\Rightarrow 2T\cos 30° = mg\sin 30° + 2kmg\sin 30°$ | A1 | Correct unsimplified moments equation |
| $\Rightarrow \dfrac{3}{5}U = \dfrac{1}{2}mg + kmg$ | A1 | Correct equation in $U$ (and $k$), $\mu$ correct if used |
| $\Rightarrow \dfrac{3}{5}(1+k) = \dfrac{1}{2} + k$ | DM1 | Solve for $k$. Dependent on preceding M |
| $k = \dfrac{1}{4}$ | A1 | |
**Total: 6 marks**
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{36cced0d-f982-4534-a3fe-13c32fb37f5b-11_513_429_123_762}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A uniform rod $A B$ has mass $m$ and length $2 a$. The end $A$ is in contact with rough horizontal ground and the end $B$ is in contact with a smooth vertical wall. The rod rests in equilibrium in a vertical plane perpendicular to the wall and makes an angle of $30 ^ { \circ }$ with the wall, as shown in Figure 2. The coefficient of friction between the rod and the ground is $\mu$.
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $m$ and $g$, the magnitude of the force exerted on the rod by the wall.
\item Show that $\mu \geqslant \frac { \sqrt { 3 } } { 6 }$.
A particle of mass $k m$ is now attached to the rod at $B$. Given that $\mu = \frac { \sqrt { 3 } } { 5 }$ and that the rod is now in limiting equilibrium,
\item find the value of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2017 Q7 [13]}}