| Exam Board | OCR MEI |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Lamina hinged at point with string support |
| Difficulty | Standard +0.3 This is a standard M2 moments question with multiple parts requiring systematic application of moment equilibrium and force resolution. Parts (i)-(iii) involve routine calculations about a hinged lamina with given geometry, while part (iv) adds friction on an inclined plane with acceleration—all standard M2 techniques without requiring novel insight or particularly complex problem-solving. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| c.w. moments about A: \(60\cos40° \times 0.3 - 60\sin40° \times 0.1\) | M1 | oe e.g. \(60(0.3 - 0.1\tan40°)\sin50°\). Method of dealing with moment of weight. Allow \(\cos \leftrightarrow \sin\). Both weight terms correct; allow wrong overall sign but not both terms with same sign |
| \(= 9.93207\ldots\) so \(9.93 \text{ N m}\) | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P\cos40° \times 0.2 - 9.93207\ldots = 0\) | M1 | Moments of all relevant forces attempted. No extra terms. Allow \(\cos \leftrightarrow \sin\) |
| \(P = 64.827\ldots\) so \(64.8\) | A1 | cao (\(64.813\ldots\) if \(9.93\) used) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| a.c. moments about A to find NR, \(R\), at B: \(R \times 0.8 = 9.93\) or \(R \times 0.8 + 60\sin40° \times 0.1 - 60\cos40° \times 0.3 = 0\) | M1 | Attempt to use moments to find \(R\). Moments of all relevant forces attempted. No extra terms. Allow \(\cos \leftrightarrow \sin\). Note mmts about B can score M1 only if mmt of horiz component of force at A is included. If \(R\) is taken as vertical, M0 |
| \(R = 12.4150\ldots\) | A1 | FT their moment of weight from (i). Not a required answer |
| Resolve vertically: \(Y - 60 + R\cos40° = 0\) | depM1 | Note second M mark must be for complete method to find \(Y\); FT their calculated \(R\) |
| \(Y = 50.489\ldots\) so \(50.5 \text{ N}\) | A1 | FT their calculated \(R\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Resolve perp to plane: \(R - 60\cos40° - 200\sin40° = 0\) | M1 | All terms present and no extra terms. Components of 60 and 200; allow \(\cos \leftrightarrow \sin\) |
| \(R = 174.52\ldots\) | A1 | Not a required answer |
| N2L up the plane: \(200\cos40° - F - 60\sin40° = \dfrac{60}{9.8} \times 1.75\) | M1 | Use of N2L with all terms present and no extras. Components of 60 and 200; allow \(\cos \leftrightarrow \sin\). Allow use of 60 for mass |
| B1 | Use of mass not weight | |
| A1 | FT use of weight and/or sign errors | |
| \(F = 103.927\ldots\) | A1 | All correct. Not a required answer |
| As friction limiting \(F = \mu R\): \(\mu = \dfrac{103.927\ldots}{174.520\ldots}\) | M1 | FT their \(F\) and their \(R\) |
| \(= 0.59550\ldots\) so \(0.596\) | A1 | cao |
# Question 3:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| c.w. moments about A: $60\cos40° \times 0.3 - 60\sin40° \times 0.1$ | M1 | oe e.g. $60(0.3 - 0.1\tan40°)\sin50°$. Method of dealing with moment of weight. Allow $\cos \leftrightarrow \sin$. Both weight terms correct; allow wrong overall sign but not both terms with same sign |
| $= 9.93207\ldots$ so $9.93 \text{ N m}$ | E1 | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P\cos40° \times 0.2 - 9.93207\ldots = 0$ | M1 | Moments of all relevant forces attempted. No extra terms. Allow $\cos \leftrightarrow \sin$ |
| $P = 64.827\ldots$ so $64.8$ | A1 | cao ($64.813\ldots$ if $9.93$ used) |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| a.c. moments about A to find NR, $R$, at B: $R \times 0.8 = 9.93$ or $R \times 0.8 + 60\sin40° \times 0.1 - 60\cos40° \times 0.3 = 0$ | M1 | Attempt to use moments to find $R$. Moments of all relevant forces attempted. No extra terms. Allow $\cos \leftrightarrow \sin$. Note mmts about B can score M1 only if mmt of horiz component of force at A is included. If $R$ is taken as vertical, M0 |
| $R = 12.4150\ldots$ | A1 | FT their moment of weight from (i). Not a required answer |
| Resolve vertically: $Y - 60 + R\cos40° = 0$ | depM1 | Note second M mark must be for complete method to find $Y$; FT their calculated $R$ |
| $Y = 50.489\ldots$ so $50.5 \text{ N}$ | A1 | FT their calculated $R$ |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Resolve perp to plane: $R - 60\cos40° - 200\sin40° = 0$ | M1 | All terms present and no extra terms. Components of 60 and 200; allow $\cos \leftrightarrow \sin$ |
| $R = 174.52\ldots$ | A1 | Not a required answer |
| N2L up the plane: $200\cos40° - F - 60\sin40° = \dfrac{60}{9.8} \times 1.75$ | M1 | Use of N2L with all terms present and no extras. Components of 60 and 200; allow $\cos \leftrightarrow \sin$. Allow use of 60 for mass |
| | B1 | Use of mass not weight |
| | A1 | FT use of weight and/or sign errors |
| $F = 103.927\ldots$ | A1 | All correct. Not a required answer |
| As friction limiting $F = \mu R$: $\mu = \dfrac{103.927\ldots}{174.520\ldots}$ | M1 | FT their $F$ and their $R$ |
| $= 0.59550\ldots$ so $0.596$ | A1 | cao |3 Fig. 3.1 shows a rigid, thin, non-uniform 20 cm by 80 cm rectangular panel ABCD of weight 60 N that is in a vertical plane. Its dimensions and the position of its centre of mass, $G$, are shown in centimetres. The panel is free to rotate about a fixed horizontal axis through A perpendicular to its plane; the panel rests on a small smooth fixed peg at B positioned so that AB is at $40 ^ { \circ }$ to the horizontal. A horizontal force in the plane of ABCD of magnitude $P \mathrm {~N}$ acts at D away from the panel.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c8f26b7e-1be1-4abf-8fea-6847185fad81-4_451_737_493_646}
\captionsetup{labelformat=empty}
\caption{Fig. 3.1}
\end{center}
\end{figure}
(i) Show that the clockwise moment of the weight about A is 9.93 Nm , correct to 3 significant figures.\\
(ii) Calculate the value of $P$ for which the panel is on the point of turning about the axis through A .\\
(iii) In the situation where $P = 0$, calculate the vertical component of the force exerted on the panel by the axis through A .
The panel is now placed on a line of greatest slope of a rough plane inclined at $40 ^ { \circ }$ to the horizontal. The panel is at all times in a vertical plane. A horizontal force in the plane ABCD of magnitude 200 N acts at D towards the panel. This situation is shown in Fig. 3.2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c8f26b7e-1be1-4abf-8fea-6847185fad81-4_497_842_1653_616}
\captionsetup{labelformat=empty}
\caption{Fig. 3.2}
\end{center}
\end{figure}
(iv) Given that the panel is moving up the plane with acceleration up the plane of $1.75 \mathrm {~ms} ^ { - 2 }$, calculate the coefficient of friction between the panel and the plane.
\hfill \mbox{\textit{OCR MEI M2 2013 Q3 [17]}}