| Exam Board | OCR MEI |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2006 |
| Session | January |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Beam on point of tilting |
| Difficulty | Standard +0.3 This is a standard M2 moments question with three routine parts: finding reactions on a horizontal beam, calculating force when one support has zero reaction, and analyzing a beam at an angle with friction. All parts use standard techniques (taking moments about a point, resolving forces) with straightforward geometry and no novel problem-solving required. Slightly easier than average due to the step-by-step structure and given answer to show in part (iii)(A). |
| Spec | 3.03t Coefficient of friction: F <= mu*R model3.04b Equilibrium: zero resultant moment and force6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Moments about C: \(240 \times 2 = 3R_D\) | M1 | Moments about C or equivalent. Allow 1 force omitted |
| \(R_D = 160\) so 160 N | A1 | |
| Resolve vertically: \(R_C + R_D = 240\) | M1 | Resolve vertically or moments about D or equivalent. All forces present. |
| \(R_C = 80\) so 80 N | F1 | FT from their \(R_D\) only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Moments about D: \(240 \times 1 = 4T\sin 40\) | M1 | Moments about D or equivalent |
| M1 | Attempt at resolution for RHS | |
| A1 | RHS correct | |
| \(T = 93.343...\) so 93.3 N (3 s.f.) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| In equilibrium so horizontal force needed to balance component of \(T\). This must be friction and cannot be at C. | E1 | Need reference to horizontal force that must come from friction at D. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Moments about B: \(3 \times 240 \times \cos 30 = 6P\) | M1 | All terms present, no extras. Any resolution required attempted. |
| \(P = 60\sqrt{3}\) (103.92...) | E1 | Accept decimal equivalent |
| \(P\) inclined at \(30°\) to vertical | B1 | Seen or equivalent or implied in (iii)(A) or (B). |
| Resolve horizontally. Friction force \(F\): \(F = P\sin 30\) | M1 | Resolve horizontally. Any resolution required attempted |
| so \(F = 30\sqrt{3}\) (51.961...) | A1 | Any form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Resolve vertically. Normal reaction \(R\): \(P\cos 30 + R = 240\) | M1 | Resolve vertically. All terms present and resolution attempted |
| A1 | ||
| Using \(F = \mu R\) | M1 | |
| \(\mu = \dfrac{30\sqrt{3}}{240 - 60\sqrt{3} \times \dfrac{\sqrt{3}}{2}}\) | A1 | Substitute their expressions for \(F\) and \(R\) |
| \(= \dfrac{30\sqrt{3}}{240 - 90} = \dfrac{\sqrt{3}}{5} = 0.34641\) so 0.346 (3 s.f.) | A1 | cao. Any form. Accept 2 s.f. or better |
# Question 2:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| Moments about C: $240 \times 2 = 3R_D$ | M1 | Moments about C or equivalent. Allow 1 force omitted |
| $R_D = 160$ so 160 N | A1 | |
| Resolve vertically: $R_C + R_D = 240$ | M1 | Resolve vertically or moments about D or equivalent. All forces present. |
| $R_C = 80$ so 80 N | F1 | FT from **their** $R_D$ only |
## Part (ii)(A)
| Answer | Mark | Guidance |
|--------|------|----------|
| Moments about D: $240 \times 1 = 4T\sin 40$ | M1 | Moments about D or equivalent |
| | M1 | Attempt at resolution for RHS |
| | A1 | RHS correct |
| $T = 93.343...$ so 93.3 N (3 s.f.) | A1 | |
## Part (ii)(B)
| Answer | Mark | Guidance |
|--------|------|----------|
| In equilibrium so horizontal force needed to balance component of $T$. This must be friction and cannot be at C. | E1 | Need reference to horizontal force that must come from friction at D. |
## Part (iii)(A)
| Answer | Mark | Guidance |
|--------|------|----------|
| Moments about B: $3 \times 240 \times \cos 30 = 6P$ | M1 | All terms present, no extras. Any resolution required attempted. |
| $P = 60\sqrt{3}$ (103.92...) | E1 | Accept decimal equivalent |
| $P$ inclined at $30°$ to vertical | B1 | Seen or equivalent or implied in (iii)(A) or (B). |
| Resolve horizontally. Friction force $F$: $F = P\sin 30$ | M1 | Resolve horizontally. Any resolution required attempted |
| so $F = 30\sqrt{3}$ (51.961...) | A1 | Any form |
## Part (iii)(B)
| Answer | Mark | Guidance |
|--------|------|----------|
| Resolve vertically. Normal reaction $R$: $P\cos 30 + R = 240$ | M1 | Resolve vertically. All terms present and resolution attempted |
| | A1 | |
| Using $F = \mu R$ | M1 | |
| $\mu = \dfrac{30\sqrt{3}}{240 - 60\sqrt{3} \times \dfrac{\sqrt{3}}{2}}$ | A1 | Substitute **their** expressions for $F$ and $R$ |
| $= \dfrac{30\sqrt{3}}{240 - 90} = \dfrac{\sqrt{3}}{5} = 0.34641$ so 0.346 (3 s.f.) | A1 | cao. Any form. Accept 2 s.f. or better |
---2 A uniform beam, AB , is 6 m long and has a weight of 240 N .\\
Initially, the beam is in equilibrium on two supports at C and D, as shown in Fig. 2.1. The beam is horizontal.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c1785fde-a6ce-4f8b-9948-4b4dd973ce84-3_200_687_486_689}
\captionsetup{labelformat=empty}
\caption{Fig. 2.1}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Calculate the forces acting on the beam from the supports at C and D .
A workman tries to move the beam by applying a force $T \mathrm {~N}$ at A at $40 ^ { \circ }$ to the beam, as shown in Fig. 2.2. The beam remains in horizontal equilibrium but the reaction of support C on the beam is zero.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c1785fde-a6ce-4f8b-9948-4b4dd973ce84-3_318_691_1119_687}
\captionsetup{labelformat=empty}
\caption{Fig. 2.2}
\end{center}
\end{figure}
\item (A) Calculate the value of $T$.\\
(B) Explain why the support at D cannot be smooth.
The beam is now supported by a light rope attached to the beam at A , with B on rough, horizontal ground. The rope is at $90 ^ { \circ }$ to the beam and the beam is at $30 ^ { \circ }$ to the horizontal, as shown in Fig. 2.3. The tension in the rope is $P \mathrm {~N}$. The beam is in equilibrium on the point of sliding.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c1785fde-a6ce-4f8b-9948-4b4dd973ce84-3_438_633_1909_708}
\captionsetup{labelformat=empty}
\caption{Fig. 2.3}
\end{center}
\end{figure}
\item (A) Show that $P = 60 \sqrt { 3 }$ and hence, or otherwise, find the frictional force between the beam and the ground.\\
(B) Calculate the coefficient of friction between the beam and the ground.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI M2 2006 Q2 [19]}}