| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2014 |
| Session | June |
| 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 M3 statics problem involving two jointed rods with multiple parts requiring systematic application of equilibrium conditions (moments and forces). While it requires careful bookkeeping of forces at three points and taking moments about strategic points, the methods are entirely standard for Further Maths M3. The geometry is given (60° angles), and part (i) guides students by providing one answer to verify. More challenging than basic C1/M1 questions but routine for students who have practiced jointed rod problems. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Take moments about \(A\) for whole body: \(Wx2L\cos 60° + 2Wx6L\cos 60° = Rx8L\cos 60°\) | M1, A1, A1 | Correct 3 terms needed; dim correct; \(\cos 60°\) may be omitted; at least 1 correct step to show given answer; allow sign errors, \(W/2W\), cos/sin; \(R\) is reaction at \(C\); \(S\) is reaction at \(A\); for less efficient methods, M1 can only be earned when equation with one unknown, \(R\), is reached |
| \(R = 1.75W\); \(S = 1.25W\) | B1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Take moments about \(B\) for equil of \(BC\): \(TxL\sin 60° + 2Wx2L\cos 60° = 1.75Wx4L\cos 60°\) | M1*, A1, *M1 | Correct 3 resolved terms needed; dim correct; or for \(BA\): \(TxL\sin 60° + Wx2L\cos 60° = 1.25Wx4L\cos 60°\); allow sign errors, \(W/2W\), cos/sin |
| \(T = \sqrt{3}W\) | A1 [4] | Accept \(T = 1.73W\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Resolve vertically for \(AB\): \(Y + 1.25W - W = 0\) | M1 | Weight and normal term must be for same rod |
| \(Y = 0.25W\), downwards; \(X = \sqrt{3}W\) to left | A1CAO, B1ft [3] | Direction must be clear; direction must be clear |
## Question 5:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Take moments about $A$ for whole body: $Wx2L\cos 60° + 2Wx6L\cos 60° = Rx8L\cos 60°$ | M1, A1, A1 | Correct 3 terms needed; dim correct; $\cos 60°$ may be omitted; at least 1 correct step to show given answer; allow sign errors, $W/2W$, cos/sin; $R$ is reaction at $C$; $S$ is reaction at $A$; for less efficient methods, M1 can only be earned when equation with one unknown, $R$, is reached |
| $R = 1.75W$; $S = 1.25W$ | B1 [4] | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Take moments about $B$ for equil of $BC$: $TxL\sin 60° + 2Wx2L\cos 60° = 1.75Wx4L\cos 60°$ | M1*, A1, *M1 | Correct 3 resolved terms needed; dim correct; or for $BA$: $TxL\sin 60° + Wx2L\cos 60° = 1.25Wx4L\cos 60°$; allow sign errors, $W/2W$, cos/sin |
| $T = \sqrt{3}W$ | A1 [4] | Accept $T = 1.73W$ |
### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Resolve vertically for $AB$: $Y + 1.25W - W = 0$ | M1 | Weight and normal term must be for same rod |
| $Y = 0.25W$, downwards; $X = \sqrt{3}W$ to left | A1CAO, B1ft [3] | Direction must be clear; direction must be clear |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{3243c326-a51c-462f-a57c-a150d0044ea9-3_510_716_662_676}
Two uniform rods $A B$ and $B C$, each of length $4 L$, are freely jointed at $B$, and rest in a vertical plane with $A$ and $C$ on a smooth horizontal surface. The weight of $A B$ is $W$ and the weight of $B C$ is $2 W$. The rods are joined by a horizontal light inextensible string fixed to each rod at a point distance $L$ from $B$, so that each rod is inclined at an angle of $60 ^ { \circ }$ to the horizontal (see diagram).\\
(i) By considering the equilibrium of the whole body, show that the force acting on $B C$ at $C$ is $1.75 W$ and find the force acting on $A B$ at $A$.\\
(ii) Find the tension in the string in terms of $W$.\\
(iii) Find the horizontal and vertical components of the force acting on $A B$ at $B$, and state the direction of the component in each case.
\hfill \mbox{\textit{OCR M3 2014 Q5 [11]}}