| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2015 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Range of equilibrium positions |
| Difficulty | Standard +0.8 This is a multi-part limiting equilibrium problem requiring resolution of forces, taking moments about strategic points, and using the friction condition μ = F/R. Part (ii) requires finding a range of values by considering two limiting cases (slipping at different points), which demands systematic problem-solving beyond routine mechanics questions. The algebraic manipulation and inequality reasoning elevate this above standard M2 fare. |
| Spec | 3.03u Static equilibrium: on rough surfaces3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2W(1\cos\theta) + W(2\cos\theta) = R_B(4\sin\theta)\) | *M1, A1, A1 | Moments about \(A\), all terms needed and no extra; dimensionally correct; each term must include \(\sin\theta\), or \(\cos\theta\) or \(\tan\theta\). May have \(Fr\) for \(R_B\) |
| \(R_A = 3W\) | B1 | |
| \(\mu = 2/3\) | dep*M1, A1 | Use of \(Fr = \mu R_A\) to get an equation in \(W\) (or \(R_A\)) only. AG; must come from exact \(\theta\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(W(2\cos\theta) + 2W(3\cos\theta) + Fr(4\sin\theta) = R_A(4\cos\theta)\) | *M1, A1, A1 | Moments about \(B\), all terms needed and no extra; dimensionally correct |
| \(R_A = 3W\) | B1 | |
| \(\mu = 2/3\) | dep*M1, A1 | Use of \(Fr = \mu R_A\); AG; must come from exact \(\theta\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((W + aW)(2\cos\theta) = R_B(4\sin\theta)\) | *M1, A1 | Moments about \(A\), all terms needed and no extra; dimensionally correct; may have \(Fr\) for \(R_B\) |
| \(R_A = 3W + aW\) | B1 | Resolve vertically |
| \(Fr \leq \mu R_A\) | dep*M1 | \(Fr \leq \mu R_A\) to get an inequality in \(W\) and \(a\) only; allow equality here |
| \(a \leq 3\) | A1 | Allow \(a < 3\). If using equality, correct inequality need not be justified. |
## Question 7:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2W(1\cos\theta) + W(2\cos\theta) = R_B(4\sin\theta)$ | *M1, A1, A1 | Moments about $A$, all terms needed and no extra; dimensionally correct; each term must include $\sin\theta$, or $\cos\theta$ or $\tan\theta$. May have $Fr$ for $R_B$ |
| $R_A = 3W$ | B1 | |
| $\mu = 2/3$ | dep*M1, A1 | Use of $Fr = \mu R_A$ to get an equation in $W$ (or $R_A$) only. **AG**; must come from exact $\theta$ |
**OR (moments about B):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $W(2\cos\theta) + 2W(3\cos\theta) + Fr(4\sin\theta) = R_A(4\cos\theta)$ | *M1, A1, A1 | Moments about $B$, all terms needed and no extra; dimensionally correct |
| $R_A = 3W$ | B1 | |
| $\mu = 2/3$ | dep*M1, A1 | Use of $Fr = \mu R_A$; **AG**; must come from exact $\theta$ |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(W + aW)(2\cos\theta) = R_B(4\sin\theta)$ | *M1, A1 | Moments about $A$, all terms needed and no extra; dimensionally correct; may have $Fr$ for $R_B$ |
| $R_A = 3W + aW$ | B1 | Resolve vertically |
| $Fr \leq \mu R_A$ | dep*M1 | $Fr \leq \mu R_A$ to get an inequality in $W$ and $a$ only; allow equality here |
| $a \leq 3$ | A1 | Allow $a < 3$. If using equality, correct inequality need not be justified. |
---(i) Show that $\mu = \frac { 2 } { 3 }$.
A small object of weight $a W \mathrm {~N}$ is placed on the ladder at its mid-point and the object $S$ of weight $2 W \mathrm {~N}$ is placed on the ladder at its lowest point $A$.\\
(ii) Given that the system is in equilibrium, find the set of possible values of $a$.
\hfill \mbox{\textit{OCR M2 2015 Q7 [11]}}