| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 10 |
| 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 forces. While it requires systematic application of equilibrium conditions (resolving forces and taking moments) across two connected bodies, the setup is straightforward with clearly defined geometry and weights. The three parts guide students through a methodical solution: verifying friction at C, finding reaction components at B, and calculating the coefficient of friction. This is more challenging than basic single-body equilibrium problems but follows standard M3 techniques without requiring novel insight or particularly complex geometry. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03r Friction: concept and vector form6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| [mg - mkv² = ma] | M1 | For using Newton's second law with a = 0 |
| (v dv/dx)/(g - kv²) = 1 | A1 AG |
| Answer | Marks | Guidance |
|---|---|---|
| [- ½ ln(g - kv²)/jk = x (+C)] | M1 | For separating variables and attempting to integrate |
| [-[dln g) /2k = C] | M1 | For using v(0) = 0 to find C |
| x = [- ½ ln[(g - kv²)/g] /jk | A1 | Any equivalent expression for x |
| [ln[(g - kv²)/g] = ln(e^{-2kx})] | M1 | For expressing in the form ln(f(v²)) = ln g(x) or equivalent |
| v² = (1 - e^{-2kx})g/k | A1 M1 | For using e^Ax → 0 for +ve A |
| Limiting value is √(g/k) | A1ft [7] AG |
| Answer | Marks | Guidance |
|---|---|---|
| [1 - e^{-600k} = 0.81] [-600k = ln(0.19)] | M1 M1 | For using v²(300) = 0.9² g/k For using logarithms to solve for k |
| k = 0.00277 | A1 [3] |
**(i)**
| [mg - mkv² = ma] | M1 | For using Newton's second law with a = 0 |
|---|---|---|
| (v dv/dx)/(g - kv²) = 1 | A1 AG | |
**(ii)**
| [- ½ ln(g - kv²)/jk = x (+C)] | M1 | For separating variables and attempting to integrate |
|---|---|---|
| [-[dln g) /2k = C] | M1 | For using v(0) = 0 to find C |
|---|---|---|
| x = [- ½ ln[(g - kv²)/g] /jk | A1 | Any equivalent expression for x |
|---|---|---|
| [ln[(g - kv²)/g] = ln(e^{-2kx})] | M1 | For expressing in the form ln(f(v²)) = ln g(x) or equivalent |
|---|---|---|
| v² = (1 - e^{-2kx})g/k | A1 M1 | For using e^Ax → 0 for +ve A |
|---|---|---|
| Limiting value is √(g/k) | A1ft [7] AG |
**(iii)**
| [1 - e^{-600k} = 0.81] [-600k = ln(0.19)] | M1 M1 | For using v²(300) = 0.9² g/k For using logarithms to solve for k |
|---|---|---|
| k = 0.00277 | A1 [3] |5\\
\includegraphics[max width=\textwidth, alt={}, center]{af1f9f1b-f6c0-4044-9864-5b9ce309d3fa-03_462_1109_283_569}
Two uniform rods $A B$ and $B C$ have weights 64 N and 40 N respectively. The rods are freely jointed to each other at $B$. The rod $A B$ is freely jointed to a fixed point on horizontal ground at $A$ and the rod $B C$ rests against a vertical wall at $C$. The rod $B C$ is 1.8 m long and is horizontal. A particle of weight 9 N is attached to the rod $B C$ at the point 0.4 m from $C$. The point $A$ is 1.2 m below the level of $B C$ and 3.8 m from the wall (see diagram). The system is in equilibrium.\\
(i) Show that the magnitude of the frictional force at $C$ is 27 N .\\
(ii) Calculate the horizontal and vertical components of the force exerted on $A B$ at $B$.\\
(iii) Given that friction is limiting at $C$, find the coefficient of friction between the $\operatorname { rod } B C$ and the wall.
\hfill \mbox{\textit{OCR M3 Q5 [10]}}