| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Friction |
| Type | Contact force magnitude and direction |
| Difficulty | Moderate -0.5 This is a straightforward limiting equilibrium problem requiring resolution of forces in two directions and application of F=μR. The setup is clearly defined with all forces given, requiring only standard techniques: resolve horizontally and vertically, use the friction law, and basic trigonometry. Slightly easier than average due to the clear diagram and direct application of standard methods without any conceptual subtlety. |
| Spec | 3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 03.03t Coefficient of friction: F <= mu*R model |
| Answer | Marks |
|---|---|
| \(F = 4\cos 30\) and \(R = 10 - 4\sin 30\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(C^2 = (4\cos 30)^2 + (10 - 4\sin 30)^2]\) | M1 | For using cosine rule or for using \(C^2 = (C \cos \phi)^2 + (C \sin \phi)^2\) or \(C^2 = F^2 + R^2\) |
| \(C = 8.72\) | A1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \([\mu = 4\cos 30/(10 - 4\sin 30)]\) | M1 | For using \(\mu = F/R = C \cos \phi / C \sin \phi\) |
| Coefficient is \(0.433\) (accept \(0.43\)) | A1 | 2 |
| Answer | Marks |
|---|---|
| \(4 \div \sin(90° + \phi) = 10 \div \sin(150° - \phi)\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \([C = \sin 120° = (4 \div \sin 156.6°\) or \(10 \div \sin 83.4°)]\) | M1 | |
| \(C = 8.72\) | A1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \([\mu = \sqrt{3} \div 4\) or \(\mu = \cos 66.6° + \sin 66.6°]\) | M1 | For using \(\mu = F/R = C \cos \phi / C \sin \phi\) |
| Coefficient is \(0.433\) (accept \(0.43\)) | A1 | 2 |
**(i)** For triangle of forces with $60°$ shown correctly, or
$C \cos \phi = 4 \cos 30$ and $C \sin \phi = 10 - 4 \sin 30$, or
$F = 4\cos 30$ and $R = 10 - 4\sin 30$ | B1 |
$[C^2 = 4^2 + 10^2 - 2 \times 4 \times 10\cos 60$ or
$C^2 = (4\cos 30)^2 + (10 - 4\sin 30)^2]$ | M1 | For using cosine rule or for using $C^2 = (C \cos \phi)^2 + (C \sin \phi)^2$ or $C^2 = F^2 + R^2$
$C = 8.72$ | A1 | 3
**(ii)**
$[\mu = 4\cos 30/(10 - 4\sin 30)]$ | M1 | For using $\mu = F/R = C \cos \phi / C \sin \phi$
Coefficient is $0.433$ (accept $0.43$) | A1 | 2
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# Question 4 - Alternative Method:
**(i)** For obtaining $\phi = 66.6°$ or
$\tan \phi = 4 + \sqrt{3}$ from
$4 \div \sin(90° + \phi) = 10 \div \sin(150° - \phi)$ | B1 |
For using C N and (4 N or 10 N) in Lami's theorem to find C
$[C = \sin 120° = (4 \div \sin 156.6°$ or $10 \div \sin 83.4°)]$ | M1 |
$C = 8.72$ | A1 | 3
**(ii)**
$[\mu = \sqrt{3} \div 4$ or $\mu = \cos 66.6° + \sin 66.6°]$ | M1 | For using $\mu = F/R = C \cos \phi / C \sin \phi$
Coefficient is $0.433$ (accept $0.43$) | A1 | 2
---4\\
\includegraphics[max width=\textwidth, alt={}, center]{2bd9f770-65b1-48c2-bf58-24e732bb6988-2_608_723_1247_712}
A particle $P$ has weight 10 N and is in limiting equilibrium on a rough horizontal table. The forces shown in the diagram represent the weight of $P$, an applied force of magnitude 4 N acting on $P$ in a direction at $30 ^ { \circ }$ above the horizontal, and the contact force exerted on $P$ by the table (the resultant of the frictional and normal components) of magnitude $C \mathrm {~N}$.\\
(i) Find the value of $C$.\\
(ii) Find the coefficient of friction between $P$ and the table.
\hfill \mbox{\textit{CAIE M1 2011 Q4 [5]}}