| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| 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 M3 statics problem involving two connected rods in equilibrium. Part (i) requires taking moments about B for rod BC (straightforward application with given values). Part (ii) requires taking moments about A for the whole system or analyzing rod AB separately. While it involves multiple steps and careful geometry, the techniques are routine for M3 students who have practiced connected rigid bodies. The calculations are methodical rather than requiring novel insight. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks |
|---|---|
| M1, A1, A1 [3] | 2 terms involving \(\sin\beta\) and \(\cos\beta\), 75 and 50; WWW (AG) |
| allow \(\sin/\cos\) error/ sign error; allow missing \(L\) |
| Answer | Marks |
|---|---|
| M1* | all (5) terms present; each term involves \(\sin/\cos\) \(\alpha\beta\). Direct: no extra terms; *M1 dep B1A1 |
| A1 | |
| B1 | dep M1A1 |
| *M1 | dep B1 also |
| A1 [5] | all 4 seen; all values substituted |
| B1 | see B1 for magnitudes if directions wrong/missing |
| B1 | 50 & 75 may be seen on diagram in (i) |
| M1 | involves \(W\), 75, 50, \(\sin \alpha\) and \(\cos \alpha\). Dimensionally correct; no extra terms with substitution for \(\alpha\) |
| A1 | L may be cancelled |
| A1 |
### Part (m)
**Answer:** Moments about $B$ for $BC$: $75L\cos\beta = 50 \times 2L/\sin\beta$
$\tan\beta = 3/4$
| M1, A1, A1 [3] | 2 terms involving $\sin\beta$ and $\cos\beta$, 75 and 50; WWW (AG) |
| | allow $\sin/\cos$ error/ sign error; allow missing $L$ |
### Part (ii)
**Answer:** Moments about $A$ for both rods:
$WL\cos\alpha + 75(2L\cos\alpha + L\cos\beta) = 50(2L\sin\alpha + 2L\sin\beta)$
correct values for $\sin/\cos$ $\alpha/\beta$
attempt to solve
$(W=)$ 90 (N)
OR '$X' = 50 N to right on $AB$ oe
'$Y$' = 75 N down on $AB$ oe
Moments about $A$ for $AB$
$WL\cos\alpha + 75 \times 2L\cos\alpha = 50 \times 2L\sin\alpha$
$(W=)$ 90 (N)
| M1* | all (5) terms present; each term involves $\sin/\cos$ $\alpha\beta$. Direct: no extra terms; *M1 dep B1A1 |
| A1 | |
| B1 | dep M1A1 |
| *M1 | dep B1 also |
| A1 [5] | all 4 seen; all values substituted |
| | |
| B1 | see B1 for magnitudes if directions wrong/missing |
| B1 | 50 & 75 may be seen on diagram in (i) |
| M1 | involves $W$, 75, 50, $\sin \alpha$ and $\cos \alpha$. Dimensionally correct; no extra terms with substitution for $\alpha$ |
| A1 | L may be cancelled |
| A1 | |\includegraphics{figure_2}
Two uniform rods $AB$ and $BC$, each of length $2L$, are freely jointed at $B$, and $AB$ is freely jointed to a fixed point at $A$. The rods are held in equilibrium in a vertical plane by a light horizontal string attached at $C$. The rods $AB$ and $BC$ make angles $\alpha$ and $\beta$ to the horizontal respectively. The weight of rod $BC$ is $75$ N, and the tension in the string is $50$ N (see diagram).
\begin{enumerate}[label=(\roman*)]
\item Show that $\tan \beta = \frac{1}{3}$. [3]
\item Given that $\tan \alpha = \frac{12}{5}$, find the weight of $AB$. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR M3 2015 Q2 [8]}}