| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2004 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Ladder on smooth wall and rough ground |
| Difficulty | Challenging +1.2 This is a standard M2 ladder equilibrium problem requiring resolution of forces, friction inequality, and taking moments about a point. While it involves multiple steps (showing R=7mg, then finding range of P using friction limits), the techniques are routine for M2 students and the setup is clearly defined with no novel insight required. The 10m mass and specific angle make calculations straightforward. Slightly above average difficulty due to the two-part nature and need to consider friction acting in both directions for the range. |
| Spec | 3.03t Coefficient of friction: F <= mu*R model3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(M(A)\): \(N \times 2a\sin\alpha = mg \times a\cos\alpha + 10mg \times 2a\cos\alpha\) | M1 A2(1,0) | |
| \(2N\tan\alpha = 21mg\) | ||
| \(N = 7mg\) | M1 A1 | cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\uparrow\) \(R = 11mg\) | B1 | |
| \(F_r = 0.6 \times 11mg = 6.6mg\) | B1 | |
| For min \(P\): \(F_r \rightarrow\), \(P_{\min} = 7mg - 6.6mg = 0.4mg\) | M1 A1 | |
| For max \(P\): \(F_r \leftarrow\), \(P_{\max} = 7mg + 6.6mg = 13.6mg\) | M1 A1 | |
| \(0.4mg \mid P \mid 13.6mg\) | A1 | cso |
## Question 6:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $M(A)$: $N \times 2a\sin\alpha = mg \times a\cos\alpha + 10mg \times 2a\cos\alpha$ | M1 A2(1,0) | |
| $2N\tan\alpha = 21mg$ | | |
| $N = 7mg$ | M1 A1 | cso |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\uparrow$ $R = 11mg$ | B1 | |
| $F_r = 0.6 \times 11mg = 6.6mg$ | B1 | |
| For min $P$: $F_r \rightarrow$, $P_{\min} = 7mg - 6.6mg = 0.4mg$ | M1 A1 | |
| For max $P$: $F_r \leftarrow$, $P_{\max} = 7mg + 6.6mg = 13.6mg$ | M1 A1 | |
| $0.4mg \mid P \mid 13.6mg$ | A1 | cso |
*Note: In (a), if moments are taken about a point other than A, a complete set of equations for finding N is needed for the first M1. If this M1 is gained, the A2(1,0) is awarded for the moments equation as it first appears.*
---6. A uniform ladder $A B$, of mass $m$ and length $2 a$, has one end $A$ on rough horizontal ground. The coefficient of friction between the ladder and the ground is 0.6 . The other end $B$ of the ladder rests against a smooth vertical wall.
A builder of mass 10 m stands at the top of the ladder. To prevent the ladder from slipping, the builder's friend pushes the bottom of the ladder horizontally towards the wall with a force of magnitude $P$. This force acts in a direction perpendicular to the wall. The ladder rests in equilibrium in a vertical plane perpendicular to the wall and makes an angle $\alpha$ with the horizontal, where $\tan \alpha = \frac { 3 } { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that the reaction of the wall on the ladder has magnitude 7 mg .
\item Find, in terms of $m$ and $g$, the range of values of $P$ for which the ladder remains in equilibrium.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2004 Q6 [12]}}