| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Friction |
| Type | Minimum/maximum force for equilibrium |
| Difficulty | Standard +0.8 This is a non-trivial mechanics problem requiring students to resolve forces in two directions on an inclined plane, apply friction laws correctly (recognizing friction can act either up or down the plane), and then minimize P by differentiating or using calculus/geometric insight. It goes beyond standard textbook exercises where friction direction is given, requiring problem-solving to determine when equilibrium breaks and which configuration gives minimum force. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| (Parallel to plane): \(P\cos 50 + F = 2g\cos 60\) | M1 A2 | First M1 for resolving parallel to plane. \(2g\) term must use \(30°\) or \(60°\) angle; allow sin/cos confusion. A1A0 if one error |
| (Perp to plane): \(R - P\sin 50 = 2g\cos 30\) | M1 A2 | Second M1 for resolving perpendicular to plane. \(2g\) term must use \(30°\) or \(60°\) angle; allow sin/cos confusion. A1A0 if one error |
| \(F = \frac{1}{4}R\) | B1 | Seen or implied |
| Attempt to eliminate \(F\) and \(R\) to give equation in \(P\) only | M1 | Independent; must have two 3 (or 4) term equations |
| Solve for \(P\) | DM1 | Dependent on third M1 |
| \(P = 6.7\) (2 SF) or \(6.66\) (3SF) | A1 (10) |
| Answer | Marks |
|---|---|
| Working | Marks |
| \((\rightarrow): R\cos 60 - F\cos 30 = P\cos 20\) | M1 A2 |
| \((\uparrow): R\cos 30 + F\cos 60 = P\cos 70 + 2g\) | M1 A2 |
# Question 2:
| Working | Marks | Guidance |
|---------|-------|----------|
| (Parallel to plane): $P\cos 50 + F = 2g\cos 60$ | M1 A2 | First M1 for resolving parallel to plane. $2g$ term must use $30°$ or $60°$ angle; allow sin/cos confusion. A1A0 if one error |
| (Perp to plane): $R - P\sin 50 = 2g\cos 30$ | M1 A2 | Second M1 for resolving perpendicular to plane. $2g$ term must use $30°$ or $60°$ angle; allow sin/cos confusion. A1A0 if one error |
| $F = \frac{1}{4}R$ | B1 | Seen or implied |
| Attempt to eliminate $F$ and $R$ to give equation in $P$ only | M1 | Independent; must have two 3 (or 4) term equations |
| Solve for $P$ | DM1 | Dependent on third M1 |
| $P = 6.7$ (2 SF) or $6.66$ (3SF) | A1 (10) | |
**Other possible equations:**
| Working | Marks |
|---------|-------|
| $(\rightarrow): R\cos 60 - F\cos 30 = P\cos 20$ | M1 A2 |
| $(\uparrow): R\cos 30 + F\cos 60 = P\cos 70 + 2g$ | M1 A2 |
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4fd21e83-0bdf-4bb1-8a3f-76beada511ae-04_333_976_287_550}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A particle of mass 2 kg lies on a rough plane. The plane is inclined to the horizontal at $30 ^ { \circ }$. The coefficient of friction between the particle and the plane is $\frac { 1 } { 4 }$. The particle is held in equilibrium by a force of magnitude $P$ newtons. The force makes an angle of $20 ^ { \circ }$ with the horizontal and acts in a vertical plane containing a line of greatest slope of the plane, as shown in Figure 1. Find the least possible value of $P$.
\hfill \mbox{\textit{Edexcel M1 2018 Q2 [10]}}