| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Coplanar forces in equilibrium |
| Difficulty | Standard +0.3 This is a standard M1 equilibrium problem requiring resolution of forces in two directions and application of friction laws. While it has multiple parts and involves two objects, each step follows routine mechanics procedures (resolving forces, using F=μR, comparing with limiting friction) without requiring novel insight or complex problem-solving strategies. |
| 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 |
|---|---|---|
| Answer | Marks | Guidance |
| \(mg = 6.4\cos40\) | M1 | One cmpt of 6.4 N force (allow \(6.4 \times \sin/\cos 40\) or 50), mg not resolved |
| \(m = 0.5(00)\) | A1 | Accept 0.5, \(g=9.81\) (0.49976..=0.5) \(g=10\) (0.49026.. = 0.49) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H = 6.4 + 6.4\sin40\) OR \(2\times6.4\cos25 = 0.5g\cos65 + H\cos25\) | M1 | Resolves horizontally, all necessary terms (allow e.g. \(6.4 \pm 6.4\cos40\)). Resolves parallel to bisector of strings, inc cmpt weight |
| \(H = 10.5\) | A1 | Accept 11 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(R = 32\cos30 - 6.4\sin30\) | M1 | Difference of Wt cmpt and Tension (not \(H\)) cmpt |
| \(R = 24.5\) | A1 | May be implied |
| \(Fr = 32\sin30 + 6.4\cos30\) | M1 | Sum of Wt cmpt and Tension (not \(H\)) cmpt |
| \(Fr = 21.5\) | A1 | May be implied |
| \(\mu = (32\sin30 + 6.4\cos30)/(32\cos30 - 6.4\sin30)\) | M1 | Either Fr or R obtained from 2 term numerical expressions, in \( |
| \(\mu = 0.879\) AG | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(F_\text{max} = 0.879 \times 32\cos30\) \((= 24.4 \text{ N})\) | B1* | May be described simply as \(F\) or friction |
| Wt cmpt down slope \(= 32\sin30\) \((= 16 \text{ N})\) | D*M1 | Finding Wt component down slope and comparing with friction |
| Remains in eqbm | A1 | Needs Wt cmpt \(= 16 < F_\text{max}\) |
| OR \(\pm ma = 32\sin30 - 0.879\times32\cos30\) | B1* | For friction calculation |
| Finds acceleration | D*M1 | Sets up and solves N2L for \(a\) |
| Remains in eqbm | A1 | Needs \(a\) clearly in direction of friction (impossible) |
| OR angle of friction \(= \tan^{-1}0.879 = 41°\), Slope is \(30°\) | B1* | Must be explicit |
| Remains in eqbm | D*M1, A1 | Values of angle of friction and slope stated in 6(iv) |
# Question 6:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $mg = 6.4\cos40$ | M1 | One cmpt of 6.4 N force (allow $6.4 \times \sin/\cos 40$ or 50), mg not resolved |
| $m = 0.5(00)$ | A1 | Accept 0.5, $g=9.81$ (0.49976..=0.5) $g=10$ (0.49026.. = 0.49) |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H = 6.4 + 6.4\sin40$ OR $2\times6.4\cos25 = 0.5g\cos65 + H\cos25$ | M1 | Resolves horizontally, all necessary terms (allow e.g. $6.4 \pm 6.4\cos40$). Resolves parallel to bisector of strings, inc cmpt weight |
| $H = 10.5$ | A1 | Accept 11 |
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $R = 32\cos30 - 6.4\sin30$ | M1 | Difference of Wt cmpt and Tension (not $H$) cmpt |
| $R = 24.5$ | A1 | May be implied |
| $Fr = 32\sin30 + 6.4\cos30$ | M1 | Sum of Wt cmpt and Tension (not $H$) cmpt |
| $Fr = 21.5$ | A1 | May be implied |
| $\mu = (32\sin30 + 6.4\cos30)/(32\cos30 - 6.4\sin30)$ | M1 | **Either** Fr **or** R obtained from 2 term numerical expressions, in $|Fr| = \mu|R|$ |
| $\mu = 0.879$ AG | A1 | |
## Part (iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $F_\text{max} = 0.879 \times 32\cos30$ $(= 24.4 \text{ N})$ | B1* | May be described simply as $F$ or friction |
| Wt cmpt down slope $= 32\sin30$ $(= 16 \text{ N})$ | D*M1 | Finding Wt component down slope and comparing with friction |
| Remains in eqbm | A1 | Needs Wt cmpt $= 16 < F_\text{max}$ |
| OR $\pm ma = 32\sin30 - 0.879\times32\cos30$ | B1* | For friction calculation |
| Finds acceleration | D*M1 | Sets up and solves N2L for $a$ |
| Remains in eqbm | A1 | Needs $a$ clearly in direction of friction (impossible) |
| OR angle of friction $= \tan^{-1}0.879 = 41°$, Slope is $30°$ | B1* | Must be explicit |
| Remains in eqbm | D*M1, A1 | Values of angle of friction and slope stated in 6(iv) |
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{f0813713-d677-4ed7-87e1-971a64bdb6ff-4_328_698_255_657}
A particle $P$ lies on a slope inclined at $30 ^ { \circ }$ to the horizontal. $P$ is attached to one end of a taut light inextensible string which passes through a small smooth ring $Q$ of mass $m \mathrm {~kg}$. The portion $P Q$ of the string is horizontal and the other portion of the string is inclined at $40 ^ { \circ }$ to the vertical. A horizontal force of magnitude $H \mathrm {~N}$, acting away from $P$, is applied to $Q$ (see diagram). The tension in the string is 6.4 N , and the string is in the vertical plane containing the line of greatest slope on which $P$ lies. Both $P$ and $Q$ are in equilibrium.\\
(i) Calculate $m$.\\
(ii) Calculate $H$.\\
(iii) Given that the weight of $P$ is 32 N , and that $P$ is in limiting equilibrium, show that the coefficient of friction between $P$ and the slope is 0.879 , correct to 3 significant figures.\\
$Q$ and the string are now removed.\\
(iv) Determine whether $P$ remains in equilibrium.
\hfill \mbox{\textit{OCR M1 2012 Q6 [13]}}