| Exam Board | OCR MEI |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Non-uniform rod hinged or with applied force |
| Difficulty | Standard +0.3 This is a standard M2 non-uniform rod equilibrium problem with three parts of increasing complexity. Part (i) is routine moments calculation with horizontal beam. Part (ii) requires taking moments about the hinge with forces at an angle—standard technique but requires careful resolution. Part (iii) involves friction at limiting equilibrium with multiple force resolution steps. All parts use well-practiced M2 techniques without requiring novel insight, making this slightly easier than average overall. |
| 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 | Mark | Guidance |
| c.w. moments about A | M1 | Moments equation |
| \(5R_B - 3 \times 85 = 0\) so \(R_B = 51\) giving \(51\) N \(\uparrow\) | A1 | Accept no direction given |
| Either a.c. moments about B or resolve \(\uparrow\) | M1 | |
| \(R_A = 34\) so \(34\) N \(\uparrow\) | F1 | Accept no direction given |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| c.w. moments about A | M1 | Moments with attempt to resolve at least one force. Allow \(s \leftrightarrow c\) |
| \(85 \times 3\cos\alpha - 27.2 \times 5\sin\alpha = 0\) | B1 | Weight term |
| B1 | Horiz force term | |
| so \(\tan\alpha = \frac{3 \times 85}{27.2 \times 5} = \frac{15}{8}\) | E1 | Must see some arrangement of terms or equiv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Diagram with all forces present and labelled | B1 | All forces present and labelled |
| a.c. moments about B | M1 | Moments with attempt to resolve forces and all relevant forces present |
| \(85 \times 2\cos\alpha + 34 \times 2.5 - 5S\sin\alpha = 0\) | B1 | \(34 \times 2.5\) |
| A1 | All other terms correct. Allow sign errors | |
| \(S = 37.4\) | A1 | All correct |
| Resolving horizontally and vertically | M1 | Either attempted |
| \(\rightarrow\; S - F - 34\sin\alpha = 0\) so \(F = 7.4\) | E1 | |
| \(\uparrow\; R - 85 - 34\cos\alpha = 0\) | A1 | \(R = 101\) need not be evaluated here. Allow A1 for two expressions if correct other than \(s \leftrightarrow c\) |
| Using \(F = \mu R\) | M1 | |
| \(\mu = \frac{7.4}{101} = 0.07326\ldots\) so \(0.0733\) (3 s.f.) | A1 | cao |
# Question 3:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| c.w. moments about A | M1 | Moments equation |
| $5R_B - 3 \times 85 = 0$ so $R_B = 51$ giving $51$ N $\uparrow$ | A1 | Accept no direction given |
| Either a.c. moments about B or resolve $\uparrow$ | M1 | |
| $R_A = 34$ so $34$ N $\uparrow$ | F1 | Accept no direction given |
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| c.w. moments about A | M1 | Moments with attempt to resolve at least one force. Allow $s \leftrightarrow c$ |
| $85 \times 3\cos\alpha - 27.2 \times 5\sin\alpha = 0$ | B1 | Weight term |
| | B1 | Horiz force term |
| so $\tan\alpha = \frac{3 \times 85}{27.2 \times 5} = \frac{15}{8}$ | E1 | Must see some arrangement of terms or equiv |
## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Diagram with all forces present and labelled | B1 | All forces present and labelled |
| a.c. moments about B | M1 | Moments with attempt to resolve forces and all relevant forces present |
| $85 \times 2\cos\alpha + 34 \times 2.5 - 5S\sin\alpha = 0$ | B1 | $34 \times 2.5$ |
| | A1 | All other terms correct. Allow sign errors |
| $S = 37.4$ | A1 | All correct |
| Resolving horizontally and vertically | M1 | Either attempted |
| $\rightarrow\; S - F - 34\sin\alpha = 0$ so $F = 7.4$ | E1 | |
| $\uparrow\; R - 85 - 34\cos\alpha = 0$ | A1 | $R = 101$ need not be evaluated here. Allow A1 for two expressions if correct other than $s \leftrightarrow c$ |
| Using $F = \mu R$ | M1 | |
| $\mu = \frac{7.4}{101} = 0.07326\ldots$ so $0.0733$ (3 s.f.) | A1 | cao |
---3 A non-uniform beam AB has weight 85 N . The length of the beam is 5 m and its centre of mass is 3 m from A . In this question all the forces act in the same vertical plane.
Fig. 3.1 shows the beam in horizontal equilibrium, supported at its ends.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{81efb50d-c89d-4ce1-94d7-592c946f6176-4_215_828_466_660}
\captionsetup{labelformat=empty}
\caption{Fig. 3.1}
\end{center}
\end{figure}
(i) Calculate the reactions of the supports on the beam.
Using a smooth hinge, the beam is now attached at A to a vertical wall. The beam is held in equilibrium at an angle $\alpha$ to the horizontal by means of a horizontal force of magnitude 27.2 N acting at B . This situation is shown in Fig. 3.2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{81efb50d-c89d-4ce1-94d7-592c946f6176-4_725_675_1153_347}
\captionsetup{labelformat=empty}
\caption{Fig. 3.2}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{81efb50d-c89d-4ce1-94d7-592c946f6176-4_732_565_1153_1231}
\captionsetup{labelformat=empty}
\caption{Fig. 3.3}
\end{center}
\end{figure}
(ii) Show that $\tan \alpha = \frac { 15 } { 8 }$.
The hinge and 27.2 N force are removed. The beam now rests with B on a rough horizontal floor and A on a smooth vertical wall, as shown in Fig. 3.3. It is at the same angle $\alpha$ to the horizontal. There is now a force of 34 N acting at right angles to the beam at its centre in the direction shown. The beam is in equilibrium and on the point of slipping.\\
(iii) Draw a diagram showing the forces acting on the beam.
Show that the frictional force acting on the beam is 7.4 N .\\
Calculate the value of the coefficient of friction between the beam and the floor.
\hfill \mbox{\textit{OCR MEI M2 2009 Q3 [18]}}