| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod with string perpendicular |
| Difficulty | Standard +0.3 This is a standard moments problem requiring taking moments about the hinge, resolving forces, and finding a resultant direction. The perpendicular rope simplifies calculations significantly. While it involves multiple steps (moments equation, resolving horizontally and vertically, finding angle), these are routine mechanics techniques with no conceptual surprises, making it slightly easier than average. |
| Spec | 3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(T \times 0.8 = 70x1\sin\alpha + 220x2\sin\alpha\) | M1 | Moments about A (3 terms) |
| \(\sin\alpha = 1.5/1.7\) | A1 | \(\cos\alpha = 0.8/1.7,\ \alpha = 61.9°\) |
| \(T = 562.5\) N | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(H = 562.5\cos\alpha = 265\) N | B1 | \(H = 264.70\) N |
| \(V = 562.5\sin\alpha - 70 - 220\) | M1 | \(V = 206.3\) N |
| \(\tan\alpha = 265/206.3\) | M1 | |
| \(\alpha = 52.1°\) (with vertical) | A1 [4] | Or \(37.9°\) (with horizontal) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(X = (70+220)\cos\alpha = 136.6\) | B1 | Resolving along the rod AB |
| \(Y = 562.5 - (70+220)\sin\alpha = 306.7\) | M1 | Resolving perpendicular to AB |
| \(\tan\theta = 306.7/136.6\) | M1 | |
| \(\theta = 65.99°\) or \(66.0°\) (with beam) | A1 [4] |
## Question 4:
### Part (i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $T \times 0.8 = 70x1\sin\alpha + 220x2\sin\alpha$ | M1 | Moments about A (3 terms) |
| $\sin\alpha = 1.5/1.7$ | A1 | $\cos\alpha = 0.8/1.7,\ \alpha = 61.9°$ |
| $T = 562.5$ N | A1 [3] | |
### Part (ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $H = 562.5\cos\alpha = 265$ N | B1 | $H = 264.70$ N |
| $V = 562.5\sin\alpha - 70 - 220$ | M1 | $V = 206.3$ N |
| $\tan\alpha = 265/206.3$ | M1 | |
| $\alpha = 52.1°$ (with vertical) | A1 [4] | Or $37.9°$ (with horizontal) |
**OR:**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $X = (70+220)\cos\alpha = 136.6$ | B1 | Resolving along the rod AB |
| $Y = 562.5 - (70+220)\sin\alpha = 306.7$ | M1 | Resolving perpendicular to AB |
| $\tan\theta = 306.7/136.6$ | M1 | |
| $\theta = 65.99°$ or $66.0°$ (with beam) | A1 [4] | |
---4\\
\includegraphics[max width=\textwidth, alt={}, center]{af7d1fc8-5552-48b8-a359-895b2b5d3d6c-3_560_894_258_628}
A uniform beam $A B$ has length 2 m and weight 70 N . The beam is hinged at $A$ to a fixed point on a vertical wall, and is held in equilibrium by a light inextensible rope. One end of the rope is attached to the wall at a point 1.7 m vertically above the hinge. The other end of the rope is attached to the beam at a point 0.8 m from $A$. The rope is at right angles to $A B$. The beam carries a load of weight 220 N at $B$ (see diagram).\\
(i) Find the tension in the rope.\\
(ii) Find the direction of the force exerted on the beam at $A$.
\hfill \mbox{\textit{CAIE M2 2010 Q4 [7]}}