| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2007 |
| 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 multi-part mechanics problem requiring moments about different points, resolution of forces at a joint, and geometric reasoning to find an unknown angle. While the individual techniques (taking moments, resolving forces) are standard M3 content, the problem requires careful systematic work through connected parts with multiple unknowns, placing it moderately above average difficulty for A-level. |
| Spec | 3.04b Equilibrium: zero resultant moment and force6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For taking moments of forces on BC about B | |
| \(80 \times 0.7\cos 60° = 1.4T\) | A1 | |
| Tension is 20N | A1 | |
| \([X = 20\cos 30°]\) | M1 | For resolving forces horizontally |
| Horizontal component is 17.3N | A1ft | ft \(X = T\cos 30°\) |
| \([Y = 80 - 20\sin 30°]\) | M1 | For resolving forces vertically |
| Vertical component is 70N | A1ft | [7 marks total] ft \(Y = 80 - T\sin 30°\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For taking moments of forces on AB, or on ABC, about A | |
| \(17.3 \times 1.4\sin\alpha = (80 \times 0.7 + 70 \times 1.4)\cos\alpha\) or \(80 \times 0.7\cos\alpha + 80(1.4\cos\alpha + 0.7\cos 60°) = 20\cos 60°(1.4\cos\alpha + 1.4\cos 60°) + 20\sin 60°(1.4\sin\alpha + 14\sin 60°)\) | A1ft | |
| \([\tan\alpha = (\frac{1}{2}\times 80 + 70)/17.3 = 11/\sqrt{3}]\) | M1 | For obtaining a numerical expression for \(\tan\alpha\) |
| \(\alpha = 81.1°\) | A1 | [4 marks total] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For taking moments of forces on BC about B | |
| \(H \times 1.4\sin 60° + V \times 1.4\cos 60° = 80 \times 0.7\cos 60°\) | A1 | Where H and V are components of T |
| M1 | For using \(H = V\sqrt{3}\) and solving simultaneous equations | |
| Tension is 20N | A1 | |
| Horizontal component is 17.3N | B1ft | ft value of H used to find T |
| \([Y = 80 - V]\) | M1 | For resolving forces vertically |
| Vertical component is 70N | A1ft | [7 marks total] ft value of V used to find T |
# Question 5:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For taking moments of forces on BC about B |
| $80 \times 0.7\cos 60° = 1.4T$ | A1 | |
| Tension is 20N | A1 | |
| $[X = 20\cos 30°]$ | M1 | For resolving forces horizontally |
| Horizontal component is 17.3N | A1ft | ft $X = T\cos 30°$ |
| $[Y = 80 - 20\sin 30°]$ | M1 | For resolving forces vertically |
| Vertical component is 70N | A1ft | [7 marks total] ft $Y = 80 - T\sin 30°$ |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For taking moments of forces on AB, or on ABC, about A |
| $17.3 \times 1.4\sin\alpha = (80 \times 0.7 + 70 \times 1.4)\cos\alpha$ or $80 \times 0.7\cos\alpha + 80(1.4\cos\alpha + 0.7\cos 60°) = 20\cos 60°(1.4\cos\alpha + 1.4\cos 60°) + 20\sin 60°(1.4\sin\alpha + 14\sin 60°)$ | A1ft | |
| $[\tan\alpha = (\frac{1}{2}\times 80 + 70)/17.3 = 11/\sqrt{3}]$ | M1 | For obtaining a numerical expression for $\tan\alpha$ |
| $\alpha = 81.1°$ | A1 | [4 marks total] |
## Alternative Method for Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For taking moments of forces on BC about B |
| $H \times 1.4\sin 60° + V \times 1.4\cos 60° = 80 \times 0.7\cos 60°$ | A1 | Where H and V are components of T |
| | M1 | For using $H = V\sqrt{3}$ and solving simultaneous equations |
| Tension is 20N | A1 | |
| Horizontal component is 17.3N | B1ft | ft value of H used to find T |
| $[Y = 80 - V]$ | M1 | For resolving forces vertically |
| Vertical component is 70N | A1ft | [7 marks total] ft value of V used to find T |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{a04e6d4e-2437-4761-87ee-43e6771fbbd9-3_549_447_253_849}
Two uniform rods $A B$ and $B C$, each of length 1.4 m and weight 80 N , are freely jointed to each other at $B$, and $A B$ is freely jointed to a fixed point at $A$. They are held in equilibrium with $A B$ at an angle $\alpha$ to the horizontal, and $B C$ at an angle of $60 ^ { \circ }$ to the horizontal, by a light string, perpendicular to $B C$, attached to $C$ (see diagram).\\
(i) By taking moments about $B$ for $B C$, calculate the tension in the string. Hence find the horizontal and vertical components of the force acting on $B C$ at $B$.\\
(ii) Find $\alpha$.\\
\includegraphics[max width=\textwidth, alt={}, center]{a04e6d4e-2437-4761-87ee-43e6771fbbd9-3_691_665_1370_740}
A circus performer $P$ of mass 80 kg is suspended from a fixed point $O$ by an elastic rope of natural length 5.25 m and modulus of elasticity $2058 \mathrm {~N} . P$ is in equilibrium at a point 5 m above a safety net. A second performer $Q$, also of mass 80 kg , falls freely under gravity from a point above $P$. $P$ catches $Q$ and together they begin to descend vertically with initial speed $3.5 \mathrm {~ms} ^ { - 1 }$ (see diagram). The performers are modelled as particles.\\
\hfill \mbox{\textit{OCR M3 2007 Q5 [11]}}