| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2013 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Two jointed rods in equilibrium |
| Difficulty | Challenging +1.8 This is a challenging M3 statics problem involving two jointed rods with multiple forces, requiring systematic application of equilibrium conditions across both rods, resolution of forces at the joint, and friction analysis. The geometry with specific angles (60° and 30°) and the joint at the midpoint creates a multi-step problem requiring careful force resolution and moment calculations about strategic points. While methodical, it demands strong spatial reasoning and algebraic manipulation beyond typical M1/M2 questions. |
| Spec | 3.03u Static equilibrium: on rough surfaces3.04b Equilibrium: zero resultant moment and force6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Moments about \(C\) for \(CD\) | M1 | Allow M if sin/cos wrong |
| \(Wl\frac{\sqrt{3}}{2}(\cos 30°) = Ql\sqrt{3}(\cos 30°)\) | A1 | |
| \((Q =)\ W/2\) | A1 AG | |
| Resolve vertically | M1 | |
| \((R =)\ \frac{3}{2}W\) | A1 [5] | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(X = 0\) | B1 | |
| Resolve vert for \(CD\) or \(AB\): \(Y = W/2\) | B1* | \(Y + Q = W\) or \(Y + W = R\) |
| Vertically downwards | *B1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Moments about \(C\) for \(AB\) | M1 | Allow M if sin/cos wrong or sign errors; need all terms |
| \(Pl\cos 30° + Fl\cos 30° = Rl\sin 30°\) | A1 | Correct |
| Use \(P\) in terms of \(F\) | M1 | \(F = P\) or other correct 2nd step; Allow if missing term above |
| Find \(F\) in terms of \(W\), or in terms of \(R\) | M1 | \(F = \frac{\sqrt{3}}{4}W\); Or getting 'their' \(F\) oe, ie putting \(F = \mu R\) in moment equation |
| \(\mu = (F/R) = \sqrt{3}/6\) | A1 [5] | Accept decimal answers from 0.288675 |
| OR: Moments about \(A\) for \(AB\) | M1 | Allow M if sin/cos wrong or sign errors; need all terms |
| \(Wl\sin 30° + (Y)l\sin 30° + F2l\cos 30° = R2l\sin 30°\) | A1 | May have \(X\) term if not 0 in (ii) |
| Write \(Y\) (and \(X\)) in terms of \(W\) | M1 | |
| Find \(F\) in terms of \(W\), or in terms of \(R\), oe | M1 | \(F = \frac{\sqrt{3}}{4}W\) |
| \(\mu = (F/R) = \sqrt{3}/6\) | A1 | Accept decimal answers from 0.288675 |
# Question 6:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Moments about $C$ for $CD$ | M1 | Allow M if sin/cos wrong |
| $Wl\frac{\sqrt{3}}{2}(\cos 30°) = Ql\sqrt{3}(\cos 30°)$ | A1 | |
| $(Q =)\ W/2$ | A1 **AG** | |
| Resolve vertically | M1 | |
| $(R =)\ \frac{3}{2}W$ | A1 **[5]** | CAO |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $X = 0$ | B1 | |
| Resolve vert for $CD$ or $AB$: $Y = W/2$ | B1* | $Y + Q = W$ or $Y + W = R$ |
| Vertically downwards | *B1 **[3]** | |
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Moments about $C$ for $AB$ | M1 | Allow M if sin/cos wrong or sign errors; need all terms |
| $Pl\cos 30° + Fl\cos 30° = Rl\sin 30°$ | A1 | Correct |
| Use $P$ in terms of $F$ | M1 | $F = P$ or other correct 2nd step; Allow if missing term above |
| Find $F$ in terms of $W$, or in terms of $R$ | M1 | $F = \frac{\sqrt{3}}{4}W$; Or getting 'their' $F$ oe, ie putting $F = \mu R$ in moment equation |
| $\mu = (F/R) = \sqrt{3}/6$ | A1 **[5]** | Accept decimal answers from 0.288675 |
| OR: Moments about $A$ for $AB$ | M1 | Allow M if sin/cos wrong or sign errors; need all terms |
| $Wl\sin 30° + (Y)l\sin 30° + F2l\cos 30° = R2l\sin 30°$ | A1 | May have $X$ term if not 0 in (ii) |
| Write $Y$ (and $X$) in terms of $W$ | M1 | |
| Find $F$ in terms of $W$, or in terms of $R$, oe | M1 | $F = \frac{\sqrt{3}}{4}W$ |
| $\mu = (F/R) = \sqrt{3}/6$ | A1 | Accept decimal answers from 0.288675 |
---6 A uniform $\operatorname { rod } A B$, of weight $W$ and length $2 l$ is in equilibrium at $60 ^ { \circ }$ to the horizontal with $A$ resting against a smooth vertical plane and $B$ resting on a rough section of a horizontal plane. Another uniform rod $C D$, of length $\sqrt { 3 } l$ and weight $W$, is freely jointed to the mid-point of $A B$ at $C$; its other end $D$ rests on a smooth section of the horizontal plane. $C D$ is inclined at $30 ^ { \circ }$ to the horizontal (see diagram).\\
\includegraphics[max width=\textwidth, alt={}, center]{dfe477d4-eae6-40e1-b704-1a97485f4c7e-4_508_1075_438_495}\\
(i) Show that the force exerted by the horizontal plane on $C D$ is $\frac { 1 } { 2 } W$. Find the normal component of the force exerted by the horizontal plane on $A B$.\\
(ii) Find the magnitude and direction of the force exerted by $C D$ on $A B$.\\
(iii) Given that $A B$ is in limiting equilibrium, find the coefficient of friction between $A B$ and the horizontal plane.
\hfill \mbox{\textit{OCR M3 2013 Q6 [13]}}