| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Two jointed rods in equilibrium |
| Difficulty | Standard +0.8 This is a challenging M3 statics problem requiring analysis of a two-rod system with multiple equilibrium conditions. Students must resolve forces in two directions for rod BC, take moments about two different points (B for part i, A for part ii), and combine equations to find the coefficient of friction. The geometry with 30° and 60° angles adds computational complexity, and coordinating the analysis of individual rods versus the whole system requires solid conceptual understanding beyond routine textbook exercises. |
| Spec | 3.03u Static equilibrium: on rough surfaces3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Moments about \(B\) for equilibrium of \(BC\) | M1 | \(2Wl\cos60° + F2l\sin60° = R2l\cos60°\); 3 moment terms, condone sin/cos errors and missing \(l\); Need trig terms for M1 |
| \(W + \sqrt{3}\,F = R\) AG | A1 | Must be formula for \(R\); correct, with sin/cos evaluated |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Moments about \(A\) for equilibrium of whole system | M1 | At least one of \(F\) and \(R\) terms must involve lengths of both rods; At least 3 moment terms, condone sin/cos errors, sign errors and \(l/2l\) confusion/missing. Wrong use of forces at \(B\) gets M0 |
| \(Wl\cos30 + 2W(2l\cos30 + l\cos60) + F(2l\sin60 + 2l\sin30) = R(2l\cos30 + 2l\cos60)\) | A1 | 4 terms, accept sin/cos errors and \(l/2l\) confusion/missing and sign errors for A1 |
| sin/cos left in, but correct | A1 | |
| \(W\!\left(\frac{5\sqrt{3}}{2}+1\right) + F\!\left(\sqrt{3}+1\right) = R\!\left(\sqrt{3}+1\right)\) | A1 | Fully correct, oe. Mark final answer; accept \(5.33W + 2.73F = 2.73R\), \(W\!\left(\frac{13}{4} - \frac{3\sqrt{3}}{4}\right) + F = R\) |
| Allow full credit for candidates who work out internal forces at \(B\) and work correctly from there. | Eg \(3R = \sqrt{3}F + 7.5W\) | |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Solving 2 sim equations to eliminate \(F\) or \(R\) | M1 | Both equations must involve \(W\), \(F\) and \(R\); allow slips in working |
| \(F = \frac{3\sqrt{3}}{4}W\) | A1 | \(F = 1.299\,W\) |
| \(R = \frac{13}{4}W\) | A1 | \(R = 3.25\,W\) |
| Use \(F = \mu R\) to find \(\mu\) | M1 | At any point |
| \((\mu =)\ \frac{3\sqrt{3}}{13}\ \ (0.39970)\) | A1 | Accept 0.4 if with correct working; \(5.33(R - 1.73F) + 2.73F = 2.73R\), \(2.6R = 6.52F\) |
| [5] |
## Question 6(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Moments about $B$ for equilibrium of $BC$ | M1 | $2Wl\cos60° + F2l\sin60° = R2l\cos60°$; 3 moment terms, condone sin/cos errors and missing $l$; Need trig terms for M1 |
| $W + \sqrt{3}\,F = R$ AG | A1 | Must be formula for $R$; correct, with sin/cos evaluated |
| **[2]** | | |
## Question 6(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Moments about $A$ for equilibrium of whole system | M1 | At least one of $F$ and $R$ terms must involve lengths of both rods; At least 3 moment terms, condone sin/cos errors, sign errors and $l/2l$ confusion/missing. Wrong use of forces at $B$ gets M0 |
| $Wl\cos30 + 2W(2l\cos30 + l\cos60) + F(2l\sin60 + 2l\sin30) = R(2l\cos30 + 2l\cos60)$ | A1 | 4 terms, accept sin/cos errors and $l/2l$ confusion/missing and sign errors for A1 |
| sin/cos left in, but correct | A1 | |
| $W\!\left(\frac{5\sqrt{3}}{2}+1\right) + F\!\left(\sqrt{3}+1\right) = R\!\left(\sqrt{3}+1\right)$ | A1 | Fully correct, oe. Mark final answer; accept $5.33W + 2.73F = 2.73R$, $W\!\left(\frac{13}{4} - \frac{3\sqrt{3}}{4}\right) + F = R$ |
| Allow full credit for candidates who work out internal forces at $B$ and work correctly from there. | | Eg $3R = \sqrt{3}F + 7.5W$ |
| **[4]** | | |
## Question 6(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Solving 2 sim equations to eliminate $F$ or $R$ | M1 | Both equations must involve $W$, $F$ and $R$; allow slips in working |
| $F = \frac{3\sqrt{3}}{4}W$ | A1 | $F = 1.299\,W$ |
| $R = \frac{13}{4}W$ | A1 | $R = 3.25\,W$ |
| Use $F = \mu R$ to find $\mu$ | M1 | At any point |
| $(\mu =)\ \frac{3\sqrt{3}}{13}\ \ (0.39970)$ | A1 | Accept 0.4 if with correct working; $5.33(R - 1.73F) + 2.73F = 2.73R$, $2.6R = 6.52F$ |
| **[5]** | | |6 Two uniform rods $A B$ and $B C$, each of length $2 l$, are freely jointed at $B$. The weight of $A B$ is $W$ and the weight of $B C$ is $2 W$. The rods are in a vertical plane with $A$ freely pivoted at a fixed point and $C$ resting in equilibrium on a rough horizontal plane. The normal and frictional components of the force acting on $B C$ at $C$ are $R$ and $F$ respectively. The rod $A B$ makes an angle $30 ^ { \circ }$ to the horizontal and the rod $B C$ makes an angle $60 ^ { \circ }$ to the horizontal (see diagram).\\
\includegraphics[max width=\textwidth, alt={}, center]{3e8248ca-74f1-443f-a5db-d7da532d2815-4_682_901_479_587}\\
(i) By considering the equilibrium of $\operatorname { rod } B C$, show that $W + \sqrt { 3 } F = R$.\\
(ii) By taking moments about $A$ for the equilibrium of the whole system, find another equation involving $W , F$ and $R$.\\
(iii) Given that the friction at $C$ is limiting, calculate the value of the coefficient of friction at $C$.
\hfill \mbox{\textit{OCR M3 2013 Q6 [11]}}