| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2008 |
| 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 two-rod equilibrium problem requiring systematic application of moments and force resolution. While it involves multiple steps (finding reactions, friction forces, and joint forces) and careful geometry, the approach is methodical and follows textbook procedures for M3. The geometry is given rather than needing to be derived, and each part builds logically on the previous one. Slightly above average difficulty due to the multi-body system and bookkeeping required, but well within standard M3 scope. |
| Spec | 3.04b Equilibrium: zero resultant moment and force6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(1.4R_A = 150\times0.95 + 130\times0.25\) or \(1.4R_B = 130\times1.15 + 150\times0.45\) or \(1.2F - 0.9(280 - R_B) + 0.45\times150 - 1.2F + 0.5R_B - 0.25\times130 = 0\) | M1 | For taking moments about B or about A for the whole, or for taking moments about X for the whole and using \(R_A + R_B = 280\) and \(F_A = F_B\) |
| A1 | ||
| \(R_A = 125\text{N}\) | A1 | AG |
| \(R_B = 155\text{N}\) | B1 | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(1.2F_A = -150\times0.45 + 0.9R_A\) or \(1.2F_B = 0.5R_B - 130\times0.25\) | M1 | For taking moments about X for XA or XB |
| A1 | ||
| \(F_A\) or \(F_B = 37.5\text{N}\) | A1ft | \(F_B = (1.25R_B - 81.25)/3\) |
| \(F_B\) or \(F_A = 37.5\text{N}\) | B1ft | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Horizontal component is \(37.5\text{N}\) to the left | B1ft | ft \(H = F\) or \(H = 56.25 - 0.75V\) or \(12H = 325 + 5V\) |
| \([Y + R_A = 150]\) | M1 | For resolving forces on XA vertically |
| Vertical component is \(25\text{N}\) upwards | A1ft | 3 |
# Question 5:
## Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $1.4R_A = 150\times0.95 + 130\times0.25$ or $1.4R_B = 130\times1.15 + 150\times0.45$ or $1.2F - 0.9(280 - R_B) + 0.45\times150 - 1.2F + 0.5R_B - 0.25\times130 = 0$ | M1 | For taking moments about B or about A for the whole, or for taking moments about X for the whole and using $R_A + R_B = 280$ and $F_A = F_B$ |
| | A1 | |
| $R_A = 125\text{N}$ | A1 | AG |
| $R_B = 155\text{N}$ | B1 | 4 | |
## Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $1.2F_A = -150\times0.45 + 0.9R_A$ or $1.2F_B = 0.5R_B - 130\times0.25$ | M1 | For taking moments about X for XA or XB |
| | A1 | |
| $F_A$ or $F_B = 37.5\text{N}$ | A1ft | $F_B = (1.25R_B - 81.25)/3$ |
| $F_B$ or $F_A = 37.5\text{N}$ | B1ft | 4 | |
## Part (iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Horizontal component is $37.5\text{N}$ to the left | B1ft | ft $H = F$ or $H = 56.25 - 0.75V$ or $12H = 325 + 5V$ |
| $[Y + R_A = 150]$ | M1 | For resolving forces on XA vertically |
| Vertical component is $25\text{N}$ upwards | A1ft | 3 | ft $3V = 225 - 4H$ or $V = 2.4H - 65$ |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{85402f4a-8d55-47d8-ba48-5b837609b0f4-3_581_903_267_621}
Two uniform rods $X A$ and $X B$ are freely jointed at $X$. The lengths of the rods are 1.5 m and 1.3 m respectively, and their weights are 150 N and 130 N respectively. The rods are in equilibrium in a vertical plane with $A$ and $B$ in contact with a rough horizontal surface. $A$ and $B$ are at distances horizontally from $X$ of 0.9 m and 0.5 m respectively, and $X$ is 1.2 m above the surface (see diagram).\\
(i) The normal components of the contact forces acting on the rods at $A$ and $B$ are $R _ { A } \mathrm {~N}$ and $R _ { B } \mathrm {~N}$ respectively. Show that $R _ { A } = 125$ and find $R _ { B }$.\\
(ii) Find the frictional components of the contact forces acting on the rods at $A$ and $B$.\\
(iii) Find the horizontal and vertical components of the force exerted on $X A$ at $X$, stating their directions.
\hfill \mbox{\textit{OCR M3 2008 Q5 [11]}}