| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2012 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Two jointed rods in equilibrium |
| Difficulty | Challenging +1.2 This is a standard two-rod equilibrium problem requiring systematic application of moments and resolution of forces. While it involves multiple steps (finding tension, forces at joint B, then angle α) and requires careful bookkeeping of directions, the techniques are all standard M3 material with no novel insights needed. The given cos β = 12/13 simplifies calculations significantly. Slightly above average difficulty due to the multi-part nature and need to work through both rods systematically. |
| Spec | 3.04b Equilibrium: zero resultant moment and force6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(84.5 \times 12L/13 = T(2L)\) | M1 | For taking moments about \(B\) for \(BC\); must be 2 terms involving \(T\), \(L\), \(84.5\) and \(\sin/\cos\beta\) |
| A1 | Must use \(12/13\) for \(\cos\beta\) | |
| Tension is \(39\) N | A1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(X = 39 \times 5/13\) | M1, A1 FT | For resolving forces on \(BC\) horiz or vert; explicit expression for \(X\); must involve their \(T\) and \(\sin/\cos\beta\) |
| \(Y = 84.5 - 39 \times 12/13\) | A1 FT | Explicit expression for \(Y\) |
| \(X\) is to the left and \(Y\) is upwards | A1cao | AG (numerical values must be correct) dep M1A1A1; accept on diagram |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(84.5 \times L\cos\alpha + 48.5 \times 2L\cos\alpha = 15 \times 2L\sin\alpha\) | M1* | For taking moments about \(A\) for \(AB\); must involve 3 terms, \(84.5\), \(48.5\), \(15\), \(\sin\alpha\) and \(\cos\alpha\); allow sign errors, \(L/2L\) |
| \(\left[\tan\alpha = \dfrac{84.5 + 97}{30}\right]\) | A1, *M1 | For obtaining a numerical expression for \(\tan\alpha\); similar scheme for those who take moments about \(A\) for whole system |
| \(\alpha = 1.41°\) or \(80.6°\) | A1 | |
| [4] |
## Question 5:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $84.5 \times 12L/13 = T(2L)$ | M1 | For taking moments about $B$ for $BC$; must be 2 terms involving $T$, $L$, $84.5$ and $\sin/\cos\beta$ |
| | A1 | Must use $12/13$ for $\cos\beta$ |
| Tension is $39$ N | A1 | |
| **[3]** | | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $X = 39 \times 5/13$ | M1, A1 FT | For resolving forces on $BC$ horiz or vert; explicit expression for $X$; must involve their $T$ and $\sin/\cos\beta$ |
| $Y = 84.5 - 39 \times 12/13$ | A1 FT | Explicit expression for $Y$ |
| $X$ is to the left and $Y$ is upwards | A1cao | AG (numerical values must be correct) dep M1A1A1; accept on diagram |
| **[4]** | | |
### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $84.5 \times L\cos\alpha + 48.5 \times 2L\cos\alpha = 15 \times 2L\sin\alpha$ | M1* | For taking moments about $A$ for $AB$; must involve 3 terms, $84.5$, $48.5$, $15$, $\sin\alpha$ and $\cos\alpha$; allow sign errors, $L/2L$ |
| $\left[\tan\alpha = \dfrac{84.5 + 97}{30}\right]$ | A1, *M1 | For obtaining a numerical expression for $\tan\alpha$; similar scheme for those who take moments about $A$ for whole system |
| $\alpha = 1.41°$ or $80.6°$ | A1 | |
| **[4]** | | |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{43ed8ec7-67f1-418a-8d4e-ee96448647fd-3_441_450_213_808}
Two uniform rods $A B$ and $B C$, each of length $2 L \mathrm {~m}$ and of weight 84.5 N , are freely jointed at $B$, and $A B$ is freely jointed to a fixed point at $A$. The rods are held in equilibrium in a vertical plane by a light string attached at $C$ and perpendicular to $B C$. The rods $A B$ and $B C$ make angles $\alpha$ and $\beta$ to the horizontal, respectively (see diagram). It is given that $\cos \beta = \frac { 12 } { 13 }$.\\
(i) Find the tension in the string.\\
(ii) Hence show that the force acting on $B C$ at $B$ has horizontal component of magnitude 15 N and vertical component of magnitude 48.5 N , and state the direction of the component in each case.\\
(iii) Find $\alpha$.
\hfill \mbox{\textit{OCR M3 2012 Q5 [11]}}