| Exam Board | OCR MEI |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Framework or multiple rod structures |
| Difficulty | Challenging +1.2 This is a multi-part moments question requiring systematic application of equilibrium conditions (sum of forces and moments equals zero) across three scenarios. While it involves multiple steps and careful bookkeeping of distances and forces, the techniques are standard M2 material with no novel insights required. The framework structure and given center of mass location make it more straightforward than problems requiring geometric reasoning to find the center of mass. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force6.04c Composite bodies: centre of mass |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| a.c. moments about B: \(10T_C - 15 \times 2 = 0\) | M1 | Moments with all forces present, no extra forces |
| so \(T_C = 3\). Tension at C is 3 N | A1 | |
| \(\uparrow T_C + T_B - 15 = 0\) | M1 | May take moments again |
| so \(T_B = 12\). Tension at B is 12 N | F1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| a.c. moments about A: \(25T\sin 30 - 15 \times 17 = 0\) | M1 | Attempt at moments with resolution; allow \(\cos \leftrightarrow \sin\) error. All forces present, no extra forces |
| so \(T = 20.4\) | A1 | cao |
| At A let force \(\uparrow\) be \(Y\) N: \(\uparrow Y + T\sin 30 - 15 = 0\) so \(Y = 4.8\) | B1 | FT (can take moments about C) |
| \(\rightarrow X = T\cos 30 = 17.6669\ldots\) N | B1 | FT. Need not be evaluated |
| \(\sqrt{4.8^2 + (T\cos 30)^2}\) | M1 | |
| \(= 18.3073755\ldots\) so 18.3 N (3 s.f.) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Let force be \(P\). a.c. moments about D: \(8 \times 15 - 12 \times P = 0\) | M1 | Moments about \(D\) with all forces present, no extra forces |
| so \(P = 10\) on point of tipping | A1 | cao |
| Using \(F_{\max} = \mu R\) on point of slipping with \(R = 15\) | M1 | |
| B1 | ||
| gives \(F_{\max} = 0.65 \times 15 = 9.75\) | A1 | cao |
| so slips first | E1 | cao and WWW |
# Question 2:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| a.c. moments about B: $10T_C - 15 \times 2 = 0$ | M1 | Moments with all forces present, no extra forces |
| so $T_C = 3$. Tension at C is 3 N | A1 | |
| $\uparrow T_C + T_B - 15 = 0$ | M1 | May take moments again |
| so $T_B = 12$. Tension at B is 12 N | F1 | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| a.c. moments about A: $25T\sin 30 - 15 \times 17 = 0$ | M1 | Attempt at moments with resolution; allow $\cos \leftrightarrow \sin$ error. All forces present, no extra forces |
| so $T = 20.4$ | A1 | cao |
| At A let force $\uparrow$ be $Y$ N: $\uparrow Y + T\sin 30 - 15 = 0$ so $Y = 4.8$ | B1 | FT (can take moments about C) |
| $\rightarrow X = T\cos 30 = 17.6669\ldots$ N | B1 | FT. Need not be evaluated |
| $\sqrt{4.8^2 + (T\cos 30)^2}$ | M1 | |
| $= 18.3073755\ldots$ so 18.3 N (3 s.f.) | A1 | cao |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Let force be $P$. a.c. moments about D: $8 \times 15 - 12 \times P = 0$ | M1 | Moments about $D$ with all forces present, no extra forces |
| so $P = 10$ on point of tipping | A1 | cao |
| Using $F_{\max} = \mu R$ on point of slipping with $R = 15$ | M1 | |
| | B1 | |
| gives $F_{\max} = 0.65 \times 15 = 9.75$ | A1 | cao |
| so slips first | E1 | cao and WWW |
---2 The rigid object shown in Fig. 2.1 is made of thin non-uniform rods. ABC is a straight line; $\mathrm { BC } , \mathrm { BE }$ and ED form three sides of a rectangle. The centre of mass of the object is at G. The lengths are in centimetres. The weight of the object is 15 N .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ea3c0177-bf3b-4475-9ab1-ae628aeb0bf0-3_273_444_397_813}
\captionsetup{labelformat=empty}
\caption{Fig. 2.1}
\end{center}
\end{figure}
Initially, the object is suspended by light vertical strings attached to B and to C and hangs in equilibrium with AC horizontal.\\
(i) Calculate the tensions in each of the strings.
In a new situation the strings are removed. The object can rotate freely in a vertical plane about a fixed horizontal axis through A and perpendicular to ABCDE. The object is held in equilibrium with AC horizontal by a force of magnitude $T \mathrm {~N}$ in the plane ABCDE acting at C at an angle of $30 ^ { \circ }$ to CA . This situation is shown in Fig. 2.2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ea3c0177-bf3b-4475-9ab1-ae628aeb0bf0-3_356_451_1292_808}
\captionsetup{labelformat=empty}
\caption{Fig. 2.2}
\end{center}
\end{figure}
(ii) Calculate $T$.
Calculate also the magnitude of the force exerted on the object by the axis at A .
The object is now placed on a rough horizontal table and is in equilibrium with ABCDE in a vertical plane and DE in contact with the table. The coefficient of friction between the edge DE and the table is 0.65 . A force of slowly increasing magnitude (starting at 0 N ) is applied at A in the direction AB . Assume that the object remains in a vertical plane.\\
(iii) Determine whether the object slips before it tips.
\hfill \mbox{\textit{OCR MEI M2 2012 Q2 [16]}}