| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Prism or block on inclined plane |
| Difficulty | Standard +0.3 This is a standard moments problem requiring taking moments about a point, resolving forces, and applying friction laws. While it involves multiple steps (moments equation, resolving horizontally and vertically, friction calculation), each step uses routine mechanics techniques with no novel insight required. The geometry is straightforward and the question guides students through part (i) by specifying to take moments about D. |
| Spec | 3.04b Equilibrium: zero resultant moment and force6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0.8T = 260 \times (DG) \times \cos\theta\) | M1 | Moments about D |
| \(DG = 1.7/2,\ \theta = (30+D)\) | M1 | Both needed |
| Angle \(BDC = 28°\) | DA1 | \(D = 28.072...\) |
| \(0.8T = 260 \times (1.7/2) \times \cos58.07\) | A1ft | \(ftcv(DG \neq 0.8, 1.5, 1.7,\ \theta \neq 30, 28)\) |
| \(T = 146\) N AG | A1 [5] | |
| OR Moment of weight | M1 | |
| \(= (260\cos30)\times 0.75 - (260\sin30)\times 0.4\) | DA1 | Difference of moments of perp components \((116.87...)\) |
| M1 | Moments about D | |
| \(0.8T = 116.87...\) | A1 | Needs no evaluation |
| \(T = 146\) N AG | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(F_r = 146\cos30\) | B1 | \(126.52...\) |
| \(R = 260 + 146\cos60\) | B1 | \(333.04...\) |
| \(\mu = (146\cos30)/(260 + 146\sin30)\) | M1 | Denominator not 260 |
| \(\mu = 0.38(0)\) | A1 [4] |
## Question 6:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.8T = 260 \times (DG) \times \cos\theta$ | M1 | Moments about D |
| $DG = 1.7/2,\ \theta = (30+D)$ | M1 | Both needed |
| Angle $BDC = 28°$ | DA1 | $D = 28.072...$ |
| $0.8T = 260 \times (1.7/2) \times \cos58.07$ | A1ft | $ftcv(DG \neq 0.8, 1.5, 1.7,\ \theta \neq 30, 28)$ |
| $T = 146$ N AG | A1 [5] | |
| **OR** Moment of weight | M1 | |
| $= (260\cos30)\times 0.75 - (260\sin30)\times 0.4$ | DA1 | Difference of moments of perp components $(116.87...)$ |
| | M1 | Moments about D |
| $0.8T = 116.87...$ | A1 | Needs no evaluation |
| $T = 146$ N AG | A1 | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F_r = 146\cos30$ | B1 | $126.52...$ |
| $R = 260 + 146\cos60$ | B1 | $333.04...$ |
| $\mu = (146\cos30)/(260 + 146\sin30)$ | M1 | Denominator not 260 |
| $\mu = 0.38(0)$ | A1 [4] | |
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{6503ebb1-5649-4ca5-9500-da4fb28009dd-3_454_1029_1379_557}\\
$A B C D$ is the cross-section through the centre of mass of a uniform rectangular block of weight 260 N . The lengths $A B$ and $B C$ are 1.5 m and 0.8 m respectively. The block rests in equilibrium with the point $D$ on a rough horizontal floor. Equilibrium is maintained by a light rope attached to the point $A$ on the block and the point $E$ on the floor. The points $E , A$ and $B$ lie in a straight line inclined at $30 ^ { \circ }$ to the horizontal (see diagram).\\
(i) By taking moments about $D$, show that the tension in the rope is 146 N , correct to 3 significant figures.\\
(ii) Given that the block is in limiting equilibrium, calculate the coefficient of friction between the block and the floor.
\hfill \mbox{\textit{CAIE M2 2013 Q6 [9]}}