| Exam Board | OCR MEI |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2007 |
| Session | June |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Non-uniform rod on supports or with strings |
| Difficulty | Standard +0.3 This is a standard M2 moments question with multiple parts building in complexity. Parts (i)-(iii) involve routine application of equilibrium conditions (sum of forces = 0, sum of moments = 0) with straightforward arithmetic. Part (iv) requires slightly more insight about friction and optimization, but follows a predictable pattern for this topic. The inclined plane adds minor complexity but the trig values are given. Overall, this is slightly easier than average as it's methodical rather than requiring novel problem-solving. |
| Spec | 3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Taking moments about B: \(R_A \times 0.6 = 200 \times 0.2\) | M1 | Moments about one support |
| Wait — distances: A is 0.2m from G, B is 0.6m from G | ||
| \(R_A \times 0.6 = 200 \times 0.6\)... re-reading: B to G = 0.6m, G to A = 0.2m | ||
| Moments about B: \(R_A \times (0.6+0.2) = 200 \times 0.6\) | M1 A1 | |
| \(R_A \times 0.8 = 120\), \(R_A = 150\) N | A1 | |
| \(R_B = 200 - 150 = 50\) N | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Taking moments about C (C is 0.8m from D, G is 0.4m from C): | M1 | |
| \(R_D \times 0.8 = 200 \times 0.4\) (downward force at G, peg at D upward or downward) | A1 | |
| Check: D is at left end region, C is support | ||
| Moments about C: \(R_D \times 0.8 = 200 \times 0.4 + ...\), careful with diagram | ||
| \(R_C - R_D = 200\) (resolve) | M1 | |
| \(R_D \times 0.8 = 200 \times 0.4\) gives \(R_D = 100\) N downward (peg pushes down) | A1 | Direction stated on diagram |
| \(R_C = 300\) N upward | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Taking moments about Q (perpendicular distances): | M1 | |
| \(R_P \times 0.8\cos\alpha = 200 \times 0.4\cos\alpha\) (or use perpendicular distances along slope) | M1 | Must use correct perpendicular distances |
| Moments about Q: \(R_P \times 0.8 = 200 \times 0.4\) (distances along plank) | A1 | |
| Wait — distances along plank: Q to P = 0.8m, P to G = 0.4m, G to end = 1.4m, Q to end = 0.2m beyond Q... | ||
| Re-reading fig 3.3: Q is 0.2m from left end, P is 0.8m from Q (i.e. 1.0m from left), G is 0.4m from P | ||
| Moments about Q: \(R_P \times 0.8 = 200\cos\alpha \times (0.4+0.8)\) ... perpendicular to plank | M1 A1 | |
| \(R_P = \frac{200 \times 0.96 \times 1.2}{0.8} = 288\) N | A1 | |
| Moments about P: \(R_Q \times 0.8 = 200\cos\alpha \times 0.4 + ...\) | M1 | |
| \(R_Q = \frac{200\times0.96\times0.4}{0.8}\)... checking signs with geometry | A1 | |
| \(R_Q = 96\) N | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Friction needed at P and Q: resolve along plane | M1 | |
| Net force along plane: \(200\sin\alpha = 200 \times 0.28 = 56\) N | A1 | |
| Friction at rough contact \(= 56\) N (if only one contact rough) | ||
| \(\mu = \frac{F}{R}\); at P: \(\mu = \frac{56}{288}\); at Q: \(\mu = \frac{56}{96}\) | M1 | |
| Smallest \(\mu\) is at P (larger normal reaction) | A1 | |
| \(\mu = \frac{56}{288} = \frac{7}{36} \approx 0.194\) | A1 |
# Question 3:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Taking moments about B: $R_A \times 0.6 = 200 \times 0.2$ | M1 | Moments about one support |
| Wait — distances: A is 0.2m from G, B is 0.6m from G | | |
| $R_A \times 0.6 = 200 \times 0.6$... re-reading: B to G = 0.6m, G to A = 0.2m | | |
| Moments about B: $R_A \times (0.6+0.2) = 200 \times 0.6$ | M1 A1 | |
| $R_A \times 0.8 = 120$, $R_A = 150$ N | A1 | |
| $R_B = 200 - 150 = 50$ N | A1 | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Taking moments about C (C is 0.8m from D, G is 0.4m from C): | M1 | |
| $R_D \times 0.8 = 200 \times 0.4$ (downward force at G, peg at D upward or downward) | A1 | |
| Check: D is at left end region, C is support | | |
| Moments about C: $R_D \times 0.8 = 200 \times 0.4 + ...$, careful with diagram | | |
| $R_C - R_D = 200$ (resolve) | M1 | |
| $R_D \times 0.8 = 200 \times 0.4$ gives $R_D = 100$ N downward (peg pushes down) | A1 | Direction stated on diagram |
| $R_C = 300$ N upward | A1 | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Taking moments about Q (perpendicular distances): | M1 | |
| $R_P \times 0.8\cos\alpha = 200 \times 0.4\cos\alpha$ (or use perpendicular distances along slope) | M1 | Must use correct perpendicular distances |
| Moments about Q: $R_P \times 0.8 = 200 \times 0.4$ (distances along plank) | A1 | |
| Wait — distances along plank: Q to P = 0.8m, P to G = 0.4m, G to end = 1.4m, Q to end = 0.2m beyond Q... | | |
| Re-reading fig 3.3: Q is 0.2m from left end, P is 0.8m from Q (i.e. 1.0m from left), G is 0.4m from P | | |
| Moments about Q: $R_P \times 0.8 = 200\cos\alpha \times (0.4+0.8)$ ... perpendicular to plank | M1 A1 | |
| $R_P = \frac{200 \times 0.96 \times 1.2}{0.8} = 288$ N | A1 | |
| Moments about P: $R_Q \times 0.8 = 200\cos\alpha \times 0.4 + ...$ | M1 | |
| $R_Q = \frac{200\times0.96\times0.4}{0.8}$... checking signs with geometry | A1 | |
| $R_Q = 96$ N | A1 | |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Friction needed at P and Q: resolve along plane | M1 | |
| Net force along plane: $200\sin\alpha = 200 \times 0.28 = 56$ N | A1 | |
| Friction at rough contact $= 56$ N (if only one contact rough) | | |
| $\mu = \frac{F}{R}$; at P: $\mu = \frac{56}{288}$; at Q: $\mu = \frac{56}{96}$ | M1 | |
| Smallest $\mu$ is at P (larger normal reaction) | A1 | |
| $\mu = \frac{56}{288} = \frac{7}{36} \approx 0.194$ | A1 | |
---3 A uniform plank is 2.8 m long and has weight 200 N . The centre of mass is G.\\
(i) Fig. 3.1 shows the plank horizontal and in equilibrium, resting on supports at A and B .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8d4aeab2-332a-442f-b1e7-0bbf8a945f0f-5_229_1125_434_459}
\captionsetup{labelformat=empty}
\caption{Fig. 3.1}
\end{center}
\end{figure}
Calculate the reactions of the supports on the plank at A and at B .\\
(ii) Fig. 3.2 shows the plank horizontal and in equilibrium between a support at C and a peg at D .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8d4aeab2-332a-442f-b1e7-0bbf8a945f0f-5_236_1141_993_461}
\captionsetup{labelformat=empty}
\caption{Fig. 3.2}
\end{center}
\end{figure}
Calculate the reactions of the support and the peg on the plank at C and at D , showing the directions of these forces on a diagram.
Fig. 3.3 shows the plank in equilibrium between a support at P and a peg at Q . The plank is inclined at $\alpha$ to the horizontal, where $\sin \alpha = 0.28$ and $\cos \alpha = 0.96$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8d4aeab2-332a-442f-b1e7-0bbf8a945f0f-5_424_1099_1692_475}
\captionsetup{labelformat=empty}
\caption{Fig. 3.3}
\end{center}
\end{figure}
(iii) Calculate the normal reactions at P and at Q .\\
(iv) Just one of the contacts is rough. Determine which one it is if the value of the coefficient of friction is as small as possible. Find this value of the coefficient of friction.
\hfill \mbox{\textit{OCR MEI M2 2007 Q3 [19]}}