| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Friction |
| Type | Particle on inclined plane - force at angle to slope |
| Difficulty | Standard +0.3 This is a standard M1 friction problem requiring resolution of forces in two directions and application of F=μR at limiting equilibrium. While it involves multiple steps (resolving parallel and perpendicular to plane, using the friction law), the setup is conventional and the algebra straightforward once tan α = 3/4 gives sin α = 3/5 and cos α = 4/5. Slightly easier than average due to the guided structure showing R=20N first. |
| Spec | 3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 03.03r Friction: concept and vector form3.03u Static equilibrium: on rough surfaces |
| Answer | Marks | Guidance |
|---|---|---|
| \(4\cos\alpha + F = W\sin\alpha\) | M1 A1 | |
| \(R = 4\sin\alpha + W\cos\alpha\) | M1 A1 | |
| \(F = 0.5R\) | B1 | |
| \(\cos\alpha = 0.8\) or \(\sin\alpha = 0.6\) | B1 | |
| \(R = 20\text{N}\) | M1 A1 | GIVEN ANSWER |
| \(W = 22\text{N}\) | A1 | |
| (9) |
| Answer | Marks | Guidance |
|---|---|---|
| \(R\sin\alpha = 4 + F\cos\alpha\) | M1 A1 | |
| \(R\cos\alpha + F\sin\alpha = W\) | M1 A1 | |
| \(F = 0.5R\) | B1 | |
| \(\cos\alpha = 0.8\) or \(\sin\alpha = 0.6\) | B1 | |
| \(R = 20\text{N}\) | M1 A1 | GIVEN ANSWER |
| \(W = 22\text{N}\) | A1 | |
| (9) | ||
| 9 |
$4\cos\alpha + F = W\sin\alpha$ | M1 A1 |
$R = 4\sin\alpha + W\cos\alpha$ | M1 A1 |
$F = 0.5R$ | B1 |
$\cos\alpha = 0.8$ or $\sin\alpha = 0.6$ | B1 |
$R = 20\text{N}$ | M1 A1 | **GIVEN ANSWER**
$W = 22\text{N}$ | A1 |
| (9) |
OR
$R\sin\alpha = 4 + F\cos\alpha$ | M1 A1 |
$R\cos\alpha + F\sin\alpha = W$ | M1 A1 |
$F = 0.5R$ | B1 |
$\cos\alpha = 0.8$ or $\sin\alpha = 0.6$ | B1 |
$R = 20\text{N}$ | M1 A1 | **GIVEN ANSWER**
$W = 22\text{N}$ | A1 |
| (9) |
| **9** |
---\includegraphics{figure_1}
A particle of weight $W$ newtons is held in equilibrium on a rough inclined plane by a horizontal force of magnitude 4 N. The force acts in a vertical plane containing a line of greatest slope of the inclined plane. The plane is inclined to the horizontal at an angle $\alpha$, where $\tan \alpha = \frac{3}{4}$, as shown in Figure 1.
The coefficient of friction between the particle and the plane is $\frac{1}{2}$.
Given that the particle is on the point of sliding down the plane,
\begin{enumerate}[label=(\roman*)]
\item show that the magnitude of the normal reaction between the particle and the plane is 20 N,
\item find the value of $W$. [9]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2011 Q3 [9]}}