| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Friction |
| Type | Contact force magnitude and direction |
| Difficulty | Moderate -0.3 This is a standard M1 friction question testing limiting equilibrium in multiple configurations. Part (i) is direct application of F=μR with horizontal forces, part (ii) requires Pythagoras for resultant contact force, and parts (iii)(a-b) involve resolving forces at different angles. All steps are routine textbook exercises requiring no novel insight, making it slightly easier than average for A-level. |
| Spec | 3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(3 = 8\mu\) | M1 | Uses \(F = \mu R\), Allow \(R\) is \(8\) or \(8g\), \(Fr = 3\) only |
| \(\mu = 0.375\) | A1 | \(3/8\) (fraction), not \(3 \div 8\) (division) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(C^2 = 3^2 + 8^2\) | M1 | Uses Pythagoras with \(3\) and \(8\) or \(8g\) |
| \(C = 8.54 \text{ N}\) | A1 | Accept \(8.5\) or \(\sqrt{73}\) |
| \(\tan\theta = 3/8\) or \(\tan\theta = 8/3\) | M1 | Uses tan with \(3\) and \(8\) or \(8g\); Or CorS with answer for \(C\) |
| \(\theta = 20.6°\) with vertical or \(69.4°\) with horizontal | A1 | Accept \(21\) or \(69\), direction clear by words or diagram; isw work after correct angle magnitude found |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(T(\cos 0) - 3 = +/-3\) | M1 | \(T(\cos 0) - 3 = 0\) is M0; \(T\cos 0 - 3 = -3\) assumes Fr direction has not changed |
| \(T = 6\) | A1 | Answer alone is sufficient for M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(R = +/-(8 - T \times \text{SorC}30)\) | M1 | Accept \(8g\) with cmpt \(T\); (This is required also in the SC case) |
| \(R = 8 - T\sin 30\) | A1 | oe |
| \(Fr = +/-(T \times \text{CorS}30 - 3)\) | M1 | Accept \(3\) with cmpt \(T\), not \(T \times \text{CorS}30 +/- 3 = 0\); SC Does not allow for change in direction of friction |
| \(Fr = T\cos 30 - 3\) | A1 | oe; \(Fr = 3 - T\cos 30\) A1 |
| \(0.375 = (T\cos 30 - 3)/(8 - T\sin 30)\) | M1 | Accept use of \(\mu\) from (i). For forming an equation in \(T\) alone; \(0.375 = (3 - T\cos 30)/(8 - T\sin 30)\) M1 |
| \(T = 5.70\) | A1 | \(T = 0\) A0 |
| *OR Alternative for last 4 marks:* | [6] | SC (Alternative) |
| \(Fr = 0.375(8 - T\sin 30)\) | - | Accept use of \(\mu\) from (i) |
| \(Fr = +/-(T \times \text{CorS}30 - 3)\) | M1 | oe |
| \(Fr = T\cos 30 - 3\) | A1 | \(Fr = 3 - T\cos 30\) A1 |
| \(0.375(8 - T\sin 30) = T\cos 30 - 3\) | M1 | For forming an equation in \(T\) alone; \(0.375(8 - T\sin 30) = (3 - T\cos 30)\) M1 |
| \(T = 5.70\) | A1 | \(T = 0\) A0 |
# Question 6:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $3 = 8\mu$ | M1 | Uses $F = \mu R$, Allow $R$ is $8$ or $8g$, $Fr = 3$ only |
| $\mu = 0.375$ | A1 | $3/8$ (fraction), not $3 \div 8$ (division) |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $C^2 = 3^2 + 8^2$ | M1 | Uses Pythagoras with $3$ and $8$ or $8g$ |
| $C = 8.54 \text{ N}$ | A1 | Accept $8.5$ or $\sqrt{73}$ |
| $\tan\theta = 3/8$ or $\tan\theta = 8/3$ | M1 | Uses tan with $3$ and $8$ or $8g$; Or CorS with answer for $C$ |
| $\theta = 20.6°$ with vertical or $69.4°$ with horizontal | A1 | Accept $21$ or $69$, direction clear by words or diagram; isw work after correct angle magnitude found |
## Part (iii)(a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $T(\cos 0) - 3 = +/-3$ | M1 | $T(\cos 0) - 3 = 0$ is M0; $T\cos 0 - 3 = -3$ assumes Fr direction has not changed |
| $T = 6$ | A1 | Answer alone is sufficient for M1A1 |
## Part (iii)(b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $R = +/-(8 - T \times \text{SorC}30)$ | M1 | Accept $8g$ with cmpt $T$; (This is required also in the SC case) |
| $R = 8 - T\sin 30$ | A1 | oe |
| $Fr = +/-(T \times \text{CorS}30 - 3)$ | M1 | Accept $3$ with cmpt $T$, not $T \times \text{CorS}30 +/- 3 = 0$; SC Does not allow for change in direction of friction |
| $Fr = T\cos 30 - 3$ | A1 | oe; $Fr = 3 - T\cos 30$ A1 |
| $0.375 = (T\cos 30 - 3)/(8 - T\sin 30)$ | M1 | Accept use of $\mu$ from (i). For forming an equation in $T$ alone; $0.375 = (3 - T\cos 30)/(8 - T\sin 30)$ M1 |
| $T = 5.70$ | A1 | $T = 0$ A0 |
| *OR Alternative for last 4 marks:* | [6] | SC (Alternative) |
| $Fr = 0.375(8 - T\sin 30)$ | - | Accept use of $\mu$ from (i) |
| $Fr = +/-(T \times \text{CorS}30 - 3)$ | M1 | oe |
| $Fr = T\cos 30 - 3$ | A1 | $Fr = 3 - T\cos 30$ A1 |
| $0.375(8 - T\sin 30) = T\cos 30 - 3$ | M1 | For forming an equation in $T$ alone; $0.375(8 - T\sin 30) = (3 - T\cos 30)$ M1 |
| $T = 5.70$ | A1 | $T = 0$ A0 |
---6 A particle $P$ of weight 8 N rests on a horizontal surface. A horizontal force of magnitude 3 N acts on $P$, and $P$ is in limiting equilibrium.\\
(i) Calculate the coefficient of friction between $P$ and the surface.\\
(ii) Find the magnitude and direction of the contact force exerted by the surface on $P$.\\
(iii)\\
\includegraphics[max width=\textwidth, alt={}, center]{66eb8290-3a80-40bf-be40-a936ed7d5a1b-4_190_579_580_598}
The initial 3 N force continues to act on $P$ in its original direction. An additional force of magnitude $T \mathrm {~N}$, acting in the same vertical plane as the 3 N force, is now applied to $P$ at an angle of $\theta ^ { \circ }$ above the horizontal (see diagram). $P$ is again in limiting equilibrium.
\begin{enumerate}[label=(\alph*)]
\item Given that $\theta = 0$, find $T$.
\item Given instead that $\theta = 30$, calculate $T$.
\end{enumerate}
\hfill \mbox{\textit{OCR M1 2014 Q6 [14]}}