| Exam Board | OCR MEI |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2006 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod hinged to wall with string support |
| Difficulty | Standard +0.3 This is a standard M2 statics problem involving two-rod systems with hinges. Part (i) requires routine moment calculations about a point with horizontal rods (straightforward). Parts (ii)-(iv) involve resolving forces and taking moments with rods at 60° to vertical—standard textbook techniques requiring trigonometry and simultaneous equations, but no novel insight. Slightly above average due to the multi-part nature and inclined geometry, but well within typical M2 scope. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Diagrams showing correct forces on each rod | B1 | |
| Taking moments about A for rod AB: | M1 | |
| \(F_B \times 3 = 90 \times 2 + F_R \times 3.5\) ... wait, moment about B for AB: | ||
| Let \(B\) exert vertical force \(F\) on AB. Taking moments about A for AB: | ||
| \(F \times 3 = 90 \times 2\) (if no support at R — need to incorporate R) | ||
| Taking moments about A for whole system: | M1 | |
| \(F_C \times 6 + F_R \times 3.5 = 90 \times 2 + 75 \times 5\) | A1 | |
| Taking moments about B for BC: \(F_C \times 3 = 75 \times 2 - F_B^{vert}\times 0\) | M1 | Moments about B for BC |
| \(F_C \times 3 = 75 \times 2\), so \(F_C = 50\) N | A1 | |
| Hinge force at B (vertical): resolve BC: \(F_B = 75 - 50 = 25\) N upward | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Taking moments about A for rod AB (inclined at \(60°\) to vertical, length 3 m): | M1 | Correct moment equation |
| Perpendicular distances correctly calculated using \(60°\) geometry | M1 | |
| \(90 \times \frac{3}{2}\sin 60° = U \times 3\sin 60° - V \times 3\cos 60°\) ... rearranged to | A1 | |
| \(60\sqrt{3} = U + V\sqrt{3}\) | A1 | Shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Diagram showing: weight 75 N downward at Q, reaction forces \(U\)N and \(V\)N (equal and opposite to those on AB) at B, normal reaction and friction at C | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Taking moments about C for rod BC | M1 | |
| \(U \times 3\sin 60° - V \times 3\cos 60° = 75 \times \frac{3}{2}\sin 60°\) giving second equation | A1 | |
| Resolving horizontally for BC: \(U = \) friction at C | M1 | |
| Resolving vertically for BC: \(V = 75 - N_C\) | ||
| Solving simultaneously: \(V = 30\sqrt{3} - 30\), \(U = 30\) (approximately) | A1 A1 | |
| Frictional force at C \(= U = 30\) N | M1 A1 |
# Question 2:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Diagrams showing correct forces on each rod | B1 | |
| Taking moments about A for rod AB: | M1 | |
| $F_B \times 3 = 90 \times 2 + F_R \times 3.5$ ... wait, moment about B for AB: | | |
| Let $B$ exert vertical force $F$ on AB. Taking moments about A for AB: | | |
| $F \times 3 = 90 \times 2$ (if no support at R — need to incorporate R) | | |
| Taking moments about A for whole system: | M1 | |
| $F_C \times 6 + F_R \times 3.5 = 90 \times 2 + 75 \times 5$ | A1 | |
| Taking moments about B for BC: $F_C \times 3 = 75 \times 2 - F_B^{vert}\times 0$ | M1 | Moments about B for BC |
| $F_C \times 3 = 75 \times 2$, so $F_C = 50$ N | A1 | |
| Hinge force at B (vertical): resolve BC: $F_B = 75 - 50 = 25$ N upward | A1 | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Taking moments about A for rod AB (inclined at $60°$ to vertical, length 3 m): | M1 | Correct moment equation |
| Perpendicular distances correctly calculated using $60°$ geometry | M1 | |
| $90 \times \frac{3}{2}\sin 60° = U \times 3\sin 60° - V \times 3\cos 60°$ ... rearranged to | A1 | |
| $60\sqrt{3} = U + V\sqrt{3}$ | A1 | Shown |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Diagram showing: weight 75 N downward at Q, reaction forces $U$N and $V$N (equal and opposite to those on AB) at B, normal reaction and friction at C | B1 | |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Taking moments about C for rod BC | M1 | |
| $U \times 3\sin 60° - V \times 3\cos 60° = 75 \times \frac{3}{2}\sin 60°$ giving second equation | A1 | |
| Resolving horizontally for BC: $U = $ friction at C | M1 | |
| Resolving vertically for BC: $V = 75 - N_C$ | | |
| Solving simultaneously: $V = 30\sqrt{3} - 30$, $U = 30$ (approximately) | A1 A1 | |
| Frictional force at C $= U = 30$ N | M1 A1 | |
---2 Two heavy rods AB and BC are freely jointed together at B and to a wall at A . AB has weight 90 N and centre of mass at $\mathrm { P } ; \mathrm { BC }$ has weight 75 N and centre of mass at Q . The lengths of the rods and the positions of P and Q are shown in Fig. 2.1, with the lengths in metres.
Initially, AB and BC are horizontal. There is a support at R , as shown in Fig. 2.1. The system is held in equilibrium by a vertical force acting at C .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{31c38a58-e9d5-4d01-90af-3b41213a9c7d-3_381_703_584_680}
\captionsetup{labelformat=empty}
\caption{Fig. 2.1}
\end{center}
\end{figure}
(i) Draw diagrams showing all the forces acting on $\operatorname { rod } \mathrm { AB }$ and on $\operatorname { rod } \mathrm { BC }$.
Calculate the force exerted on AB by the hinge at B and hence the force required at C .
The rods are now set up as shown in Fig. 2.2. AB and BC are each inclined at $60 ^ { \circ }$ to the vertical and C rests on a rough horizontal table. Fig. 2.3 shows all the forces acting on AB , including the forces $X \mathrm {~N}$ and $Y \mathrm {~N}$ due to the hinge at A and the forces $U \mathrm {~N}$ and $V \mathrm {~N}$ in the hinge at B . The rods are in equilibrium.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{31c38a58-e9d5-4d01-90af-3b41213a9c7d-3_393_661_1615_429}
\captionsetup{labelformat=empty}
\caption{Fig. 2.2}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{31c38a58-e9d5-4d01-90af-3b41213a9c7d-3_355_438_1530_1178}
\captionsetup{labelformat=empty}
\caption{Fig. 2.3}
\end{center}
\end{figure}
(ii) By considering the equilibrium of $\operatorname { rod } \mathrm { AB }$, show that $60 \sqrt { 3 } = U + V \sqrt { 3 }$.\\
(iii) Draw a diagram showing all the forces acting on rod BC .\\
(iv) Find a further equation connecting $U$ and $V$ and hence find their values. Find also the frictional force at C .
\hfill \mbox{\textit{OCR MEI M2 2006 Q2 [18]}}