| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Horizontal force on slope |
| Difficulty | Standard +0.3 This is a standard M1 mechanics problem involving resolving forces on an inclined plane with friction. Part (a) is a 'show that' requiring straightforward resolution perpendicular to the plane. Parts (b) and (c) require understanding limiting friction and determining friction direction, which are routine M1 skills. The question involves multiple parts and careful resolution, but uses standard techniques without requiring novel insight, making it slightly easier than average. |
| Spec | 3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces |
| Answer | Marks | Guidance |
|---|---|---|
| Part | Answer/Working | Marks |
| (a) | Resolving perpendicular to the plane: \(S = 120\cos\alpha + 30\sin\alpha = 114\) * | M1 A1 A1 A1 |
| (b) | Resolving perpendicular to the plane: \(R = 120\cos\alpha = 96\) | M1 A1 |
| \(F_{\max} = \frac{1}{2}R\) | M1 | |
| Resolving parallel to the plane: In equilibrium: \(F_{\max} = F_{\max} + 120\sin\alpha = 48 + 72 = 120\) | M1 A(2,1,0) A1 | (8 marks) |
| (c) | \(30 + F = 120\sin\alpha\) OR \(30 - F = 120\sin\alpha\) | M1 A1 |
| So \(F = 42N\) acting up the plane | A1 | (3 marks) |
| [15 marks total] |
| **Part** | **Answer/Working** | **Marks** | **Guidance** |
|----------|-------------------|----------|-------------|
| (a) | Resolving perpendicular to the plane: $S = 120\cos\alpha + 30\sin\alpha = 114$ * | M1 A1 A1 A1 | (4 marks) |
| (b) | Resolving perpendicular to the plane: $R = 120\cos\alpha = 96$ | M1 A1 | |
| | $F_{\max} = \frac{1}{2}R$ | M1 | |
| | Resolving parallel to the plane: In equilibrium: $F_{\max} = F_{\max} + 120\sin\alpha = 48 + 72 = 120$ | M1 A(2,1,0) A1 | (8 marks) |
| (c) | $30 + F = 120\sin\alpha$ OR $30 - F = 120\sin\alpha$ | M1 A1 | |
| | So $F = 42N$ acting up the plane | A1 | (3 marks) |
| | | | **[15 marks total]** |6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4878b6c2-0c62-4398-8a8f-913139bc8a14-10_426_768_239_653}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A particle of weight 120 N is placed on a fixed rough plane which is inclined at an angle $\alpha$ to the horizontal, where $\tan \alpha = \frac { 3 } { 4 }$.\\
The coefficient of friction between the particle and the plane is $\frac { 1 } { 2 }$.\\
The particle is held at rest in equilibrium by a horizontal force of magnitude 30 N , which acts in the vertical plane containing the line of greatest slope of the plane through the particle, as shown in Figure 2.
\begin{enumerate}[label=(\alph*)]
\item Show that the normal reaction between the particle and the plane has magnitude 114 N .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4878b6c2-0c62-4398-8a8f-913139bc8a14-10_433_774_1464_604}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
The horizontal force is removed and replaced by a force of magnitude $P$ newtons acting up the slope along the line of greatest slope of the plane through the particle, as shown in Figure 3. The particle remains in equilibrium.
\item Find the greatest possible value of $P$.
\item Find the magnitude and direction of the frictional force acting on the particle when $P = 30$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2011 Q6 [15]}}