| Exam Board | OCR MEI |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | January |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Framework or multiple rod structures |
| Difficulty | Standard +0.3 This is a standard M2 moments and frameworks question with routine calculations. Part (a) involves straightforward equilibrium and moments about a point to find reactions and tipping conditions. Part (b) requires resolving forces and taking moments on a pin-jointed framework - all standard techniques with no novel insight required. The 'show that R=600' is a confidence-building step. While multi-part with several marks, each step follows textbook methods making it slightly easier than average. |
| 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/Working | Mark | Guidance |
| c.w. moments about P: \(2 \times 600 - 3R_Q = 0\) so force of \(400\) N ↑ at Q | M1, A1 | Moments taken about a named point. |
| a.c. moments about Q or resolve: \(R_P = 200\) so force of \(200\) N ↑ at P | M1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(R_P = 0\) | B1 | Clearly recognised or used. |
| c.w. moments about Q | M1 | Moments attempted with all forces. Dep on \(R_P = 0\) or \(R_P\) not evaluated. |
| \(2L - 1 \times 600 = 0\) so \(L = 300\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\cos\alpha = \frac{15}{17}\) or \(\sin\alpha = \frac{8}{17}\) or \(\tan\alpha = \frac{8}{15}\) | B1 | Seen here or below or implied by use. |
| c.w. moments about A | M1 | Moments. All forces must be present and appropriate resolution attempted. |
| \(16 \times 340\cos\alpha - 8R = 0\) | A1 | |
| so \(R = 600\) | E1 | Evidence of evaluation. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Diagram | B1 | Must have \(600\) (or \(R\)) and \(340\) N and reactions at A. |
| B1 | All internal forces clearly marked as tension or thrust. Allow mixture. [Max of B1 if extra forces present] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| B↓: \(340\cos\alpha + T_{BC}\cos\alpha = 0\) so \(T_{BC} = -340\) (Thrust of \(340\) N in BC) | M1, A1 | Equilibrium at a pin-joint |
| C→: \(T_{BC}\sin\alpha - T_{AC}\sin\alpha = 0\) so \(T_{AC} = -340\) (Thrust of \(340\) N in AC) | F1 | |
| B←: \(T_{AB} + T_{BC}\sin\alpha - 340\sin\alpha = 0\) so \(T_{AB} = 320\) (Tension of \(320\) N in AB) | M1, A1 | Method for \(T_{AB}\) |
| Tension/Thrust all consistent with working | F1 | [Award a max of 4/6 if working inconsistent with diagram] |
# Question 4:
## Part (a)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| c.w. moments about P: $2 \times 600 - 3R_Q = 0$ so force of $400$ N ↑ at Q | M1, A1 | Moments taken about a named point. |
| a.c. moments about Q or resolve: $R_P = 200$ so force of $200$ N ↑ at P | M1, A1 | |
## Part (a)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R_P = 0$ | B1 | Clearly recognised or used. |
| c.w. moments about Q | M1 | Moments attempted with all forces. Dep on $R_P = 0$ or $R_P$ not evaluated. |
| $2L - 1 \times 600 = 0$ so $L = 300$ | A1 | |
## Part (b)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cos\alpha = \frac{15}{17}$ or $\sin\alpha = \frac{8}{17}$ or $\tan\alpha = \frac{8}{15}$ | B1 | Seen here or below or implied by use. |
| c.w. moments about A | M1 | Moments. All forces must be present and appropriate resolution attempted. |
| $16 \times 340\cos\alpha - 8R = 0$ | A1 | |
| so $R = 600$ | E1 | Evidence of evaluation. |
## Part (b)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Diagram | B1 | Must have $600$ (or $R$) and $340$ N and reactions at A. |
| | B1 | All internal forces clearly marked as tension or thrust. Allow mixture. [Max of B1 if extra forces present] |
## Part (b)(iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| B↓: $340\cos\alpha + T_{BC}\cos\alpha = 0$ so $T_{BC} = -340$ (Thrust of $340$ N in BC) | M1, A1 | Equilibrium at a pin-joint |
| C→: $T_{BC}\sin\alpha - T_{AC}\sin\alpha = 0$ so $T_{AC} = -340$ (Thrust of $340$ N in AC) | F1 | |
| B←: $T_{AB} + T_{BC}\sin\alpha - 340\sin\alpha = 0$ so $T_{AB} = 320$ (Tension of $320$ N in AB) | M1, A1 | Method for $T_{AB}$ |
| Tension/Thrust all consistent with working | F1 | [Award a max of 4/6 if working inconsistent with diagram] |4
\begin{enumerate}[label=(\alph*)]
\item A uniform, rigid beam, AB , has a weight of 600 N . It is horizontal and in equilibrium resting on two small smooth pegs at P and Q . Fig. 4.1 shows the positions of the pegs; lengths are in metres.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3865b4b3-97c7-412b-aabd-2705a954a847-5_229_647_404_790}
\captionsetup{labelformat=empty}
\caption{Fig. 4.1}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Calculate the forces exerted by the pegs on the beam.
A force of $L \mathrm {~N}$ is applied vertically downwards at B . The beam is in equilibrium but is now on the point of tipping.
\item Calculate the value of $L$.
\end{enumerate}\item Fig. 4.2 shows a framework in a vertical plane constructed of light, rigid rods $\mathrm { AB } , \mathrm { BC }$ and CA . The rods are freely pin-jointed to each other at $\mathrm { A } , \mathrm { B }$ and C and to a fixed point at A . The pin-joint at C rests on a smooth, horizontal support. The dimensions of the framework are shown in metres. There is a force of 340 N acting at B in the plane of the framework. This force and the $\operatorname { rod } \mathrm { BC }$ are both inclined to the vertical at an angle $\alpha$, which is defined in triangle BCX . The force on the framework exerted by the support at C is $R \mathrm {~N}$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3865b4b3-97c7-412b-aabd-2705a954a847-5_675_869_1434_678}
\captionsetup{labelformat=empty}
\caption{Fig. 4.2}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Show that $R = 600$.
\item Draw a diagram showing all the forces acting on the framework and also the internal forces in the rods.\\[0pt]
\item Calculate the internal forces in the three rods, indicating whether each rod is in tension or in compression (thrust). [Your working in this part should correspond to your diagram in part (ii).]
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI M2 2009 Q4 [19]}}